Exercise 4.5.1 (If $P \leq H \leq G$, then $P \in Syl_p(G) \Rightarrow P \in Syl_p(H)$)

Proof. If $G=p^{\alpha }m$, where $p\nmid m$ and $P\in \mathit{𝑆𝑦}{l}_p(G)$, then $|P|=p^{\alpha }$. Since $P\le H\le G$, by Lagrange’s Theorem, $|P|$ divides $|H|$ and $|H|$ divides $|G|$, so $p^{\alpha }\mid |H|$ and $|H|\mid p^{\alpha }m$, thus $|H|=p^{\alpha }l$ for some integer $l$, and $p^{\alpha }l\mid p^{\alpha }m$, so $l\mid m$. Since $p\nmid m$, then $p\nmid l$, so

Since $P\le H$ and $|P|=p^{\alpha }$, this shows that $P$ is a $p$-Sylow of $H$.

If $P\le H\le G$, then

To give a counterexample of the converse, consider $G=Z_{12}=⟨x⟩$, and $H=⟨x^2⟩$, $P=⟨x^6⟩$. Then $P\le H\le G$, where $|P|=2,|H|=6$ and $|G|=12$. Therefore $P\in \mathit{𝑆𝑦}{l}_2(H)$, but $P\notin \mathit{𝑆𝑦}{l}_2(G)$.

In general, a Sylow $p$-subgroup of a subgroup of $G$ need not be a Sylow $p$-subgroup of $G$. □