Exercise 4.5.34 (If $P \in Syl_p(G)$ and $N \unlhd G$ then $PN/N$ is a Sylow $p$-subgroup of $G/N$ )

Proof. Let $Q$ be some fixed Sylow $p$-subgroup of $N$. Note that $P\cap N$ is a $p$-subgroup of $N$. By the conjugacy part of Sylow’s Theorem, we know that there is some $g\in N$ such that $P\cap N\le \mathit{𝑔𝑄}{g}^{-1}$. Put $R=\mathit{𝑔𝑄}{g}^{-1}$. Then $R$ is a Sylow $p$-subgroup of $N$ such that

Since $R$ is a $p$-subgroup, the same conjugacy part of Sylow’s Theorem shows that there is some $h\in G$ such that $R\le \mathit{h𝑃}{h}^{-1}$, so $h^{-1}\mathit{𝑅h}\le P$. Moreover $N⊴G$, and $R\le N$, thus $h^{-1}\mathit{𝑅h}\le N$, so

By (1) and (2), we obtain $|P\cap N|\le |R|$, and $|R|=|h^{-1}\mathit{𝑅h}|\le |P\cap N|$, thus $|P\cap N|=|R|$, where $P\cap N\le R$, so

This shows that $P\cap N$ is a Sylow $p$-subgroup of $N$.

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

Since $N⊴G$, $\mathit{𝑃𝑁}$ is a subgroup of $G$, and $|\mathit{𝑃𝑁}|=\frac{|P||N|}{|P\cap N|}$. Therefore

and

where $m∕m^{\prime }$ is an integer: since $|N|\mid |G|$, then $m^{\prime }\mid p^{\alpha }m$, where $g.c.d(m^{\prime },p)=1$, thus $m^{\prime }\mid m$. Moreover, $p\nmid m$, thus $p\nmid (m∕m^{\prime })$. This shows that $\mathit{𝑃𝑁}∕N$ is a Sylow $p$-subgroup of $G∕N$ (see another proof in Exercise 3.3.9). □