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.
$\therefore$ $n_p(G) \geq n_p(G/N)$ $\blacksquare$