Stein 3.12

$\textbf{Exercise 12}$ : Consider the function $f(x)=x^2\sin(\frac{1}{x^2})$ for $x \neq 0$ and $f(0)=0$. Show that $f'$ exists for every $x\in[-1,1]$ but $f' \not \in{L^1[-1,1]}$
$\textit{Proof}$ : For $x \neq 0$, $f'(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$...an algebraic combination of continuous functions defined for all nonzero $x$. To check $f'(0)$, we may look at the difference quotient directly: $$\lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \rightarrow 0} \frac{f(h)}{h}=\lim_{h \rightarrow 0} h\sin\left(\frac{1}{h^2}\right)=0$$ So, $f'$ exists everywhere in $[-1,1]$.
Now consider a collection of closed intervals of $(0,1]$ defined by the points $\{\alpha_k\}_{k=1}^\infty$ and $\{\beta_k\}_{k=1}^\infty$ where $\alpha_i \neq \beta_j \quad \forall i,j \in{\mathbb{Z}^+}$ On any one of these closed intervals, $f'$ is continuous, so we may apply the fundamental theorem of calculus which gives $$\int_\alpha^\beta |f'(x)|dx=x^2\sin\left(\frac{1}{x^2}\right)_{\alpha}^{\beta}$$ Let $\alpha_k$ be s.t. $f(\alpha_k)=\alpha_k$ and $\beta_{k}$ s.t. $f(\beta_{k})=0$ i.e. the roots and local maximum values of $f$. Then, $\alpha_k= \pm \sqrt{\frac{2}{\pi (2k-1)}}$ and $\beta_k = \pm \sqrt{\frac{1}{\pi k}}$. Then, $$\int_{-1}^1 |f'(t)|dt \geq \int_0^1 |f'(t)|dt \geq \sum_{k=1}^\infty \int_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}|f'(t)|dt = \sum_{k=1}^\infty x^2\sin\left(\frac{1}{x^2}\right)_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}$$ $$\geq \sum_{k=1}^\infty\frac{2}{\pi (2k-1)}= \frac{2}{\pi} \sum_{k=1}^\infty \frac{1}{2k-1} \geq \frac{2}{\pi}\sum_{k=2}^\infty\frac{1}{3k} = \frac{2}{3\pi} \sum_{k=2}^\infty \frac{1}{k} = +\infty$$ $\therefore$ $f' \not \in{L^1}$ $\blacksquare$

Stein 1.26

$\textbf{Exercise 1.26}$ Given the conditions that $A \subseteq E \subseteq B$, $A,B \in{\mathcal{M}}$ where $\mathcal{M}$ is the $\sigma$-algebra of Lebesgue measurable sets, $m(A)=m(B) < \infty$ then $E\in{\mathcal{M}}$

Stategy: We wish to write $E$ as the union or intersection of some measurable sets.

$\textit{Proof}$ : Consider $B-A=B \cap A^c$. Since $A,B\in{\mathcal{M}}$, then by the $\sigma$-algebra structure of $\mathcal{M}$, $A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under compliments) and $B \cap A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under countable intersections). We may write $B=A \cup (B-A)$ and so $A \cap (B-A) = \emptyset$. The measure of a union of disjoint measurable sets is the sum of their measures (this is theorem 3.2) so, $m(A)+m(B-A)=m(B)$. Because of our assumption that $m(A)=m(B) < \infty$, then $m(B-A)=0$. Since $A \subseteq E \subseteq B$ then $E-A \subseteq B-A$. By the monotinicity of $m_*$, $m_*(E-A) \leq m_*(B-A)=0$. So, $m_*(E-A)=0 \Longrightarrow E-A \in{\mathcal{M}}$ since sets of outer measure 0 are measurable. Again using the $\sigma$-algebra structure of $\mathcal{M}$, $E= (E-A) \cup A$...a union of 2 measurable sets. $\therefore$ $E \in{\mathcal{M}} \quad \blacksquare$

Stein 3.32

$\textbf{Exercise 32}$ : Let $f: \mathbb{R} \longrightarrow \mathbb{R}$. Prove that $f$ is Lipschitz continuous i.e. $\exists M>0$ s.t. $$\frac{|f(x)-f(z)|}{|x-z|} \leq M \quad \forall x,z\in{\mathbb{R}} \quad (\star)$$ iff $f$ is absolutely continuous $\textit{and}$ $|f'(x)| \leq M$ for a.e. $x$.

$\textit{Proof}$
"$(\Longrightarrow)$" Assume that $f$ has the above Lipschitz condition.
Remark: we should first establish that $f$ is absolutely continuous and then used the fact that an absolutely continuous function has a first derivative almost everywhere. This first derivative is then bounded by the Lipschitz constant $M$.
Recall that $f$ is $\textit{absolutely continuous}$ if $\forall \epsilon>0$, $\exists \delta(\epsilon)>0$ s.t. for any collection of disjoint intervals $\{(a_i,b_i)\}_{i=1}^n$ defined by a partition that $\sum_{i=1}^n(b_i-a_i) \leq \delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq \epsilon$. With the Lipschitz condition $(\star)$ let $\delta=\frac{\epsilon}{M}$. Then, $|f(b_i)-f(a_i)|\leq M |b_i-a_i|=M \frac{\epsilon}{M}=\epsilon$. Let $|b_i-a_i| \leq \frac{\delta}{n}$. Then, $|f(b_i)-f(a_i)|\leq \frac{\epsilon}{n}$ and so $$\sum_{i=1}^n(b_i-a_i) \leq n \cdot \frac{\delta}{n}=\delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq n \cdot \frac{\epsilon}{n} =\epsilon$$ So, $f$ is absolutely continuous.
Now, $f$ has a bounded first derivative, which is defined a.e. This can be seen as follows: $\forall x,h \in{\mathbb{R}}$, $$\left|\dfrac{f(x+h)-f(x)}{(x+h)-x}\right| \leq M$$ This statement is true for all $x$ and $h$, so it is therefore true in the limit $$\lim_{h \rightarrow 0}\left|\dfrac{f(x+h)-f(x)}{h}\right| =|f'(x)| \leq M$$ "$(\Longleftarrow)$" Assume that $|f'(x)| \leq M$ and $f$ is absolutely continuous. By Theorem 3.11, an absolutely continuous function $f$ has the properties that $f'$ exists a.e., $f'\in{L^1}$ and $f$ can be represented with the integral of its derivative: $$f(b)-f(a)=\int_a^bf'(t)dt \quad \forall a,b \in{\mathbb{R}}$$ Using the triangle inequality, the assumption $|f'(t)| \leq M$, and the linearity of the Lebesgue integral, $$|f(b)-f(a)| = \left|\int_a^bf'(t)dt\right| \leq \int_a^b|f'(t)|dt \leq \int_a^b M dt = M\int_a^b dt=M|b-a|$$ $\therefore$ $f$ is Lipschitz continuous. $\blacksquare$

Stein 2.19

$\textbf{Exercise 2.19}$ Let $f\in{L^1(\mathbb{R}^d)}$. For each $\alpha>0$ define $E_\alpha=\{x : |f(x)|>\alpha\}$. Prove that $$\int_{\mathbb{R}^d}|f(x)|dx = \int_0^\infty m(E_\alpha)d\alpha$$ Strategy: this proof uses some techniques similar to the proof of corollary 3.8. The first trick is relate function value to measure of an interval: $|f(x)|=m([0,|f(x)|))$.

$\textit{Proof}$: Since $f$ integrable, the following equalities hold in the non-extended sense. Again relating integration and measure, we have $m([0,|f(x)|))=\int_\mathbb{R}\chi_{[0,|f(x)|)}$ $$\int_{\mathbb{R}^d}|f(x)|dx=\int_{\mathbb{R}^d}m[0,|f(x)|)dx =\int_{\mathbb{R}^d} \left(\int_\mathbb{R}\chi_{[0,|f(x)|)}(\alpha)d\alpha\right)dx \quad (\star)$$ with $\alpha$ acting as a dummy variable in the above right integral.
Observe that for $\alpha\in{\mathbb{R}}$ (or even $\alpha>0$) $$ \chi_{[0,|f(x)|)}(\alpha) = \left\{ \begin{array}{lr} 1 & : 0<\alpha< f(x) \\ 0 & : \alpha > |f(x)| \\ \end{array} \right. $$ $$ \chi_{E_\alpha}(x) = \left\{ \begin{array}{lr} 1 & : x\in{E_\alpha} \Longleftrightarrow \alpha< f(x) \\ 0 & : x\not\in{E_\alpha} \Longleftrightarrow f(x) \leq \alpha \\ \end{array} \right. $$ Note that since $f\in{L^1}$ then $f$ is measurable, implying by definition of measurable functions that the set $E_\alpha$ is measurable, which is iff $\chi_{E_\alpha}$ is measurable.
By Tonelli's theorem, and $\alpha>0$, $(\star)$ becomes: $$=\int_\mathbb{R} \left( \int_{\mathbb{R}^d}\chi_{E_\alpha}(x)dx \right) d \alpha =\int_\mathbb{R}m(E_\alpha)d\alpha = \int_0^\infty m(E_\alpha)d \alpha \quad \blacksquare$$

Stein 3.7

$\textbf{Exercise 3.7}$ Prove that if a measurable subset $E$ of $[0,1]$ satisfies $$m(E \cap I) \geq \alpha m(I) \quad (\star)$$ for some $\alpha > 0$ and all intervals $I \in{[0,1]}$ then $m(E) = 1$

Strategy: Instead of appealing directly to corollary 1.5 , we consider $\chi_E$. If we can then show that $\chi_E=1$ a.e. for $E\in{[0,1]}$ then this implies that $m(E)=1$.

$\textit{Proof}$ : Since $I$ is an interval, then $m(I)>0$, so in $(\star)$ above, we can divide through to get $$0<\alpha \leq \frac{m(E \cap I)}{m(I)}$$ Consider $\chi_E$, which is of course integrable on [0,1]. Then, the Lebesgue differentiation theorem gives us that $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_I\chi_E(x)dx=\chi_E \quad a.e. x\in{[0,1]}$$ Again, one of the key themes in these types of problems is to relate integration and measure. We may write $\int \chi_E = \int_E 1 = m(E)$. Further, we can write $\int_I \chi_E = \int_{I \cap E} 1 = m(E \cap I)$. So, our above expression becomes $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_{(I \cap E)} 1 dx= \lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} = \chi_E \quad a.e. x\in{[0,1]}$$ $$\lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} \geq \alpha > 0$$ So, $$\chi_E > 0 \quad a.e.$$ However, characteristic functions take on only values of 0 and 1. So, $$\chi_E>0 \Longrightarrow \chi_E = 1 \quad a.e. \Longrightarrow \int \chi_E = 1 = m(E) \quad \blacksquare$$

Stein 2.11

Prove that if $f\in{L^1(\mathbb{R}^d)}$, real-valued, and $\int_E f(x)dx \geq 0$ for every $E\in{\mathcal{M}}$ then $f(x) \geq 0$ a.e. $x$. Similarly, if $\int_E f(x)dx=0$ for every $E\in{\mathcal{M}}$ then $f(x) = 0$ a.e.

$\textit{Proof}$ : This problem can be tackled in various ways using the tools so far. One way is as follows:

Define $A=\{x \mid f(x) < 0 \}$. $f$ is measurable so $ \{f < a \} \in{\mathcal{M}} $, $\forall a\in{\mathbb{R}}$. Letting $a=0$ makes the set $A\in{\mathcal{M}}$. By assumption, $$\int_Af(x)dx \geq 0$$ By the way $A$ is defined, $$f\chi_A(x) \leq 0$$ $$\int_Af(x)dx = \int_\mathbb{R}f(x) \chi_A(x)dx \leq \int_\mathbb{R}0=0$$ Therefore, $$ \int_Af(x)dx=0$$ We have $f(x) < 0$ $\forall x\in{A}$ and $\int_Af(x)dx=0$. Now assume that to the contrary that $m(A) > 0$. Let $Q_n $ be the compact cube centered at the origin with side lengths $n$. So, $A \cap Q_n$ is measurable and bounded and $A\cap Q_n \nearrow_{n\rightarrow \infty} A$. [Then, by corollary 3.3, $m(A)=\lim_{n\rightarrow \infty}m(A \cap Q_n)$.] On $A \cap Q_n$ we have conditions for Lusin's theorem met. So, on a contained closed set $F_\epsilon \subseteq A \cap Q_n$ $f\in{C(F_\epsilon)}$. Additionally, $F_\epsilon$ has positive measure. With $f(x_0) <0$ for some $x_0$ on a set of positive measure, then the integral of $f$ around a neighborhood of $x_0$ must be negative since $f$ is continuous...a contradiction.
$\therefore \quad m(A)=0$ i.e. $f \geq 0$ a.e.

Alternatively, a very quick and direct way of proof is to use one of the main results of chapter 3: the Lebesgue differentiation theorem. This states that if $f$ is integrable (or even just locally integrable) then $$\lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt = f(x) \quad a.e.$$ Balls in $\mathbb{R}^d$ are measurable, so over any ball, $\int_B f(t)dt \geq 0 $. Since $f$ is integrable, we apply the Lebesgue differentiation theorem. So, $$f(x) = \lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt \geq 0 \quad a.e.$$ i.e. $f \geq 0$ a.e. Done.

Similarly, if $\int_E f(x)dx=0$, then we get $$f(x) = \lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt=0 \quad a.e. \quad \quad \blacksquare$$

Stein 2.9 (Chebyshev's inequality)

(Chebyshev's inequality)

Let $f\in{L^1(\mathbb{R}^d)}$, $f \geq 0$, $\alpha > 0$ and $E_\alpha = \{x | f(x) > \alpha \}$.

Then, $$m(E_\alpha) \leq \frac{1}{\alpha} \int f $$

$\textit{Proof}$ : $$E_\alpha = \{x | f(x) > \alpha \} = \{ x | \frac{f(x)}{\alpha} > 1 \}$$ Integration can be considered a way of measuring a set, so we can write $$m(E_\alpha) = \int \chi_{E_\alpha} = \int_{E_\alpha} 1 $$ By the monotinicity and linearity of the Lebesgue integral, $$\leq \int_{E_\alpha} \frac{f(x)}{\alpha} = \frac{1}{\alpha}\int_{E_\alpha}f \leq \frac{1}{\alpha} \int f \quad \quad \blacksquare $$

Commutativity condition forces diagonalization

$\textit{exercise}$ : Let $A,B,C\in{M_{n \times n}(\mathbb{R})}$ and suppose the characteristic polynomial of $A$ splits and separates in $\mathbb{C}$. Suppose also that $AB=BA$ and $AC=CA$. Prove that $BC=CB$

$\textit{proof}$: If the characteristic polynomial $p_A(t)$ splits and separates in $\mathbb{C}$ i.e. has no repeated roots in $\mathbb{C}$ then $A$ will diagonalize in $\mathbb{C}$. We first show that if $D$ is diagonal in $\mathbb{C}$ and $DB=BD$ then this implies that $B$ must also be diagonal. \\ Let $D\in{M_{n \times n}(\mathbb{C})}$ where $D$ is diagonal. Since $D$ separates in $\mathbb{C}$ then for the diagonal entries, $a_{ii} \neq a_{jj} \quad \forall i \neq j$. If $DB=BD$ then $B$ is diagonal since $$ [DB]_{ij}= \left\{ \begin{array}{lr} d_{ii}b_{ii} : i=j \\ d_{ii}b_{ij} : i \neq j \end{array} \right. $$ $$ [BD]_{ij}= \left\{ \begin{array}{lr} b_{ii}d_{ii} : i=j \\ b_{ij}d_{jj} : i \neq j \end{array} \right. $$ Since $DB=BD$, then $d_{ii}b_{ij}=b_{ij}d_{jj}$. But with each $d_{ii} \neq d_{jj}$ and $\mathbb{C}$ being a field (commutative, inverses exist, and also no zero divisors in case $d_{ii}=0$ for some $i$) this forces $b_{ij}=0$. Similarly, $b_{ji}=0$, so all off-diagonal entries of $B$ are zero i.e. $B$ is diagonal.

We have shown that if given a diagonal matrix with unique entries commuting with another matrix, then the other matrix must also be diagonal; regardless of the splitting field of the characteristic polynomial. This commutativity condition is once which can be employed in certain problems to create simultaneous diagonalization of linear operators. This technique is sometimes used in solving PDE's.

Now, given the conditions above, show that $BC=CB$.
$A$ diagonalizes in $\mathbb{C}$, so there exists an invertible matrix (change of basis matrix) $Q$ and a diagonal matrix $D$ s.t. $D=Q^{-1}AQ$. Since $AB=BA$ it follows that $$(Q^{-1}AQ)(Q^{-1}BQ)=Q^{-1}ABQ=Q^{-1}BAQ=(Q^{-1}BQ)(Q^{-1}AQ)$$ Thus, by the previous result, the matrix $Q^{-1}BQ$ is diagonal. Similarly, $Q^{-1}CQ$ must be diagonal. Using the fact that diagonal matrices commute with one another, we get, $$(Q^{-1}BQ)(Q^{-1}CQ)=(Q^{-1}CQ)(Q^{-1}BQ)$$ $$Q^{-1}BCQ=Q^{-1}CBQ \quad \therefore BC = CB$$

Jordan Canonical Form (example)

Let $$A=\begin{pmatrix} 3&0&1\\ 2&2&2\\ -1&0&1\\ \end{pmatrix}$$ Determine a change of basis matrix $Q$ and a matrix $J$ such that $J=Q^{-1}AQ$ where $J$ is the Jordan canonical form of $A$.

$\textit{Solution:}$ We are typically interested in $\lambda$, $v$ such that $Av=\lambda v$. Choosing vectors in our space $V$ that are purely scaled under the linear transformation represented by the matrix $A$ can give us a subspace that is dilated by $A$. We also insist that $v \neq 0$, since 0 adds nothing to a basis.
$Av= \lambda v$
$Av-\lambda v = 0$
$(A-\lambda I)v=0$
So, we wish to look at $N(A-\lambda I) = \{v\in{V} | (A-\lambda I)v=0 \}$. With the assumption that $v \neq 0$, then this means that $N(A-\lambda I)$ is non-trivial $\Longleftrightarrow (A-\lambda I)$ is linearly dependent $\Longleftrightarrow det(A- \lambda I)=0$ by the invertible matrix theorem. To compute $det(A- \lambda I)$ we choose a cofactor expansion down the second column: $$det(A- \lambda I)=det \begin{pmatrix} 3-\lambda &0&1\\ 2&2-\lambda &2\\ -1&0&1-\lambda \end{pmatrix} =(2-\lambda)((3-\lambda)(1-\lambda)-(-1)(1)) $$ $$=(2-\lambda)(\lambda^2-4\lambda+4)=(2-\lambda)(\lambda-2)(\lambda-2)=p(\lambda)$$ So, the characteristic polynomial $p$ of $A$ has eigenvalue $\lambda=2$ with a multiplicity of 3. To check $A$ for diagonalizability, we can look at the dimension of the eigenspace corresponding to $\lambda=2$ and/or we can look at the minimal polynomial of $A$. The minimal polynomial $m$ of $A$ has the following characteristics: monic, irreducible, minimal degree, contains all the roots/eigenvalues, divides the characteristic polynomial, $m(A)=0$. With these constraints, the minimal polynomial is either $m(t)=t-2,(t-2)^2,(t-2)^3=p(t)$. $A-2I \neq 0$ so we now check $(A-2I)^2$: $$\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix}= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}$$ So, $m(t)=(t-2)^2$, which is not separable, i.e. has repeated roots, thus $A$ does not diagonalize. We may also note that the second and third rows of $A-2I$ are multiples of the first, so $rank(A-2I)=1$. So, the dimension of the eigenspace corresponding to $\lambda=2$ is $$dim(E_{\lambda=2})=3-rank(A-2I)=3-1=2$$ which is consistent with the minimal polynomial having a repeated root. However, the dimension of the generalized eigenspace, $dim(K_{\lambda=2})=3$. So, the Jordan canonical form for $A$ must have a 2-cycle and a 1-cycle. $$J=\begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix}$$ The desired cycle structure can be diagrammed as follows:
$A-2I:v' \mapsto v \mapsto 0$
$A-2I: w \mapsto 0$
Our change of basis matrix $Q$ then lines up these cycles (which go right to left) with the Jordan blocks of $J$ $$Q=\begin{pmatrix} |&|&|\\ w&v&v'\\ |&|&| \end{pmatrix}$$ Now to find $w,v$ and $v'$:
$$(A-2I) =\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \sim \begin{pmatrix} 1&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}$$ So with two rows of 0's we can have 2 free variables, so we can thus make 2 "regular" (non-generalized) eigenvectors. This is consistent with the previous observation that $dim(E_{\lambda=2})=3-1=2$. In a matrix-vector form, we have $a+c=0$ and 2 free variables. Letting $a=c=0$ and $b=1$ gives us $(0,1,0)=e_2$ and we can check that $(A-2I)e_2=0$: $$\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$ So our eigenvector $w=e_2$ gives us a 1-cycle. To obtain $v'$ (our starting 2-cycle vector) we need to pick it so that it is outside our eigenspace $E_{\lambda=2}$. Even if we don't know exactly what this space is, we can often get lucky just by trying vectors in the standard basis $\{e_1,e_2,e_3\}$.
$(A-2I)e_1=(1,2,-1)$, so $$(A-2I)^2e_1=\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 1\\ 2\\ -1\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$ This gives us our desired 2-cycle. If we are not lucky enough to find a starting cycle vector in the standard basis, we can always force a starting cycle vector by looking at $N((A-\lambda I)^2)$.
Our change of basis matrix is then $$Q=\begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix}$$ To check that indeed $J=Q^{-1}AQ$ we could find $Q^{-1}$. $Q^{-1}$ exists since a generalized eigenbasis always consists of linearly independent vectors that make up the columns of $Q$, so $Q$ is invertible. However, we don't need to actually do this. If we simply multiply the above equation through by $Q$ on the left, then we get $QJ=AQ$. $$QJ=\begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$ $$AQ = \begin{pmatrix} 3&0&1\\ 2&2&2\\ -1&0&1 \end{pmatrix} \begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$

D+F 7.3.12

$\textbf{Exercise}[7.3.12]$ (Parts (a) and (b)) Let $D \in{\mathbb{Z}}$ that is not a perfect square and let $$S=\{\begin{pmatrix} a&b\\ Db&a \end{pmatrix} \mid a,b\in{\mathbb{Z}}\}$$
(a) Prove that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$
(b) If $D$ is not a perfect square in $\mathbb{Z}$ prove that the map $\varphi: \mathbb{Z}[\sqrt{D}] \longrightarrow S$ defined by $$\varphi(a+b\sqrt{D})=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$$ is a ring isomorphism.
$\textit{Proof:}$ To show that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$, we need to show that $S \neq \emptyset $, $S$ is closed under subtraction, and $S$ is closed under multiplication. $S$ is clearly non-empty. For example, $\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} \in{S}$
Now let $A,B\in{S}$. $$A=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}, B=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}$$ Then, $$A-B=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}-\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} a-c&b-d\\ D(b-d)&a-c \end{pmatrix} \in{S}$$ The product, $$AB=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ cbD+adD&Dbd+ac \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}\in{S}$$ Therefore, $S$ is a subring.

We now need to show that $\varphi$ is well defined, a bijection, and a ring homomorphism. To check that $\varphi$ is well defined, we pick two equal representatives from $\mathbb{Z}[\sqrt{D}]$ and show that their images in $S$ are equal. So, let $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$. Note that the assumption that $D$ is square-free is necessary. For example, let $D=4=2^2$. Consider $0+1 \cdot \sqrt{4}=2$ and $2+0\cdot \sqrt{4}=2$. These are 2 elements in $\mathbb{Z}[\sqrt{4}]$ that are equal, yet have a different representation. Then, $\varphi(0+1 \cdot \sqrt{4})=\begin{pmatrix} 0&1\\ 2&0 \end{pmatrix}$ however, $\varphi(1+0 \cdot \sqrt{4})=\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}$. This gives us 2 different matrices being mapped from the same element. So, if $D$ has a square in its factorization, then $\varphi$ is ill-defined.
Let $D$ be square-free. Now, $\varphi(x)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$ and $\varphi(y)=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}$. Since $a,b,c,d\in{\mathbb{Z}}$ and $D$ is square-free, $x=y \Longleftrightarrow a=c$ and $b=d$ thus $\varphi(x)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\varphi(y)$. Reversing this process, if $\varphi(x)=\varphi(y)$ then each entry in the above matrices must be equal, and we get that $a=c$ and $b=d$ so $\varphi$ is an injective map.
$\varphi$ is surjective: Let $M\in{S}$ where $M=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$. Picking $x\in{\mathbb{Z}[\sqrt{D}]}$ where $x=a+b\sqrt{D}$, then $\varphi(x)=M$.
$\varphi$ is a ring homomorphism since it preserves both operations. Again letting $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$, we see $\varphi$ preserves addition since: $$\varphi(x+y)=\varphi((a+b\sqrt{D})+(c+d\sqrt{D}))=\varphi((a+c)+(b+d)\sqrt{D})$$ $$=\begin{pmatrix} a+c&b+d\\ D(b+d)&a+c \end{pmatrix}=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}+\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\varphi(x)+\varphi(y)$$ $$\varphi(xy)=\varphi((a+b\sqrt{D})(c+d\sqrt{D}))=\varphi(ac+ad\sqrt{D}+bc\sqrt{D}+bdD)=\varphi(ac+(ad+bc)\sqrt{D}+bdD)$$ $$=\varphi(ac+bdD+(ad+bc)\sqrt{D})=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}$$ $$\varphi(x)\varphi(y)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix} \begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}=\varphi(xy)$$ Thus, $\varphi$ is a well-defined, bijective ring homomorphism i.e. a ring isomorphism.
$\therefore \quad \mathbb{Z}[\sqrt{D}] \cong S$ $\blacksquare$

Q[\sqrt{2}] not isomorphic to Q[\sqrt{3}]

Show that $\mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$

$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}|a,b\in{\mathbb{Q}}\}$ and $\mathbb{Q}[\sqrt{3}]= \{c+d\sqrt{3}|c,d\in{\mathbb{Q}}\}$.

$\textit{Proof}$: Assume to the contrary that $\mathbb{Q}[\sqrt{2}] \cong \mathbb{Q}[\sqrt{3}]$. So, there exists an isomorphism between these two fields. First, we justify but do not prove that there is no problem of identification with elements between these two fields. In other words the number say 2 in $\mathbb{Q}[\sqrt{2}]$ is the same as the number 2 in $\mathbb{Q}[\sqrt{3}]$. With an isomorphism between them, $$\varphi(1)=1$$ $$\varphi(1+1)=\varphi(1)+\varphi(1)=1+1=2$$ And so on, so the integers must be fixed by this isomorphism. With the integers fixed, then the field of fraction of the integers, which is $\mathbb{Q}$ must be fixed. In other words, the ground field of these two field extensions must be fixed. This is consistent with the idea in Galois theory that automorphisms of field extensions fix the ground field.
Isomorphisms fix structures. In $\mathbb{Q}[\sqrt{2}]$ there is an element, namely $\sqrt{2}$ such that when squared, we get the number 2. Therefore, there must exist an element in $\mathbb{Q}[\sqrt{3}]$ such that when squared, we get 3.
$(c+d\sqrt{3})^2=2$
$c^2+2cd\sqrt{3}+3d^2=2$. $(\star)$
$c,d\in{\mathbb{Q}}$ and $cd\sqrt{3} \not \in{\mathbb{Q}}$, so $cd\sqrt{3}=0 \Longrightarrow c=0$ or $d=0$ since we are in a field.
If $c=0$ then $(\star)$ becomes
$3d^2=2 \Longrightarrow d=\pm \sqrt{\frac{2}{3}} \not \in {\mathbb{Q}}$...a contradiction
Id $d=0$, then $(\star)$ becomes
$c^2=2 \Longrightarrow c=\pm \sqrt{2} \not \in{\mathbb{Q}}$...a contradiction.
In any case, an isomorphism between $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ leads to a contradiction.
$\therefore \quad \mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$

D+F 7.4.4

Assume $R$ is a commutative ring with $1 \neq 0$. Prove that $R$ is a field iff $0$ is a maximal ideal.

Recall that an ideal $M$ of $R$ is maximal if $M \neq R$ and the only ideals containing $M$ are $R$ and $M$.
Recall that one way of defining a field is that it is a commutative ring with identity in which every nonzero element has an inverse (with respect to both $+$ and $\times$ of course)
This problem is another way of establishing that fields have trivial ideal structures. Note the analogy here with groups: Simple groups are special types of groups that have trivial normal subgroup structures. Fields are special types of rings that have trivial ideal structures.
$\textit{Proof}$ : "$\Longrightarrow$": Let $I$ be an ideal of $R$ s.t. $(0) \subset I \subset R$. So $\exists r \neq 0$ s.t. $r\in{I}$. Since $R$ is a field, multiplicative inverses exist, so $rr^{-1}\in{I} \Longrightarrow 1 \in{I}$. With the identity being in $I$, this means that $s \cdot 1 \in{I}$ $\forall s \in{R}$ since $I$ is, in particular, a left ideal. So, $s\in{I}$. But $s$ was arbitrary, so all elements lie in $I$, i.e. $I=R$.
$\therefore \quad (0)$ is a maximal ideal of $R$.

"$\Longleftarrow$": Assume that $(0)\in{R}$ is a maximal ideal. So, the only ideals containing $(0)$ are $(0)$ and $R$. So, $\forall r \neq 0$, the ideal generated by $0$ and $r$, $(0,r)=R$. In particular, if $r \neq 0$ then $1 \in{(0,r)}$. So, $1=a \cdot 0 + b\cdot r$ for some $a,b\in{R}$.
$br=1$
$b=r^{-1}$. Since $b$ and $r$ were arbitrary, inverses exist in general in $R$. So, $R$ is a commutative ring with identity $1 \neq 0$ in which every element has an inverse.
$\therefore \quad R$ is a field. $\blacksquare$

D+F prop. 14.1.2

[(Prop. 14.1.2)] Given the conditions that $K/F$ is a field extension, $\alpha \in{K}$ is algebraic over $F$ and $\sigma \in{Aut(K/F)}$ then prove that any polynomial with coefficients in $F$ having $\alpha$ as a root also has $\sigma(\alpha)=\sigma\alpha$ as a root.

Some notes: $\alpha\in{K}$ albegraic over $F$ means that $\alpha$ is the root of some nonzero polynomial in $F$.
$\sigma \in{Aut(K/F)}$ means that $\sigma$ is an operation preserving permutation on the elements of $K$ and $\sigma$ fixes $F$ i.e. $\sigma(a)=a \quad \forall a\in{F}$.


$\textit{Proof}$ :
Let $\alpha$ be a root $f(x)\in{F}[x]$. Dividing a polynomial through by a nonzero constant to make it monic does not change the roots. So, we may write $$\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=0 \quad a_i\in{F}$$ Now applying $\sigma$ to both sides, we get $$\sigma(\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=\sigma(0)=0 \quad (\star)$$ $$=\sigma(\alpha^n) + \sigma(a_{n-1}\alpha^{n-1} + \cdots + \sigma(a_0)=0$$ since $\sigma$ is a homomorphism with respect to addition. $$=(\sigma(\alpha))^n+ \sigma(a_{n-1})\sigma(\alpha)^{n-1} + \cdots + \sigma(a_0)$$ since $\sigma$ is also a homomorphism with respect to multiplication. Now, since $\sigma$ fixes $F$, $$=(\sigma\alpha)^n + a_{n-1}(\sigma\alpha)^{n-1}+ \cdots + a_0$$ Which states exactly that $\sigma\alpha$ is also a root of $f(x)$.

$(\star): \sigma(0)=0$ since $\sigma$ is an automorphism: $\sigma(0)=\sigma(0+0)=\sigma(0)+\sigma(0) \Longrightarrow \sigma(0)=0$. $\blacksquare$

D+F 14.6.4

Determine the Galois group of $x^4-25$

Recall that the Galois group of a separable polynomial $f(x) \in{F[x]}$ is defined to be the Galois group of the splitting field of $f(x)$ over $F$.

$x^4-25$ partially factors over $\mathbb{Q}$ as $$x^4-25=(x^2-5)(x^2+5)$$ and the four roots are $\pm\sqrt{5} \pm\sqrt{5}i$. So, this polynomial has no repeated roots, and is thus separable. We first determine the splitting field and then use the fundamental theorem of Galois to flip the diagram of known subfields to get the Galois group lattice. We need to adjoin $\sqrt{5}$ to $\mathbb{Q}$, but notice that the two complex roots are not contained in $\mathbb{Q}[\sqrt{5}]$. So, we need to adjoin both $\sqrt{5}$ and $i$ for our splitting field to contain all 4 roots. Thus, the splitting field of $x^4-25$ should be $\mathbb{Q}[\sqrt{5},i]$. The lattice of known subfields is $\mathbb{Q}$ as the base field, then $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[i]$ and $\mathbb{Q}[\sqrt{5}i]$ as degree 2 extensions over $\mathbb{Q}$. Then, $\mathbb{Q}[\sqrt{5},i]$ on top as a degree 2 extension of the 3 intermediate subfields. By the fundamental theorem of Galois, the flipped image of this subfield lattice is exactly the lattice of the Galois group of $x^4-25$. This lattice is the lattice for the Klein 4-group.

D+F 13.5.8

Prove that $f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]}$.

$\textit{Proof}$ : First note that since we are in a field of charachteristic $p$, then we may use the properties of the Frobenius automorphism: $$(a+b)^p \equiv a^p + b^p \quad and \quad (ab)^p \equiv a^pb^p$$ where we have used "$\equiv$" to remind ourselves that we are working in charachteristic $p$.
Similarly for more terms, we have $$(a+b+c)^p \equiv ((a+b)+c)^p \equiv (a+b)^p+c^p \equiv a^p + b^p + c^p$$ $$(abc)^p \equiv ((ab)c)^p \equiv (ab)^pc^p \equiv a^pb^pc^p$$ Let $$f(x)=a_nx^n+ \cdots +a_0 \quad a_i\in{\mathbb{F}_p}$$ Then by induction, $$f(x)^p \equiv (a_nx^n+ \cdots + a_0)^p \equiv (a_nx^n)^p + \cdots + (a_0)^p \equiv (a_n)^p(x^n)^p + \cdots + a_0^p$$ $$ \equiv a_n(x^n)^p+ \cdots + a_0 \equiv f(x^p)$$ Since $a^p \equiv a (modp)$.
$\therefore \quad f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]} \quad \blacksquare$

D+F 13.4.1

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4-2$.

$\textit{Proof}$ : The four roots of $x^4-2$ can found by factoring: $$x^4-2=(x^2)^2-(\sqrt{2})^2=(x^2-\sqrt{2})(x^2+\sqrt{2})$$ $$=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-\sqrt[4]{2}i)(x+\sqrt[4]{2}i)$$ These four roots can be thought of as the 4th roots of unity scaled by $\sqrt[4]{2}$.
In this factorization, we need to adjoin $\sqrt[4]{2}$ and $i$. Now, $$\mathbb{Q}[\sqrt[4]{2}]=\{a_0+a_1\sqrt[4]{2}+a_2\sqrt{2}+a_3\sqrt[4]{2}^3|a_i\in{\mathbb{Q}}\}$$ so, $$|\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}|=4$$ But the 2 complex roots $\sqrt[4]{2}i$ and $-\sqrt[4]{2}i$ are not in $\mathbb{Q}[\sqrt[4]{2}]$, so we need to also adjoin $i$. Now, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}[\sqrt[4]{2}]|=2$$ So, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}|=2 \cdot 4 = 8$$ Thus, the splitting field of $x^4-2$ is a degree 8 extension.

D+F 13.4.4

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^6-4$.

Splitting fields are the smallest field extensions which allow $f(x)$ to factor/split completely. $$x^6-4=(x^3)^2-(2)^2=(x^3+2)(x^3-2)$$ $$=(x+\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6)(x-\sqrt[3]{2}\zeta_6^5)(x-\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6^2)(x-\sqrt[3]{2}\zeta_6^4)$$ Where $\zeta_6=\frac{1+\sqrt{-3}}{2}$. These are the 6th roots of unity scaled by $\sqrt[3]{2}$. So, in order for all 6 of these roots to be contained in an extension of $\mathbb{Q}$, we need to adjoin $\sqrt{-3}$ and $\sqrt[3]{2}$. So, the splitting field of $x^6-4$ is $\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]$. Now, $|\mathbb{Q}[\sqrt{-3}]:\mathbb{Q}|=2$ and $|\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}|=3$. By corollary 22 p. 529, $$|\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]:\mathbb{Q}|=2 \cdot 3 = 6$$

D+F 7.4.9

Let $R$ be the ring of continuous functions on $[0,1]$ and let $I$ be the collection of functions $f(x)\in{R}$ s.t. $$f(\frac{1}{3})=f(\frac{1}{2})=0$$ Prove that $I$ is an ideal of $R$ but not a prime ideal.

$\textit{Proof}$ : To show that $I$ is an ideal, we need to show that $I \neq \emptyset$, $I$ is closed under subtraction and $I$ is closed under left and right multiplication by arbitrary elements of $R$.
$0$, the zero function is in $I$, so $I \neq \emptyset$
Let $f(x),g(x) \in{I}$. Then $f(\frac{1}{3})=f(\frac{1}{2})=g(\frac{1}{3})=g(\frac{1}{2})=0$. $$(f-g)(\frac{1}{3})=f(\frac{1}{3})-g(\frac{1}{3})=0-0=0$$ and $$(f-g)(\frac{1}{2})=f(\frac{1}{2})-g(\frac{1}{2})=0-0=0$$ Or also $$f(\frac{1}{2})-g(\frac{1}{3})=0-0=0$$ $$f(\frac{1}{3})-g(\frac{1}{2})=0-0=0$$ So, $I$ is closed under subtraction.
Now let $f(x)\in{I}$ and $h(x)\in{R}$. $$(fh)(\frac{1}{2})=f(\frac{1}{2}) \cdot h(\frac{1}{2})=0 \cdot a = 0$$ Similarly, $(hf)(\frac{1}{2})=0=(fh)(\frac{1}{3})=(hf)(\frac{1}{3})=0$.
So, $I$ is an ideal of $R$.

Since the ring of continuous functions is a commutative ring, we may have the notion of a prime ideal. An ideal $P$ is prime in the commutative ring $R$ if $P \neq R$ and $$fg \in{P} \Longrightarrow f\in{P} \quad or \quad g\in{P}$$ We would like to find $f,g \in{I}$ s.t. $fg\in{I}$ but $f \not \in{I}$ and $g \not \in {I}$.
Consider the linear functions $f(x)=x-\frac{1}{3}$ and $g(x)=x-\frac{1}{2}$.
$$f(\frac{1}{2}) \neq 0 \Longrightarrow f\not \in {I}$$ $$g(\frac{1}{3}) \neq 0 \Longrightarrow g \not \in{I}$$ However, $$(fg)(x)=(x-\frac{1}{3})(x-\frac{1}{2})$$ and so $$(fg)(\frac{1}{3})=(fg)(\frac{1}{2})=0 \Longrightarrow fg\in{I}$$ which amounts to a counterexample to $I$ being prime. So, $I$ is an ideal of $R$ but not a prime ideal. $\blacksquare$

Prime implies Irreducible in integral domains

Proposition 8.3.10:
In an integral domain, a prime element is always irreducible.

First recall some definitions:
Definition: An integral domain is a commutative ring with no zero divisors.
The lack of zero divisors gives integral domains a useful cancellation property:
$$ab=ac \Longrightarrow b=c \quad or \quad a=0$$ A nonzero elemnent $p$ is prime in the ring $R$ if the ideal generated by $p$, $(p)$ is a prime ideal.
An ideal $P$ is prime if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
An element $r\in{R}-\{0\}-R^\times$ is irreducible if $r=xy \Longrightarrow x\in{R^\times}$ or $y\in{R^\times}$ where $R^\times$ is the group of units of $R$.

$\textit{Proof}$ : Let $R$ be an integral domain and let $p\in{R}$ be a prime element. So, $(p)$ is a nonzero prime ideal generated by $p$ and $p=ab$. We want to show that $a\in{R^\times}$ or $b\in{R^\times}$.
So, $ab=p\in{(p)}$. Since $(p)$ is a prime ideal this means that $(p) \neq R$ and $ab\in{(p)} \Longrightarrow a\in{(p)}$ or $b\in{(p)}$. Say $a\in{(p)}$. Then, $\exists r\in{R}$ s.t. $a=pr$. So, we have:
$$p=ab=pr \cdot b$$ $$p=p \cdot rb$$ $$p \cdot 1 = p \cdot rb$$ Since $R$ is an integral domain, this imples
$$p=0 \quad or \quad 1 = rb$$ But $p \neq 0$ since prime elements are nonzero. Thus, $1=rb \Longrightarrow b\in{R^\times}$.
Swapping the roles of $a$ and $b$, if $b\in{(p)}$ then $a\in{R^\times}$. In either case, $p$ is irreducible. $\blacksquare$

D+F 7.4.19

Let $R$ be a finite commutative ring with identity. Prove that every prime ideal of $R$ is a maximal ideal.

$\textit{Proof}$ : Let $I$ be an ideal of $R$.
$I$ maximal $\Longleftrightarrow$ $R/I$ is a field (prop. 12)
$R/I$ a field $\Longrightarrow$ $R/I$ an integral domain. Conversely,
$R/I$ an integral domain $\Longrightarrow$ $R/I$ a field since $|R| < \infty$. (cor. 3 p. 228)
$\Longleftrightarrow$ $I$ is a prime ideal (prop. 13)
$\therefore$ in finite commutative rings with identity, maximal iff prime. $\blacksquare$

D+F 7.4.13

Let $\varphi:R \rightarrow S$ be a homomorphism of commutative rings.
(a) Prove that if $P$ is a prime ideal of $S$ then either $\varphi^{-1}(P)=R$ or $\varphi^{-1}(P)$ is a prime ideal of $R$.
(b) Prove that if $M$ is a maximal ideal of $S$ and $\varphi$ is surjective then $\varphi^{-1}(M)$ is a maximal ideal of $R$. Give an example to show that this may not be true if $\varphi$ is not surjective.

$\textit{Proof}$ : Exercise 7.3.24 establishes that the preimage or pullback of an ideal is indeed an ideal. So $\varphi^{-1}(P)$ is an ideal of $R$.

Recall that an ideal of a commutative ring is called a prime ideal if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.

Let $r_1r_2 \in{\varphi^{-1}(P)} \Longrightarrow \varphi(r_1r_2)\in{P}$. $\varphi(r_1r_2)=\varphi(r_1)\varphi(r_2)\in{P}$ since $\varphi$ is a ring homomorphism. Since $P$ is a prime ideal in $R$, this means $\varphi(r_1)\in{P}$ or $\varphi(r_2)\in{P}$ which implies
$r_1\in{\varphi^{-1}(P)}$ or $r_2\in{\varphi^{-1}(P)}$ So, $\varphi^{-1}(P)$ is a prime ideal in $R$. Now, if $\varphi^{-1}(P)$ happened to contain a unit (and there is no apparent reason why it cannot), then by proposition 9 . 253, $\varphi^{-1}(P)=R$. In any case, $\varphi^{-1}(P)$ is either a prime ideal or all of $R$.

For part (b), recall that an ideal $M$ of $S$ is a maximal ideal if $M \neq S$ and the only ideals containing $M$ are $M$ and $S$.

Let $M$ be a maximal ideal of $S$. Since $M$ is not all of $S$, then $\exists s \in{S}$ and $s\not \in{M}$. Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and $r\not \in \varphi^{-1}(M)$. Thus, $\varphi^{-1}(M) \neq R$.
Let $L \subset R$ be an ideal of $R$ containing $\varphi^{-1}(M)$. We want to show that $L=R$ or $L=\varphi^{-1}(M)$.
$L$ contains $\varphi^{-1}(M) \Longrightarrow \varphi(L)$ contains $M$. Then, $\varphi(L)=M$ or $S$ since $M$ is maximal in $S$. Therefore, $L=\varphi^{-1}(M)$ or $R$. $\blacksquare$.

If we relax the condition that $\varphi$ is surjective, then consider the following example:
$\phi: R \rightarrow \mathbb{F}_2$
$\phi: r \mapsto 0$
$(0)=0$ is maximal in $\mathbb{F}_2$. But, $\phi^{-1}(\mathbb{F}_2)=R$, the entire ring, so is not maximal.

D+F 7.3.2

Prove that the rings $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.

$\textit{Proof}$ : In order for $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ to be isomorphic, they must have exactly the same algebraic structure. Now, the units of $\mathbb{Z}$ are $$\mathbb{Z}^*=\{1,-1\}$$ The units of $\mathbb{Q}$ are $$\mathbb{Q}^*=\mathbb{Q}-\{0\}$$ By proposition 4, p. 235, the above are the units of $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$, respectively. With completely different subsets of units, $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ cannot be isomorphic. $\blacksquare$

7.3.1

Prove that $2\mathbb{Z} \not\cong 3\mathbb{Z}$.

$\textit{Proof}$ : Suppose to the contrary that $2\mathbb{Z} \cong 3\mathbb{Z}$. Then $\exists \varphi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}$ s.t. $\varphi$ is a ring isomorphism. $$\varphi(0)=\varphi(0+0)=\varphi(0)+\varphi(0)$$ We can subtract $\varphi(0)$ from both sides since $(3 \mathbb{Z},+)$ is an Abelian group. So, $$\varphi(0)=0$$ Now, $$\varphi(0)= \varphi(1+1)=\varphi(1)+\varphi(1)=0$$ What is $\varphi(1)$? In $3 \mathbb{Z}$, $0+0=0$, $1+1=1$ and $2+2 \equiv 1$. From above, $\varphi(1)+\varphi(1)=0$ so the only possibility is $\varphi(1)=0$.

$\varphi(0)=0$ and $\varphi(1)=0 \Longrightarrow \varphi$ is not an injective map, which contradicts the assumption that $\varphi$ was an isomorphism.

$\therefore$ $2 \mathbb{Z} \not \cong 3 \mathbb{Z}$ $\blacksquare$

D+F 4.5.18

Prove that a group of order 200 has a normal Sylow 5-subgroup.

$\textit{Proof}$ : Note that $200=2^3 \cdot 5^2$. By Sylow (iii), we know that $$n_5 \equiv 1(mod5)$$ and $$n_5 | 2^3$$ So the possible values of $n_5$ are $n_5=1,6$. But $6 \nmid 8$. So, we are forced that $$n_5=1$$ By Sylow (ii), Sylow p-subgroups are conjugate to each other. So, this Sylow 5-subgroup, $P$, which has order $5^2=25$ is conjugate to itself i.e. $$gPg^{-1}=P \quad \forall g\in{G} \Longrightarrow P \unlhd G$$ This is corollary 20, p. 142. So, for groups of order 200, they must always contain a normal Sylow 5-subgroup. $\blacksquare$

D+F 4.5.36

Prove that if $N \unlhd G$ then $n_p(G) \geq n_p(G/N)$.

In this proof, we are comparing sizes of sets. We are comparing the number of Sylow $p$-subgroups of $G$ to the number of Sylow $p$-subgroups of the quotient group $G/N$. To do this, we create a map $\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$. To show that $n_p(G) \geq n_p(G/N)$, we proceed by showing that $\varphi$ is a surjective (onto) map. This implies that the domain of $\varphi$ has cardinality greater than or equal to the cardinality of its codomain $Syl_p(G/N)$.

$\textit{Proof}$ : Let $|G|=p^am$, $|N|=p^bn$. Then $|G/N|=p^{a-b}(\frac{m}{n})$.
By exercise 4.5.34, we are going to choose our target elements in $Syl_p(G/N)$ to be of the form $PN/N$. So, $$\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$$ $$P \mapsto PN/N$$ This will allow us to use the 2nd and 4th isomorphism theorems easily.
Pick $R/N\in{Syl_p(G/N)}$. We want to show $\exists P \in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$.
If $R/N \in{Syl_p(G/N)}$ then $R/N \leq G/N$ and $|R/N|=p^{a-b}$.
By the 4th isomorphism theorem, $$R/N \leq G/N$$ where $N \leq R\leq G$
Now, $|R|=|R/N| \cdot |N|=p^{a-b} \cdot p^b\cdot n = p^a \cdot n$. So, applying Sylow's theorem to $|R|$, $\exists P \leq R$ s.t. $P\in{Syl_p(R)}$ where $|P|=p^a$.
$|P|=p^a \Longrightarrow P\in{Syl_p(G)}$.
Also, by the 2nd isomorphism theorem, $$PN/N \cong R/N$$ which can also be viewed as Sylow p-subgroups of $G/N$ being conjugate to each other, and thus isomorphic since conjugation is a group automorphism. Combining the conditions:
$N \leq R$
$P \leq R$
$P\in{Syl_p(R)}$
$PN/N \cong R/N$
we get that $$PN \leq R$$ By the 4th isomorphism theorem, $$PN/N \leq R/N$$ And since $|PN/N|=|R/N|$, then $$PN/N = R/N$$ Given arbitrary $R/N \in{Syl_p(G/N)}$ we have identified $P\in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$. Thus, $\varphi$ is a surjective mapping.
$\therefore$ $n_p(G) \geq n_p(G/N)$ $\blacksquare$

Friedburg 5.4.18

Let $A$ be an $n \times n$ matrix with charachteristic polynomial $$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t+a_0$$ (a) Prove that $A$ is invertible iff $a_0 \neq 0$.
(b) Prove that if $A$ is invertible, then $$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$ (c) Use (b) to compute $A^{-1}$ for $A=\begin{pmatrix} 1&2&1\\ 0&2&3\\ 0&0&-1\\ \end{pmatrix}$
This exercise provides an interesting result. We get a formula for the inverse of a matrix $A$.

$\textit{Proof}$ : Suppose to the contrary that $a_0=0$. We will show that this implies $A^{-1} \not \exists$
If $a_0=0$ then the charachteristic polynomial of $A$ becomes $$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t=t[(-1)^nt^{n-1}+a_{n-1}t^{n-2}+ \cdots +a_1]$$ $\Longleftrightarrow t$ is a factor of $f(t)$ $\Longleftrightarrow t=0$ is a root of $f(t) \Longleftrightarrow 0$ is an eigenvalue of $A$ $\Longleftrightarrow$ (invertible matrix theorem) $A^{-1} \not \exists$.
Each of the above steps was "iff" so,
$\therefore$ $A$ invertible iff $a_0 \neq 0$

If $A$ is invertible, then by part (a) above, $a_0 \neq 0$ in $f(t)$. We may write the charachteristic polynomial $f(t)$ in matrix form as $f(A)$. By the Cayley-Hamilton theorem, a matrix "satisfies" its own charachteristic polynomial. So, $$f(A)=0=(-1)^nA^n+ \cdots + a_1A+a_0I$$ $$-a_0I=(-1)^nA^n+ \cdots a_1A$$ Since $A$ is invertible, we may apply $A^{-1}$ to both sides of the above equation to get $$-a_0IA^{-1}=[(-1)^nA^n+\cdots +a_1A]A^{-1}$$ $$-a_0A^{-1}=(-1)^nA^{n-1}+\cdots+a_1$$ $$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$ $\blacksquare$
For part (c), we compute $$|A-tI|=(1-t)(2-t)(-1-t)=-t^3+2t^2+t-2$$ So, $a_0=-2$ and our formula gives $$A^{-1}=\frac{-1}{-2}(-A^2+2A+I)$$ Performing martrix operations, we get $$A^{-1}=\begin{pmatrix} 1&-1&-2\\ 0&\frac{1}{2}&\frac{3}{2}\\ 0&0&-1\\ \end{pmatrix}$$

4.5.23

$|G|=462 \Longrightarrow G$ is not simple.

$\textit{Proof}$ : Note that $462=2 \cdot 3 \cdot 5 \cdot 7$. By Sylow (iii), we know that $n_{11} \equiv 1 (mod11)$ and $n_{11} \mid 2 \cdot 3 \cdot 7 = 42$. So, $$n_{11}\in{\{1,12,23,34\}}$$ But, only the number 1 from this list divides 42. So, this forces $n_{11}=1$. By the conjugacy portion (ii) of Sylow's Theorem, Sylow p-subgroups are conjugate to each other. But if there is only 1, then it is conjugate to itself, thus normal (this is corollary 20, p. 142). So, we have identified a non-trivial, normal subgroup of $G$.
$\therefore$ $G$ is not simple. $\blacksquare$

D+F 7.3.10

Decide which of the following are ideals in $\mathbb{Z}[x]$

(a) The set of all polynomials whose constant term is a multiple of 3.

$\textit{Answer}$ : This set, call it $J$, is a subring. $0$ is a multiple of 3 i.e. $0 \equiv 0 (mod3)$. So, $0\in{J}$. Also, polynomial subtraction leaves the difference of 2 polynomial's constant terms in the form $3n-3m=3(n-m) \equiv 0 (mod3) \quad n,m,\in{\mathbb{Z}}$.
Multiplication of polynomials is done by continued use of the distributive law. The only purely constant term in the product of 2 polynomials in $\mathbb{Z}[x]$ will be the last term. Let $f(x)$ be a polynomial with constant term $3k, k\in{\mathbb{Z}}$. Let $g(x)\in{\mathbb{Z}[x]}$. Then, the product polynomial $f(x)g(x)=g(x)f(x)$ will have constant term $3\cdot k \cdot n \equiv 0 (mod 3)$. Thus, the set of polynomials with a constant term being a multiple of 3 is closed under left and right multiplication in the ring $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of the ring $\mathbb{Z}[x]$.

(b) The set of all polynomials whose coefficient of $x^2$ is a multiple of 3.

$\textit{Answer}$ : Consider the polynomial $3x^2+x$ which is in the given set. Then, $(3x^2+x)\cdot (3x^2+x)=9x^4+6x^3+x^2$. Thus, this set is not even closed under multiplication by its own elements. So, it is not an ideal of $\mathbb{Z}[x]$.

(c) The set of all polynomials whose constant term, coefficient of $x$ and coefficient of $x^2$ are zero.

$\textit{Answer}$ : This set, J, is a subring since $0\in{J}$ and subtraction does not change degrees of terms.
Multiplication of polynomials only adds degree to terms. Polynomials in $J$ either have degree 0 (the zero element) or degree greater than 2. So, the product of an arbitrary polynomial with one in $J$ must have either degree 0 or have degree greater than 2. Thus, this set is closed under right and left multiplication by polynomials in $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of $\mathbb{Z}[x]$.

(d) $\mathbb{Z}[x^2]$ i.e. the polynomials in which only even powers of $x$ appear.

$\textit{Answer}$ : This set, $J$, is a subring since $0\in{J}$ and subtraction of polynomials does not change the degree of any individual terms. However, $J$ is not an ideal. Consider $x^2\in{J}$. $x^2 \cdot x = x^3 \not \in{J}$.
$\therefore $ $J$ is not an ideal of $\mathbb{Z}[x]$.

(e) the set of polynomials whose coefficients sum to zero.

$\textit{Answer}$ : $p(x)=0\in{J}$. If $f(x),g(x) \in{J}$ then $f(1)-g(1)=(f-g)(1)=0-0=0$. Now let $f(x)\in{J}$ and $h(x)\in{\mathbb{Z}[x]}$. Then, $f(1)h(1)=0 \cdot c = (fh)(1)= 0$ Thus, $fh \in{J}$. Polynomial multiplication is commutative, so $hf\in{J}$. So, $J$ contains the additive identity, is closed under subtraction and is closed under left and right multiplication by arbitrary elements in $\mathbb{Z}[x]$.
$\therefore \quad J$ is an ideal of $\mathbb{Z}[x]$.

(f) the set of polynomials $p(x)$ s.t. $p'(0)=0$, where $p'(x)$ is the usual first derivative of $p(x)$ with respect to $x$.

$\textit{Answer}$ : Let $f(x)=x^2+1 \in{J}$ and $g(x)=x^2+x\in{\mathbb{Z}[x]}$. Then, $$(fg)'(0)=f(0)g'(0)+g(0)f'(0)=1 \cdot 1 + 0 \cdot 0 = 1$$ So, $fg \not \in{J}$.
$\therefore \quad J$ is not an ideal of $\mathbb{Z}[x]$.

7.1.21

Let $X$ be a nonempty set and let $\mathcal{P}(X)$ be the set of all subsets of $X$ (the power set of $X$). Define addition and multiplication on $\mathcal{P}(X)$ by: $$A+B \equiv (A-B) \cup (B-A)$$ and $$A \cdot B \equiv A \cap B$$ i.e. addition is the symmetric difference of sets and multiplication is intersection of sets.
(a) Prove that $\mathcal{P}(X)$ is a ring under these operations. Note: $\mathcal{P}(X)$ and its subrings are often referred to as rings of sets.
(b) Prove that this ring is commutative, has an identity and is a Boolean ring.

$\textit{Proof}$ : To show that $\mathcal{P}(X)$ is a ring, it suffices to show that ($\mathcal{P}(X),+,\cdot)$ satisfies the 3 axions of rings. Using basic set theory, we see that
(i) $(\mathcal{P}(X),+)$ is an Abelian group since $$A+B=(A-B)\cup (B-A)=(B-A) \cup (A-B)=B+A$$ (ii) $\cdot$ is associative since $$A\cdot(B\cdot C)=A \cap (B \cap C)=(A \cap B) \cap C= (A \cdot B) \cdot C$$ (iii) ($\mathcal{P}(X),+,\cdot$) also satisfies distributive laws since $$(A+B) \cdot C=((A-B) \cup (B-A)) \cap C$$ $$=((A\cap C)-(B\cap C)) \cup ((B \cap C)-(A \cap C))$$ $$=(A \cap C) + (B \cap C) = A \cdot C + B \cdot C$$ Right distribution is shown similarly.
So, ($\mathcal{P}(X),+,\cdot$) is a ring.

$\cdot$ is a commutative operation since $$A \cdot B=A \cap B = B \cap A = B \cdot A$$ The universe or entire set $X$ is s.t. $$A \cdot X= A \cap X= X \cap A = A \quad \forall A \in{\mathcal{P}(X)}$$ So, $X$ is the (multiplicative) identity. As a note, $\emptyset$ is the additive identity since $$A + \emptyset = (A-\emptyset) \cup (\emptyset - A) = A \cup \emptyset = \emptyset + A=A$$ Finally, ($\mathcal{P}(X),+,\cdot$) is a Boolean ring since $$A^2=A \cdot A = A \cap A = A$$ $ \therefore$ ($\mathcal{P}(X),+,\cdot$) is a Boolean ring (automatically commutative also by exercise 7.1.15) with identity. $\blacksquare$

D+F 14.6.5

Determine the Galois group of $x^4+4$.

$\textit{Solution}$ : By problem 13.2.3, we know that the splitting field for $x^4+4$ is only $\mathbb{Q}[i]$, which is a degree 2 extension. The 4 roots of $x^4+4$ are $\pm 1 \pm i$ so $x^4+4$ is separable since none of the roots are repeated. By prop. 5 p.562, this means that $$|Aut(\mathbb{Q}[i])/\mathbb{Q})|=|\mathbb{Q}[i]:\mathbb{Q}|=2$$ $\Longrightarrow \mathbb{Q}[i]$ is Galois over $\mathbb{Q}$ and the Galois group is $\mathbb{Z}_2$.

D+F 13.4.3

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4+x^2+1$.

$\textit{Solution}$ : First note that $x^4+x^2+1$ partially factors over $\mathbb{Q}$ as follows: $$x^4+x^2+1=x^2(x^2+1)+(x^2+1)-x^2=(x^2+1)^2-x^2$$ $$=((x^2+1)-(x))((x^2+1)+(x))=(x^2+x+1)(x^2-x+1)$$ Solving for the roots of these 2 quaratics, we obtain 4 solutions: $$\zeta_6,\zeta_6^2,\zeta_6^4,\zeta_6^5$$ Where $\zeta_6=\dfrac{1+\sqrt{-3}}{2}$.
So, the only element that we need to adjoin to $\mathbb{Q}$ to make $x^4+x^2+1$ split completely is $\sqrt{-3}$. So, our splitting field is $\mathbb{Q}[\sqrt{-3}]$ ...a degree 2 extension.

D+F 13.2.3

Determine the minimal polynomial over $\mathbb{Q}$ for the element $1+i$.

$\textit{Solution}$ : We want a monic, irreducible polynomial of minimal degree over $\mathbb{Q}$ s.t. $m(1+i)=0$. $$(1+i)(1+i)=2i$$ $$(2i)^2=-4$$ So, $$f(x)=x^4+4$$ satisfies $$f(1+i)=0$$ However, $x^4+4$ actually factors partially over $\mathbb{Q}$ as follows: $$x^4+4=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2$$ $$=((x^2+2)-(2x))((x^2+2)+(2x))=(x^2-2x+2)(x^2+2x+2)$$ The first of the above quadratics has the property that $$(1+i)^2-2(1+i)+2=2i-2-2i+2=0$$ ...a smaller degree polynomial with $1+i$ as a root. This is consistent with the observation that $$\mathbb{Q}[1+i]=\{a+b(1+i):a,b\in{\mathbb{Q}}\}=\{a+b+bi:a,b\in{\mathbb{Q}}\}$$ $$=\{c+bi:c,b\in{\mathbb{Q}}=\mathbb{Q}[i]$$ ...a degree 2 extension.
By Prop. 11 p. 521, we know that the degree of our field extension should be equal to the degree of our minimal polynomial. So, we may safely conclude that the minimal polynomial for the element $1+i$ with $\mathbb{Q}$ as our base field is $m(x)=x^2-2x+2$.

D+F 14.2.1

Determine the minimal polynomial over $\mathbb{Q}$ for the element $\sqrt{2} + \sqrt{5}$.

$\textit{Solution}$ : We wish to find a monic, irreducible polynomial of minimal degree denoted $m_{\alpha,\mathbb{Q}}$ s.t. $m_{\alpha,\mathbb{Q}}(\alpha)=0$ where $\alpha \in{\mathbb{Q}[\sqrt{2}+\sqrt{5}]}$.
$\mathbb{Q}[\sqrt{2}+\sqrt{5}]$ will be the splitting field for $m_{\alpha,\mathbb{Q}}$.

Label $\alpha = \sqrt{2} + \sqrt{5}$. Then, $$\alpha-\sqrt{2} = \sqrt{5} $$ Squaring both sides, $$\alpha^2-2\sqrt{2}\alpha+2=5$$ $$\alpha^2-3=2\sqrt{2}\alpha$$ Squaring both sides again, $$\alpha^4-6\alpha^2+9=8\alpha^2$$ $$\alpha^4-14\alpha^2+9=0$$ ...Is the minimal polynomial for the element $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$. Note that it is irreducible over $\mathbb{Q}[x]$ by the rational roots theorem.

D+F 4.5.22

$|G|=132 \Longrightarrow G$ is not simple.

$\textit{Proof}$ : The prime factorization of 132 is $132=2^2 \cdot 3 \cdot 11$. We will again use the 3rd part of Sylow's theorem to count the possible numbers of Sylow p-subgroups.
Writing $132=11^1 \cdot 12$ we note that $11 \nmid 12$ so Sylow (iii) gives $$n_{11} \equiv 1 (mod11)$$ and, $$n_{11} \mid 12$$ So, $n_{11} = 1,12$. Similarly, $$n_3 \equiv 1(mod3)$$ $$n_3 \mid 44$$ So, the only positive integers satisfying these constraints is 1 and 4, so $n_3=1,4$.
$$n_2 \equiv 1(mod2)$$ $$n_2 \mid 33$$ So, $n_3 = 1,3,11,33$
Now, suppose to the contrary that $G$ is a simple group. So, this means that $G$ has no non-trivial normal subgroups. By Corollary 20 p.142, this means that $n_p \neq 1 \quad \forall p$ in the prime factorization of $|G|=132$. So, this forces $n_{11}=12$, $n_3=4$ and $n_2=3,11,33$. The Sylow 11-subgroups are subgroups of order 11, thus isomorphic to $\mathbb{Z}_{11}$, which has 10 elements of order 11. So, with $n_{11}=12$, this means that in $G$, there are $12 \cdot 10 = 120$ elements of order 11. The Sylow 3-subgroups have order 3, so they are isomorphic to $\mathbb{Z}_3$, which has 2 elements of order 3. With $n_3 = 4$, this means that there are $4 \cdot 2 = 8$ elements of order 3 in $G$.
So far, we have counted $120+8=128$ non-identity elements of $G$. This leaves only 4 remaining. However, the Sylow 2-subgroups have order $2^2=4$. Thus, the remaining 4 elements must be exactly the Sylow 2-subgroup (which is either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$, but it doesn't actually matter at this point). So, $n_2=1$ giving us a non-trivial, normal subgroup by Corollary 20 p.142. This contradicts the assumption that $G$ was simple. In any case, we must have that $n_p=1$ for at least one $p$ in the factorization of 132.
So, any group of order 132 is not simple. $\blacksquare$

D+F 7.3.18

(a) If $I$ and $J$ are ideals of $R$, prove that their intersection $I \cap J$ is also an ideal of $R$.
(b) Prove that the intersection of an arbitrary nonempty collection of ideals is again an ideal (do not assume that the collection is countable)

$\textit{Proof}$ : For (a), we wish to show that $I \cap J$ is a nonempty subring of $R$ which is closed under left and right multiplication. With $I$ and $J$ being ideals of $R$, $0\in{I}$ and $0\in{J}$ so, $0\in{I \cap J}$ so $I \cap J \neq \emptyset$.
By exercise 7.1.4, we know that the intersection of subrings is a subring. We just need to show that $I \cap J$ is closed under left and right multiplication. Let $x\in{I \cap J}$ and $r\in{R}$. We want to show that $rx,xr\in{J} \quad \forall r\in{R}$.
$$x\in{I \cap J} \Longleftrightarrow x\in{I} \wedge x\in{J}$$ $$x\in{I} \Longrightarrow rx,xr\in{I}$$ $$x\in{J} \Longrightarrow rx,xr \in{J}$$ Thus, $xr,rx \in{I \cap J}$.
$\therefore \quad I \cap J$ is an ideal of $R$

For (b) let $$\bigcap_{k\in{\Lambda}}I_k$$ be the intersection of ideals in $R$ where $\Lambda$ is an arbitrary index of ideals.
$$0\in{I_k} \quad \forall k \in{\Lambda} \Longrightarrow 0\in{\bigcap_{k\in{\Lambda}}I_k} \Longrightarrow \bigcap_{k\in{\Lambda}}I_k \neq \emptyset$$ Let $$x\in{\bigcap_{k\in{\Lambda}}I_k}$$ and $r\in{R}$. Then, $rx,xr\in{I_k}$ $\forall k\in{\Lambda}$ since each $I_k$ is an ideal. This implies that $$rx,xr \in{\bigcap_{k\in{\Lambda}}I_k}$$ $$\Longrightarrow \bigcap_{k\in{\Lambda}}I_k$$ is an ideal of $R \quad \blacksquare$

D+F 4.5.34

Let $P\in{Syl_p(G)}$ and let $N \unlhd G$.

(a) Use the conjugacy part of Sylow's Theorem to prove that $P \cap N\in{Syl_p(N)}$

(b) Prove that $PN/N\in{Syl_p(G/N)}$

The second part of Sylow's Theorem states: Let $P$ be a Sylow p-subgroup of $G$ and $Q$ a p-subgroup of $G$. Then, $\exists g\in{G}$ s.t. $Q \leq gPg^{-1}$, i.e. $Q$ is contained in some conjugate of $P$.

$\textit{Proof}$ : Let $|G|=p^am$ where $p \nmid m$. $|N|$ divides $|G|$ by Lagrange, so let $|N|=p^bn$ where $b \leq a$, $p \nmid n$ and $n \leq m$. To show that $P \cap N \in{Syl_p(N)}$ we need to show that $|P \cap N|=p^b$.
Observe that $P \cap N$ is a p-group: $$|P \cap N| \mid |P|=p^a$$ $$|P \cap N| \mid |N|=p^bn$$ Since $p$ is prime, $$\Longrightarrow |P \cap N|=p^c \quad (c \leq b)$$ We want to now show that $b \leq c$.
Corollary 15, p. 94 states that if $H,K \leq G$ and $H \leq N_G(K)$ then the subset product, $HK \leq G$. In this case, $N \lhd G \Longrightarrow N_G(N) = G$, so the subset product $PN \leq G$. By the 4th isomorphism theorem, we may mod out by $N$, so we get that $$PN/N \leq G/N \quad (\star)$$ The order of the subset product is given by the formula $$|PN| = \frac{|P| \cdot |N|}{|P \cap N|} \Longrightarrow \frac{|PN|}{|N|}=\frac{|P|}{|P \cap N|}$$ Combining with $(\star)$ we get $$\frac{|P|}{|P \cap N|} \mid \frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}\frac{m}{n}$$ $$\frac{p^a}{p^c} \mid p^{a-b}\frac{m}{n}$$ $p^{a-c}$ is a power of a prime and $p^{a-b}$ is a power of a prime, so $$p^{a-c} \mid p^{a-b} \Longrightarrow a-c \leq a-b \Longrightarrow b \leq c$$ Thus, $ |P \cap N| = p^b \quad \therefore P \cap N \in{Syl_p(N)}$.

To prove that $PN/N\in{Syl_p(G/N)}$, we make a similar argument based on subgroup orders. $$|G/N|=\frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}(\frac{m}{n})$$ Where we can say that $\frac{m}{n}$ is in lowest terms and is a positive integer.
Again, the order of subset products is given by $$|PN|=\frac{|P| \cdot |N|}{|P \cap N|}$$ Rearranging, we get $$|PN/N|=\frac{|P|}{|P \cap N|}=\frac{p^a}{p^b}=p^{a-b}$$ $$\Longrightarrow PN/N \in{Syl_p(G/N)} \quad \blacksquare$$

D+F 7.3.24

Let $\varphi:R \rightarrow S$ be a ring homomorphism.

(a) Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$. In other words, the preimage of an ideal is an ideal.
$\textit{Proof}$ : Ideals are subrings that are closed under left and right multiplication by arbitrary ring elements. To show that $\varphi^{-1}(J)$ is a subring, we show that it is nonempty and closed under both ring operations. Since $J$ is an ideal in $S$, $$0\in{J} \Longrightarrow \varphi^{-1}(0)\in{\varphi^{-1}(J)} \Longrightarrow \varphi^{-1}(J) \neq \emptyset$$ Let $r_1$, $r_2 \in{\varphi^{-1}(J)}$. So, $\varphi(r_1)$, $\varphi(r_2) \in{J}$.
$J$ is an ideal, so $$\varphi(r_1)\varphi(r_2)=\varphi(r_1r_2)\in{J} \Longrightarrow r_1r_2\in{\varphi^{-1}(J)}$$ So, $\varphi^{-1}(J)$ is closed under multiplication making $\varphi^{-1}(J)$ a subring of $R$.
To show that $\varphi^{-1}(J)$ is an ideal, we show that it is closed under multiplication by all elements of $R$. Let $x\in{R}$. $$\varphi(xr)=\varphi(x)\varphi(r)$$ $J$ is an ideal, so, $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under left multiplication. Similarly, $$\varphi(rx)=\varphi(r)\varphi(x)$$ $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under right multiplication. $\therefore \varphi^{-1}(J)$ is an ideal of $R$ $\blacksquare$

Note: we could have skipped showing that $\varphi^{-1}(J)$ is closed under multiplication by its own elements, since this is taken care of in showing that $\varphi^{-1}(J)$ is closed with left and right multiplication by arbitrary elements in $R$.

(b) Prove that if $\varphi$ is surjective (onto) and $I$ is an ideal of $R$, then $\varphi(I)$ is an ideal of $S$. Give an example where this fails if $\varphi$ is not surjective. In other words, images of ideals are ideals under surjective ring homomorphisms.
$\textit{Proof}$ : Since $I$ is an ideal of $R$, $$0\in{I} \Longrightarrow \varphi(0)\in{\varphi(I)} \Longrightarrow \varphi(I) \neq \emptyset$$ Let $a,b\in{I}$. $I$ is an ideal, so $a+b\in{I}$. $$\varphi(a+b)=\varphi(a)+\varphi(b)\in{\varphi(I)}$$ So, $\varphi(I)$ is closed under addition.
Now, let $y\in{I}$ and $s\in{S}$ . Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and also $\varphi(x)=y$ with $x\in{I}$. $$xr\in{I} \Longrightarrow \varphi(xr)=\varphi(x)\varphi(r)\in{\varphi(I)}$$ $$rx\in{I} \Longrightarrow \varphi(rx)=\varphi(r)\varphi(x)\in{\varphi(I)}$$ So, $\varphi(I)$ is a left and right ideal, and thus an ideal of $S$.

If we relax the condition that $\varphi$ is surjective, consider the following example:
Let $\phi:\mathbb{Z} \rightarrow \mathbb{Q}$ be the identity map. $\mathbb{Z}$ is trivially an ideal of itself, but its image , which is still just $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$ since $\frac{1}{2} \cdot 1 = \frac{1}{2} \not \in{\mathbb{Z}}$ i.e. $\mathbb{Z}$ is not closed under multiplication by elements in $\mathbb{Q}$.

D+F 7.3.17

Let $R$ and $S$ be nonzero rings with identities $1_R$ and $1_S$. Let $\varphi:R \rightarrow S$ be a nonzero ring homomorphism.

(a) Prove that if $\varphi(1_R) \neq 1_S$ then $\varphi(1_R)$ is a zero divisor in $S$. Deduce that if $S$ is an integral domain, then every ring homomorphism from $R$ to $S$ sends the identity of $R$ to the identity of $S$.

(b) Prove that if $u$ is a unit in $R$ and $\varphi(1_R)=1_S$ then $\varphi(u)$ is a unit in $S$ and $\varphi(u^{-1})=\varphi(u)^{-1}$.

$\textit{Proof}$ :
(a): Let $\varphi(1_R)=y\in{S}-\{0\}$. (The identity in $R$ must be sent to a nonzero element of $S$ for $\varphi$ to be nonzero) $$y=\varphi(1_R)=\varphi(1_R1_R)=\varphi(1_R)\varphi(1_R)=y^2$$ $y=y^2$
$y-y^2=0$ (additive inverses exist because rings with respect to the addition operation are abelian groups)
$y(1_S-y)=0$ (distributive law for rings)
$ \therefore $ $y$ is a zero divisor.

If $S$ is an integral domain, then $S$ contains no zero divisors. So, in particular, $\varphi(1_R)$ is not a zero divisor. By the contrapositive of part (a), $\varphi(1_R) = 1_S$

(b) Assume that $u$ is a unit in $R$ and $\varphi(1_R)=1_S$. With $u$ being a unit of $R$, this means that
$\exists v \in{R}$ s.t. $uv=vu=1_R$. Since $\varphi$ is a ring homomorphism, $$\varphi(uv)=\varphi(vu)=\varphi(1_R)=1_S$$ $$\Longrightarrow \varphi(u)\varphi(v)=\varphi(v)\varphi(u)=1_S$$ Therefore, $ \varphi(u)$ is a unit in $S$.

We now want to show that $\varphi(u^{-1})=\varphi(u)^{-1}$. Note that $$1_S=\varphi(1_R)=\varphi(uu^{-1})=\varphi(u)\varphi(u^{-1})$$ $$\varphi(u)\varphi(u^{-1})=1_S$$ Since we have established that $\varphi(u)$ is a unit, it has an inverse. So, $$\varphi(u)^{-1}\varphi(u)\varphi(u^{-1})=\varphi(u)^{-1} \cdot 1_S$$ $$\therefore \quad \varphi(u^{-1})=\varphi(u)^{-1} \quad \blacksquare$$

D+F 7.3.16

Given $\varphi: R \rightarrow S$ be a surjective ring homomorphism. Prove that the image of the center of $R$ is contained in the center of $S$.

$\textit{Proof}$ : Denote the center of $R$ as $Z(R)$ and the center of $S$ as $Z(S)$. Let $x \in{Z(R)}$. Then, $xr=rx \quad \forall r\in{Z(R)}$.
We want to show that $\varphi(x)s=s\varphi(x) \quad \forall s\in{S}$. With $s\in{S}$, then since $\varphi$ is surjective, $\exists a\in{R}$ s.t. $\varphi(a)=s$. Then, $$s\varphi(x)=\varphi(a)\varphi(x)=\varphi(ax)=\varphi(xa)$$ since $x\in{Z(R)}$. $$\varphi(xa)=\varphi(x)\varphi(a)=\varphi(x)s \Longrightarrow \varphi(x)\in{Z(S)}$$ We have shown that the image of an arbitrary element in $Z(R)$ commutes with any element in $S$.
$\therefore \varphi(Z(R)) \subseteq Z(S) \quad \blacksquare$

commutative rings with prime ideals are fields

Let $R$ be a commutative ring. If every ideal of $R$ is prime, then $R$ is a field.

A commutative ring with identity is an integral domain if it has no zero divisors. We proceed by establishing that $R$ is an integral domain and then showing that inverses of elements exist, this $R$ is a field. Recall that an ideal $P$ is a prime ideal if $P \neq R$ and $ab \in{P} \Longrightarrow a \in{P}$ or $b \in{P}$.

$\textit{Proof}$ : Consider the ideal generated by the additive identity $(0)=0 \subseteq R$. $ab \in{(0)} \Longleftrightarrow ab=0 \Longrightarrow a=0$ or $b=0$. This is exactly the definition of not having zero divisors. So, $R$ is an integral domain.
Now let $x \neq 0$. $x^2 \in (x^2) \Longrightarrow x \cdot x \in{(x^2)}$. So, if $x \in{(x^2)} \Longrightarrow \exists a\in{R}$ s.t. $x=ax^2$. Since we have cancellation laws in an integral domain, then $xx^{-1}=ax^2x^{-1}$ so $1=ax$. So, $x$ has an inverse. $\therefore R$ is a field. $\blacksquare$

D+F 6.2.3

$|G|=380 \Longrightarrow $ $G$ is not simple.

$\textit{Proof}$ : First note that $380=2^2 \cdot 5 \cdot 19$. By Sylow's theorem (iii), we get that for $p=19$, $n_{19}=1$(mod 19) and also $n_{19} \mid 2^2 \cdot 5 = 20$. So, the possible values of $n_{19}$ are $1,20$.
For $p=5$, we get that $n_5=1$(mod 5) and also $n_5 \mid 2^2 \cdot 19 = 76$. Given these two constraints, the possible values for $n_5$ are $n_5=1,76$.
For $G$ to be simple, we cannot have any non-trivial normal subgroups. This means that by corollary 20 p.142, we must have $n_p=1$ for all $p$ in the factorization of $|G|$.

Suppose that it is the case that $n_{19}=20$ and $n_5=76$. The Sylow-19 subgroups have order 19, (the 19 factor in 380 is square-free) and the only group of order 19 is $\mathbb{Z}_{19}$, which has 18 elements of order 19. With $n_{19}=20$, this would mean that our entire group $G$ has $20 \cdot 18=360$ elements of order 19. The Sylow 5-subgroups have order 5, and the only group of order 5 is $\mathbb{Z}_5$, which has 4 elements of order 5. With $n_5=76$, this means our entire group $G$ would have $76 \cdot 4 = 304$ elements of order 5. This is an immediate contradiction on the size of $G$. Thus, it must be the case that either $n_5=1$ or $n_{19}=1$. In each case, we would have a non-trivial normal subgroup making $G$ not simple. $\blacksquare$

D+F 4.5.24

Prove that if $G$ is a group of order 231 then $Z(G)$ contains a Sylow 11-subgroup of $G$. Also, show that there is a normal Sylow 7-subgroup in $G$.

$\textit{Proof}$ : Begin by showing there is a normal Sylow 7-subgroup in $G$: By Sylow's Theorem (iii), $$n_p \equiv 1(modp)$$ and $$n_p \mid m$$ when $|G| = p^am$.
With $p=7$, we can write $|G|=231=7 \cdot 33$. The possibilities for $n_7$ are 1,8,15,22, and 29. Only the number 1 divides 33, thus $$n_7=1$$ Since Sylow p-subgroups are conjugate to each other, the Sylow subgroup of order 7 (231 has only 1 factor of 7) is normal in $G$ by corollary 20 p.142.

Now, to show that $Z(G)$ contains a Sylow-11 subgroup we initially proceed as above. The possibilities for $n_{11}$ are 1,12, and 23. But, only the number 1 also divides 21. Therefore, there is a unique Sylow 11 subgroup, $H$ in $G$ that is also normal. By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ Since $H \lhd G$, $N_G(H)=G$. Also, $Aut(H)=Aut(\mathbb{Z}_{11})$ since $\mathbb{Z}_{11}$ is the only group of order 11. So $(\star)$ above becomes $$G/C_G(H) \lesssim Aut(\mathbb{Z}_{11}) \cong \mathbb{Z}_{11}^* \cong \mathbb{Z}_{10} $$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 10$$ So, $$\frac{231}{|C_G(H)|} \mid 10 \Longrightarrow \frac{231}{|C_G(H)|}=1,2,5,10 \Longrightarrow |C_G(H)|=231$$ So, $H$ commutes with all 231 of the elements of $G$. In other words, $$C_G(H) \leq Z(G) \quad \blacksquare$$

D+F 4.5.35

Let $P \in{Syl_p(G)}$ and $H \leq G$. Show that $\exists g\in{G}$ s.t. $$gPg^{-1} \cap H \in{Syl_p(H)}$$ $\textit{Proof}$ : Let $Q \in Syl_p(H)$.

Then, $Q \leq H$ and $\exists g \in{G}$ and $\exists P \in{Syl_p(G)}$ s.t. $Q \leq gPg^{-1}$ by Sylow (ii). This means that $$Q \subseteq gPg^{-1} \cap H$$ We have found a $p$-group of $H$ that contains $Q$. Thus, $$Q \leq gPg^{-1} \cap H \in{Syl_p(H)} \quad \blacksquare$$

D+F 5.5.2

Let $H,K$ be groups, $\varphi: K \longrightarrow Aut(H)$ and $H,K \leq H \rtimes_\varphi K$.
Prove that $C_H(K)=N_H(K)$
Note: we may also assume that $H \unlhd G$ and $H \cap K = \{e\}$ by the construction of the semi-direct product $H \rtimes_\varphi K$.

$\textit{Proof}$ :
"$\subseteq$" If $h\in{C_H(K)}$ then $hk=kh \quad \forall k\in{K}$. Rearranging, $hkh^{-1}=h \Longrightarrow hKh^{-1}=H \Longrightarrow h\in{N_H(K)}$

"$\supseteq$" Let $h\in{N_H(K)}$. Look at the commutator $[h,k]=hkh^{-1}k^{-1}$ :

$h(kh^{-1}k^{-1}) \in{H}$ since $H \unlhd G=H \rtimes_\varphi K$
$(hkh^{-1})k^{-1} \in{K}$ since $h \in{N_H(K)}$. We also have that $H \unlhd G$ and $H \cap K = \{e\}$. Thus, $$hkh^{-1}k^{-1}=e \Longrightarrow hk=kh \Longrightarrow h\in{C_H(K)} \quad \blacksquare$$

D+F 5.5.1

Let $H,K$ be groups, $\varphi: K \longrightarrow Aut(H)$ and $H,K \leq H \rtimes_\varphi K$.
Prove $C_K(H)=ker\varphi$.

$\textit{Proof}$ : $$k \in{ker\varphi} \Longleftrightarrow \varphi_k=\sigma_{id} \Longleftrightarrow \varphi_k(h)=h \quad \forall h\in{H} \Longleftrightarrow$$ $$k \cdot h = h \Longleftrightarrow khk^{-1} = h \Longleftrightarrow kh=hk \Longleftrightarrow k\in{C_K(H)} \blacksquare$$

D+F 4.5.44

Given $p$ is the smallest prime dividing $|G|$, $P\in{Syl_p(G)}$ and $P$ is cyclic, prove $$N_G(P)=C_G(P)$$ $\textit{Proof}$ : By corollary 15 p.134, $$N_G(P)/C_G(P) \lesssim Aut(P)$$ Since $P$ is cyclic, then by proposition 16 p.135, $$Aut(P) \cong \mathbb{Z}_p^* \cong \mathbb{Z}_{p-1}$$ By Lagrange,
$|N_G(P)/C_G(P)|$ divides $|G|$. Also,
$|N_G(P)/C_G(P)|$ divides $p-1$.
Since $p$ is the smallest prime dividing $|G|$, this forces $|N_G(P)/C_G(P)|=1 \Longrightarrow N_G(P)=C_G(P) \quad \blacksquare $

D+F 4.5.32

Given that $P \in{Syl_p(H)}$ and $P \unlhd H \unlhd K$, then $P \unlhd K$

In general, normality of groups is not a transitive relationship. This exercise establishes this transitivity when there is an added condition.

$\textit{Proof}$: Since $H \unlhd K$, $$kHk^{-1} = H \quad \forall k\in{K}$$ Since $P \subseteq H$, $$kPk^{-1} \subseteq kHk^{-1} = H$$ Since $P \unlhd H$ and $P \in{Syl_p(H)}$, then by corollary 20 p.142, $P$ is the only $p$-Sylow subgroup of $H$. Sylow $p$-subgroups are conjugate to each other (part (ii) of Sylow's Theorem). So, if we conjugate $P$, then the only possibility is that this conjugate is sent to $P$ itself. $$kPk^{-1}=P \Longrightarrow P \unlhd K \quad \blacksquare$$

D+F 4.5.27 (modified)

Let $G$ be a group with order 315 and suppose $G$ has a normal subgroup $H$ that is cyclic of order 9. Prove that $G$ is cyclic.

$\textit{Proof}$: By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ In this case, since $H \lhd G$, $H$ normalizes all of $G$, i.e. $N_G(H)=G$. Also, $H$ being cyclic and $|H|=9 \Longrightarrow H \cong \mathbb{Z}_9$. By Proposition 16 p. 135, $$Aut(H) \cong \mathbb{Z}_9^* \cong \mathbb{Z}_6 $$ Thus, $(\star)$ becomes: $$G/C_G(H) \lesssim \mathbb{Z}_6$$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 6$$ This means that $$|G/C_G(H)| = 1,2,3,6 \Longrightarrow |C_G(H)|=315, \frac{315}{2}, \frac{315}{3}, \frac{315}{6}$$ But the order of a subgroup must be an integer, so $$|C_G(H)|=315, 105 \quad (\star \star)$$ By problem 2.2.6(b), $H \leq C_G(H) \Longleftrightarrow H$ is Abelian. Since $H \cong \mathbb{Z}_9$, it is cyclic, and thus Abelian. So, $H \leq C_G(H)$ and $|H| \mid |C_G(H)|$ by Lagrange.

$9 \mid |C_G(H)|$ but also $C_G(H)$ is always a subgroup of $G$. So, the possible values for the order of $C_G(H)$ are multiples of 9 that divide 315. $$|C_G(H)|=9,45,63,315$$ The only value above that is also consistent with $(\star \star)$ is 315. So, $|C_G(H)|=315$. So, every element of $G$ commutes with elements of $H$ i.e. $H \leq Z(G)$.

We now establish that $G$ is Abelian:
Since $H \cong \mathbb{Z}_9 \leq Z(G)$ and $Z(G) \leq G$, then by Lagrange, $$9 \mid |Z(G)| \mid 315$$ Thus, $|Z(G)|=9,45,63,315$ (The multiples of 9 that divide 315) So then, $$|G/Z(G)|=\frac{315}{315}, \frac{315}{63}, \frac{315}{45}, \frac{315}{9} = 1,5,7,35$$ Groups of order 1,5,7, and 35 are all cyclic. (We can show the 35 case by the method on p.143) By exercise 3.2.36, We get that $G$ is Abelian. Finally, by the Fundamental Theorem of Finitely Generated Abelian Groups, $$G \cong \mathbb{Z}_9 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_7$$ $gcd(9,5,7)=1$. So, by Proposition 6(1) P. 163, $G$ is cyclic. $\blacksquare$

D+F 4.5.15

Prove that a group of order 351 has a normal Sylow p-subgroup for some prime $p$ dividing its order.

$|G|=351 \Longrightarrow \exists H \lhd G$ and $\exists p$ prime s.t. $H\in{Syl_p(G)}$

$\textit{Proof}$: Note that $351=3^3 \cdot 13$. Writing $|G|=351=p^am$, then part (iii) of Sylow's Theorem implies:
$n_p \equiv 1$(mod p) and $n_p \mid m$
First, look at $n_{13}$. Picking larger primes first is sometimes advantageous because there tends to be less possibilities when generating a list.
$351=13^1 \cdot 27$. So, $n_{13} \equiv 1$(mod 13) and $n_{13} \mid 27$
So, $n_{13}=1,27$.
If $G$ is simple, then $n_{13} \neq 1$ as this would make the Sylow 13-subgroup normal by Cor.20 p. 142. Thus, if $n_{13}=27$ this means that there are $27 \cdot (13-1) = 324$ elements of order 13. Since $|G|=351$, this leaves us with $351-324=27$ elements remaining. But the Sylow 3-subgroups have order $3^3=27$. So in this case this lack of freedom forces $n_3=1$. Thus, we have a non-trivial normal subgroup, making $G$ not simple. In any case, it must be that $n_p=1$ for $p=3$ or $p=13$.
$\therefore \quad G$ is not simple. $\blacksquare$

Note: this problem also solves one part of exercise 6.2.4

D+F 4.5.14

Prove that a group of order 312 has a normal Sylow p-subgroup for some prime $p$ dividing its order.

$|G|=312 \Longrightarrow \exists H < G$ and $\exists p$ prime s.t. $H\in{Syl_p(G)}$

$\textit{Proof}$: Note that $312=2^3 \cdot 3 \cdot 13$. Writing $|G|=312=p^am$, then part (iii) of Sylow's Theorem implies:
$n_p \equiv 1$(mod p) and $n_p \mid m$
First, look at $n_{13}$. Picking larger primes first is sometimes advantageous because there tend to be less possibilities when generating a list.
$312=13^1 \cdot 24$. $n_{13} \equiv 1$(mod 13) and $n_{13} \mid 24$.
So, $n_{13} = 1$, $14$. But $n_{13} \neq 14$ since $14 \nmid 24$. We are thus forced that $n_{13} = 1$. By Corollary 20 p.142, the unique Sylow $p$-subgroup of order $13^1=13$ is normal in $G$. $\blacksquare$

D+F 4.5.13

Prove that a group of order 56 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.

$\textit{Proof}$: First note that $56=8\cdot 7 = 2^3 \cdot 7$. By part (iii) of Sylow's Theorem, the number of Sylow $p$-subgroups of $G$ is $$n_p \equiv 1(modp)$$ Additionally, $$n_p \mid m $$ The prime components of 56 are 2 and 7. So, $$n_2 \equiv 1(mod2)$$ and $$n_2 \mid 7$$ If we generate a list of possible values for $n_2$, we get $$n_2 = 1,7$$ If we generate a list of possible values for $n_7$, we get $$n_7 = 1,8$$ We will proceed by a counting argument, then using corollary 20 on page 142. Consider the case when $n_2=7$ and $n_7=8$. In other words, there are 7 subgroups of order 2 and 8 subgroups of order 7. Groups of order 7 and 2 are isomorphic to the cyclic groups $\mathbb{Z}_7$ and $\mathbb{Z}_2$, respectively. $\mathbb{Z}_7$ has 6 non-identity elements (of order 7) and $\mathbb{Z}_2$ has 1 non-identity element. In total, there are $8 \cdot 6 = 48$ elements of order 7 and $7 \cdot 1 = 7$ elements of order 2. Counting up, this gives us $48+7+1=56$ elements in $G$. However, when multiplying elements together in $G$, if $|a|=7$ and $|b|=2$, then $|ab|=lcm(7,2)=14$. So, there must be more elements in $G$ with order 14. This is a contradiction since $|G|=56$. Therefore, it must be the case that either $n_2=1$ or $n_7=1$. By the corollary on p. 142, this is equivalent to one of these Sylow $p$-subgroups being normal.
So, by a counting argument, we have forced that there must exist a normal Sylow $p$-subgroup of $G$ when $|G|=56$. $\blacksquare$

Sylow's Theorem

$\textbf{4.5 Theorem 18}$

(i) A (sub)group of order $p^a$ for some $a\in{\mathbb{N}}$ is called a $\textit{p-(sub)group}$.
(ii) Let $|G|=p^am$ where $p \nmid m$. (This can be done for any finite group by the fundamental theorem of arithmetic). A subgroup of order $p^a$ is called a $\textit{p-Sylow subgroup}$ of G.
(iii) The set of Sylow p-subgroups of G is denoted $Syl_p(G)$. The number of Sylow p-subgroups of G is denoted $n_p(G)$ or $n_p$.

$\textit{(Sylow's Theorem)}$
Let $G$ be a group of order $p^am$, where p is a prime and $p \nmid m$.

1. Sylow p-subgroups of $G$ exists, i.e. $Syl_p(G) \neq \emptyset$.

2. Let $P$ be a Sylow p-subgroup of $G$ and $Q$ a p-subgroup of $G$. Then, $\exists g\in{G}$ s.t. $Q \leq gPg^{-1}$, i.e. $Q$ is contained in some conjugate of $P$.

3. $n_p \equiv 1$(mod p).
Additionally, $|G:N_G(P)|=n_p$ for any Sylow p-subgroup $P$, thus $n_p \mid m$.
We can see that $n_p \mid m$ as follows:
In general, $P \leq N_G(P) \leq G$, so by Lagrange, $|P| \mid |N_G(P)|$ So with $|G|=p^am$ then $|N_G(P)|=p^bk$, $k \leq m$
$n_p = |G:N_G(P)|=\frac{|G|}{|N_G(P)|}=\frac{p^am}{p^ak}=\frac{m}{k}$. So, $n_p \cdot k = m$ i.e. $n_p \mid m$

We may note that Sylow's Theorem is a partial converse to Lagrange's Theorem. Previously, we took information about subgroups (such as normality or order) and deduced information about the structure of the whole group. Now, we take information about the whole group (such as it's order) and deduce properties about the internal structure of the group.

Regarding (ii): In words, this says that $Q$ is contained in some conjugate of $P$. In particular, any two Sylow p-subgroups are conjugate to each other. In fact, they are isomorphic to each other by Corollary 14 p.134: $K \cong gKg^{-1}$ i.e. the group action of conjugation is an automorphism. In particular, Sylow p-subgroups are conjugate to each other. So, if $n_p=1$, then that Sylow p-subgroup is conjugate to itself thus is a normal subgroup.

$\textit{Proof}$ (of item (i)) We proceed by using strong induction on the size of $G$. The base case $|G|=1$ holds trivially since $|G|=1=p^0$ $\forall p$ prime. So, $G$ is its own Sylow p-subgroup, $G \in{Syl_p(G)}$.

Assume that $\forall$ $|G| < n\in{\mathbb{N}}$, G contains a Sylow p-subgroup, i.e. $Syl_p(G) \neq \emptyset$.

Now consider $|G|=n=p^am$. We look at 2 cases: either $p\mid Z(G)$ or $p\nmid Z(G)$.

$\underline{Case 1}$: $p \mid Z(G)$. Since $Z(G)$ is an Abelian group, we may apply Cauchy's Theorem (Prop 21.3.4). Thus, $Z(G)$ contains a subgroup, $N$, of order $p$. Let $\bar{G}=G/N$. Then, $$|\bar{G}|=|G/N|=\frac{|G|}{|N|}=\frac{p^am}{p}=p^{a-1}m < n$$ By our induction assumption, $\bar{G}$ has a subgroup, $\bar{P}$, of order $p^{a-1}$.
$\bar{P} \leq G/N$, so let $\bar{P} = P/N$ where $N \leq P \leq G$. $$|\bar{P}|=|P/N|=\frac{|P|}{|N|}=p^{a-1}$$ So, $$|P|=|N| \cdot p^{a-1}=p \cdot p^{a-1}=p^a$$ We have identified a subgroup of $G$ that has order $p^a$. So, $P\in{Syl_p(G)}$.

$\underline{Case 2}$: $p \nmid Z(G)$. Recall the class equation, which counts the orbits of $G$ (remember conjugacy classes partition $G$): $$|G|=|Z(G)|+ \sum_{i=1}^r |G:C_G(g_i)|$$ where $g_1,...,g_r$ are representatives of non-central conjugacy classes of $G$. If it were the case that $p \mid |G:C_G(g_i)|$ $\forall i$, then since $|G|=p^am$, we can re-write the class equation as $$|Z(G)|=|G|-\sum_{i=1}^r |G:C_G(g_i)|=p^am-pk=p(p^{a-1}-k) $$ $$\Longrightarrow p \mid |Z(G)|$$ ...A contradiction to our case 2 assumption. Thus, it must be that $\exists j\in{\{1,...,r\}}$ s.t. $p \nmid |G:C_G(g_j)|$. Let $H=C_G(g_j)$. Note that $|H|<|G|=n$ since $g_j \not\in{Z(G)}$, say $$|H|=p^as, \quad p \nmid s, s

Prime index subgroups are maximal

Given $H < G$ and $|G:H|=p$, $p$ prime, then $H$ is a maximal subgroup.

$\textit{Proof}$: Assume to the contrary that $H$ is not a maximal subgroup. With this, there then exists another subgroup, say, $M$ s.t. $H < M < G$. By 3.2.11, $|G:H|=|G:M| \cdot |M:H|=mn \rightarrow \leftarrow $ since $p$ is prime. $\therefore H $ is maximal $ \blacksquare$

First Isomorphism Theorem (longer version)

$\textbf{First Isomorphism Theorem}$ i.e. The Fundamental Theorem of Homomorphisms.

Statement: Given $\varphi: G \longrightarrow H$ a group homomorphism, then:
(i) $ker \varphi \unlhd G$
(ii) $G/ker\varphi \cong im\varphi$
Where $ker \varphi$ is the kernel of the homomorphism $\varphi$ and $im\varphi$ is the image of $\varphi$, which is contained in $H$.

$\textit{Proof:}$ For part (i), we take 2 approaches. First, we note that problem 3.1.1 in Dummit and Foote establishes that with the given condition that $\varphi: G \rightarrow H$ is a group homomorphism, then the pre-image or pullback of $E\leq{H}$ is a subgroup of $G$. In symbols, $\varphi^{-1}(E) \leq G$. Additionally, $\varphi^{-1}(E) \unlhd G$ whenever $E \unlhd H$. $<1_H> \unlhd H$ since $1_H H 1_H^{-1}=H$. $$\varphi^{-1}(<1_H>)=\{g\in{G}|\varphi(g)=1_H \} = ker\varphi \Longrightarrow ker\varphi \unlhd G$$ We may also proceed as follows: Let $g\in{G}$ and let $x\in{ker\varphi} \subseteq G$. Then, since $\varphi$ is a homomorphism, $$\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g^{-1})$$ $$=\varphi(g)\cdot1_H\cdot\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=\varphi(gg^{-1})=\varphi(1_G)=1_H$$
So, $gxg^{-1}\in{ker\varphi}$ which implies $\ker\varphi \unlhd G$ We have shown that $ker\varphi$ is invariant under conjugation, so it is thus a normal subgroup.

For part (ii), we need to construct an isomorphism between $G/ker\varphi$ and $ \ im\varphi$.
Define $\Phi: G/ker\varphi \longrightarrow im\varphi$ s.t. $\Phi(aK)=\varphi(a)$.
So, $\Phi$ maps left (could also choose right) cosets in $G/ker\varphi$ to elements in $H$ that are images of elements in $G$ under $\varphi$.
Note that $G/ker\varphi$ is indeed a quotient group since $ker\varphi \unlhd G$.
Our goal is to show that $\Phi$ is an isomorphism. First, we show that $\Phi$ is actually well-defined:
For convenience of notation , let $K=ker\varphi$. Now, choose equal representatives from $G/ker\varphi$. Let $aK=bK$. Now, we want to show that these representatives are mapped to the same element, namely that $\Phi(aK)=\Phi(bK)$. If $aK=bK,$ then by operations on cosets, $$aK=bK \Longrightarrow aK(bK)^{-1}=(aK)(b^{-1}K)=(ab^{-1})K=K$$ Since $ab^{-1}K=K$, this implies that $ab^{-1}\in{K}$. With $ab^{-1}$ in the kernel of $\varphi$ we get that $\varphi(ab^{-1})=1_H$. $\varphi(ab^{-1})=\varphi(a)\varphi(b^{-1})=1_H \Longrightarrow \varphi(a)=\varphi(b)$. By our construction, $\varphi(a)= \Phi(aK)$ and $\varphi(b)=\Phi(bK)$, thus $\Phi(aK)=\Phi(bK)$. So, $\Phi$ is well-defined.
Since each of the steps in showing that $\Phi$ is well-defined can be reversed, we immediately get that $\Phi$ is injective (one-to-one). This is because the statement that $\Phi$ is one-to-one is the converse of the statement that $\Phi$ is well-defined. Lets demonstrate this anyways for completeness:
Let $\Phi(aK)=\Phi(bK)$. By construction, this means $\varphi(a)=\varphi(b) \Longrightarrow\varphi(a)\varphi(b)^{-1}=1_H \Longrightarrow \varphi(ab^{-1})=1_H \Longrightarrow ab^{-1}\in{K} \Longrightarrow ab^{-1}K=1_G K \Longrightarrow aK=bK $. Hence, $\Phi$ is one-to-one.

To show that $\Phi$ is surjective (onto), we pick and element $y\in{im\varphi}$ and show that there is a corresponding element in $G/K$ that maps to $y$. Let $y=\varphi(x)\in{im\varphi}$. Then, $\Phi(xK)=\varphi(x)=y$. Thus, $\Phi$ is onto.
To show that $\Phi$ is a homomorphism, we just need to show that $\Phi$ is operation preserving. This is the case since $$\Phi(aKbK)=\Phi(abK)=\varphi(ab)=\varphi(a)\varphi(b)=\Phi(aK)\Phi(bK)$$
So, $\Phi$ is an isomorphism, which shows that $G/K \cong im\varphi$. $\quad \blacksquare$

D+F 3.1.24

Show that the intersection of a subgroup $H$ and a normal subgroup $N$ is normal in $H$. i.e.
$H \leq G$ and $N \unlhd G \Longrightarrow H \cap N \unlhd H$.

Note: this is one of the results of the second/diamond isomorphism theorem.

$\textit{Proof:}$ Let $x\in{H \cap N}$. Since $H \leq G$, it is closed, so $\forall h\in{H}$, $hxh^{-1} \in{H}$. $N$ is normal, so it is invariant under conjugation. So when $x\in{H \cap N}$, $hxh^{-1} \in{N}$. We have $hxh^{-1}\in{H}$ and $hxh^{-1}\in{N}$, so $hxh^{-1} \in{H \cap N}$ $\therefore H \cap N \unlhd H \quad \blacksquare$

Dummit Foote 3.1.36 (iff)

$\textit{Theorem}:$ $G/Z$ cyclic $\Longleftrightarrow G$ Abelian. ($Z$ is the center of $G$)

Note: $G/Z$ is indeed a group since $Z \unlhd G$

$\textit{Proof}: \quad$ "$\Longleftarrow$" $G$ Abelian $\Longrightarrow Z=G \Longrightarrow G/Z=G/G=<1>$ which is trivially cyclic.
"$\Longrightarrow$"
$G/Z$ cyclic $\Longrightarrow$ $$\forall gZ \in{G/Z}, \quad \exists aZ\in{G/Z} \quad s.t. \quad gZ=(aZ)^n=a^nZ$$ So, $g$ and $a^n$ are in the same fiber of a homomorphism from $G$ to $G/Z$. $\Longrightarrow \quad \exists z_1\in{Z}$ s.t. $ g=a^kz_1$

Let $hZ \in{G/Z}$. So, $hZ=a^kZ$ for some $k\in{\mathbb{Z}}$. Then, $$gh=a^nz_1 \cdot a^kz_2 = a^na^kz_1z_2=a^{n+k}z_1z_2=a^{k+n}z_2z_1=a^ka^nz_2z_1=a^kz_2a^nz_1=hg$$ since $z_1$ and $z_2$ commute with all elements of $G$.

$\therefore G$ is Abelian $\quad \blacksquare$

(Generalization) A subgroup of smallest prime index where p dives |G|, is normal

The next theorem is a generalization of the previous theorem that a subgroup of index 2 is normal.

$\textit{Corollary 5} \quad $ (D+F p. 120): Let $H < G$. If $|G:H|=p$ where $p$ is the smallest prime divisor of $G$, then $H \lhd G$.

$\textit{Proof:}\quad$ Let ${G/H}^{\star}$ represent the set of left cosets of $H$ in $G$ (which is not necessarily a group since we have not yet established that $H$ is normal). Define a group action as a map $$\varphi: G \times {G/H}^{\star} \rightarrow {G/H}^{\star} $$ Group actions induce homomorphisms, and in this case, the induced homomorphism is $$\phi: G \rightarrow S_{|{G/H}^{\star}|} \cong S_p$$ By the first isomorphism theorem, $ker\phi \unlhd H < G$. If we can show that $ker\phi = H$, then $H \lhd G$ as desired.
Say $|H:ker\phi|=k$.
$$|G:ker\phi|=|G:H|\cdot |H:ker\phi|=pk$$
Also from the first isomorphism theorem, $G/\ker\phi \cong im\phi \lesssim S_p $. So, $$pk|p!$$ Dividing by $p$, $$k|(p-1)!$$ But, by assumption, $p$ is the smallest prime divisor, so this forces $k=1$. $\Longrightarrow ker\phi =H \therefore H \lhd G \quad \blacksquare$

A subgroup of index 2 is normal

$\textbf{Theorem}$ A subgroup of index 2 is normal. i.e.
If $H < G$ and $|G:H|=2$ then $H \lhd G$

$\textbf{Proof}$: The cosets of $G$ partition $G$ by the equivalence relation $x\sim y \Longleftrightarrow x^{-1}y \in{H}$. The equivalence class of $x$ is $[x]=xH$. Since $|G:H|=2$ there are only 2 equivalence classes, $1H$ and $aH$. But $1H=H1=H$ so the only possibility for the right coset $Ha$ is that it equals $aH$. Therefore, $H \lhd G$ $\blacksquare$

Note: a "blob" picture is useful here.

Review: Normal Subgroups

$\textbf{Definition}$: A subgroup $N \leq G$ is normal if it is invariant under conjugation. i.e. $$\forall x\in{N}, \quad \forall g\in{G}, \quad gxg^{-1}\in{N}$$ [Remember: invariance is an important concept ubiquitous in mathematics.]
The following are some of the conditions equivalent to $N \unlhd G$:

(i) $ \forall g\in{G}, \quad gNg^{-1}= \{gxg^{-1}|g\in{G},x\in{N} \} \subseteq N$
(ii) $ \forall g\in{G}, \quad gNg^{-1}= N$
(iii) $ \forall g\in{G}, gN=Ng$ (left and right cosets are equal)
(iv) $N_G(N)=\{g\in{G}|gNg^{-1}=N\}=G$
(v) the operation of left (right) cosets of $N$ in $G$ makes the set of left (right) cosets into a group
(vi) $N$ is a union of conjugacy classes of $G$
(vii) $\exists \varphi:G \rightarrow H$, a homomorphism for which $N=ker\varphi$

Item (vii) demonstrates some of the importance of normal subgroups; they are a way to internally classify all homomorphisms defined on a group. For example, a non-trivial group is $\textit{simple}$ iff it is isomorphic to all of its non-trivial homomorphic images. This can be seen by means of the following:
Let $\varphi : G \rightarrow H$. By the first isomorphism theorem, $$im\varphi \cong G/ker\varphi $$ If $G$ is simple, then $ker \varphi$ (which is normal) is either $G$ or $\{e\}$. Thus, $G$ will be isomorphic to all of its non-trivial homomorphic images.

Item (vi) is justified because conjugation defines an equivalence relation; so conjugacy classes partition $G$. (This appears later when deriving the class equation). Normal subgroups are invariant under conjugation, so they can be viewed as a union of conjuacy classes of $G$.
In practice, it seems to be most useful to show item (i) or (ii). Pick $x\in{N}$, then show that for arbitraty $g\in{G}$, $gxg^{-1}\in{N}$. We can think of normality as a sort of partial commutativity that says: upon conjugating an element, the possible resulting elements will be bounded within the subgroup $N$. For readers familiar with linear algebra, notice that in the group of linear transformations $GL_n(F)$, conjugation is the same as a change of basis $A \mapsto QAQ^{-1}$.

Dummit and Foote 4.4.13

Let $G$ be a group of order 203. Prove that if $H$ is a normal subgroup of order 7 in $G$ then $H \leq Z(G)$. Deduce that $G$ is Abelian.
In symbols, we may represent this problem as follows:

$\Bigg\lbrace \begin{array}{cc} |G|=203 & & \\ H \unlhd G & \Longrightarrow H \leq Z(G) \Longrightarrow G \quad Abelian \\ |H|=7 \\ \end{array}$

Proof: First note that $203=7 \cdot 29$ a square-free prime factorization. By Corollary 15, we have $N_G(H)/C_G(H) \lesssim Aut(H)$. Since $H \lhd G$, $N_G(H)=G$. With $|H|=7$, $H \cong \mathbb{Z_7}$.
Corollary 16 gives $Aut(H) \cong (\mathbb{Z}/7\mathbb{Z})^{\times} \cong \mathbb{Z}_7^* \cong \mathbb{Z}_6$. So, $G/C_G(H) \lesssim \mathbb{Z}_6 \Longrightarrow |G/C_G(H)|\huge|$ $6$ by Lagrange. So, $\dfrac{203}{C_G(H)} \huge|$ $6$. $\Longrightarrow |C_G(H)|=203 \Longrightarrow H \leq Z(G)$.

We have that $H \leq Z(G) \leq G$. Since $G$ has the square-free prime factorization $7\cdot29$, $|H|=7$ means that $H$ is a maximal (and normal) subgroup of $G$. So, either $H=Z(G)$ or $Z(G)=G$. However, if it is true that $H=Z(G)$, then $G/Z(G) = G/H$ is of order $\frac{203}{7}=29$ implying $G/H$ is cyclic. Problem 3.1.36 implies that $G$ is Abelian $\Longrightarrow Z(G)=G$ contradicting the assumption that $|Z(G)|=|H|=7$. Thus, is must be the case above that $H < Z(G) = G \Longrightarrow G$ is Abelian.
Additionally, by the Fundamental Theorem of Finite Abelian Groups, $G \cong \mathbb{Z}_7 \bigoplus \mathbb{Z}_{29} \cong \mathbb{Z}_{203} $ $\blacksquare$

Dummit and Foote 4.4.12

Let $G$ be a group of order 3825. Prove that if $H$ is a normal subgroup of order 17 in $G$ then $H \leq Z(G)$

Proof: By Corollary 15, $N_G(H)/C_G(H) \lesssim Aut(H)$. The only group of order 17 is $\mathbb{Z_{17}}$, which is cyclic. By proposition 16, $Aut(H) \cong \mathbb{Z_{17}}^* \cong \mathbb{Z_{16}}$. So, $\frac{|N_G(H)|}{|C_G(H)|}$ divides 16 by Lagrange. Since $H \unlhd G$, $N_G(H)=G$, so $\frac{|N_G(H)|}{|C_G(H)|}=\frac{3825}{|C_G(H)|}$. Since $3825=3^2\cdot 5^2 \cdot 17$, the only possibility for the order of $C_G(H)$ is 3825 i.e. $C_G(H)=G$ which means that $H$ commutes with all of $G$ $\therefore H \leq Z(G)$ $\blacksquare$

First Isomorphism Theorem (shorter version)
Given $\varphi: G \longrightarrow H$ a group homomorphism, then:

$\hspace{2cm}$ i.) $ker \varphi \unlhd G$
$\hspace{2cm}$ ii.) $G/ker\varphi \cong im\varphi$

Proof:
item(i): Let $g\in{G}$ and let $x\in{ker\varphi} \leq G$. Then, since $\varphi$ is a homomorphism, $\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g^{-1})=\varphi(g)\cdot1_H\cdot\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=\varphi(gg^{-1})=\varphi(1_G)=1_H$.
item (ii): Let $K=ker\varphi$. Define $\Phi:G/K \longrightarrow H$ by $\Phi(aK)=\varphi(a)$. $aK=bK \Longleftrightarrow ab^{-1}\in{K} \Longleftrightarrow \varphi(ab^{-1})=1_H \Longleftrightarrow \varphi(a)=\varphi(b) \Longrightarrow \Phi$ is well-defined and 1-1. $\Phi$ is a homomorphism since $\Phi(aKbK)=\Phi(abK)=\varphi(ab)=\varphi(a)\varphi(b)=\Phi(aK)\Phi(bK)$. Also, $im\Phi=im\varphi$ $\therefore \Phi$ is an isomorphism. $\blacksquare$