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. $\blacksquare$