Exercise 4.5.36 (If $N \unlhd G$, then $n_p(G/N) \leq n_p(G)$)

Proof. We suppose that $N⊴G$.

By Exercise 34, if $P\in \mathit{𝑆𝑦}{l}_p(G)$, then $\mathit{𝑃𝑁}∕N\in \mathit{𝑆𝑦}{l}_p(G∕N)$. Consider the map

We show that $\phi$ is surjective.

Let $\bar{Q}$ be any Sylow $p$-subgroup of $G∕N$, and let $Q={\pi }^{-1}(\bar{Q})$, where $\pi :G\to G∕N$ is the natural projection, defined by $\pi (g)=\mathit{𝑔𝑁}$. Then $Q\supseteq N$, and $\bar{Q}=Q∕N$.

Let $P$ be some fixed $p$-Sylow subgroup of $G$.

As in Exercise 34, write $|G|=p^{\alpha }m$, where $p\nmid m$, and $|N|=p^{\beta }{m}^{\prime }$, where $p\nmid m^{\prime }$. Since $P$ is a Sylow $p$-subgroup of $G$ and $P\cap N$ is a Sylow $p$-subgroup of $H$, then

By the solution of Exercise 34,

and $\mathit{𝑃𝑁}∕N$ is a Sylow $p$-subgroup of $G∕N$. Since $\bar{Q}$ is another subgroup of $G∕N$,

Then

This shows that $Q\in \mathit{𝑆𝑦}{l}_p(G)$.

Moreover $N\le Q$, thus $\mathit{𝑁𝑄}=Q$, so $\phi (Q)=\mathit{𝑁𝑄}∕N=Q∕N=\bar{Q}$. This proves that $\phi$ is surjective. Therefore $|\mathit{𝑆𝑦}{l}_p(G∕N)|\le |\mathit{𝑆𝑦}{l}_p(G)|$, so

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