Exercise 4.5.33 (If $P \in Syl_p(G)$ and $H \leq G$, then $P \cap H \in Syl_p(H)$)

Question

Let $P$ be a normal Sylow $p$-subgroup of $G$ and let $H$ be any subgroup of $G$. Prove that $P \cap H$ is the unique Sylow $p$-subgroup of $H$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $P \in Syl_p(G)$, where $P \unlhd G$ and $H \leq G$.

Then $P\cap H$ is a $p$-subgroup of $H$. Let $Q$ be some fixed $p$-Sylow of $H$. By the second part of Sylow's Theorem, there exists some $g \in H$ such that $P\cap H  \leq g Q g^{-1}$, so \begin{align}
P \cap H \leq R,
\end{align}
where $R =g Q g^{-1}$ is a Sylow $p$-subgroup of $H$.

Moreover $R$ is a $p$-subgroup of $G$. By the same second part of Sylow's Theorem, there is some $h \in G$ such that  $R \leq h P h^{-1} = P$, where $ R \leq H$, so
\begin{align}
 R  \leq P \cap H.
\end{align}

By (1) and (2), we obtain
$$ P \cap H = R = g Qg^{-1}.$$
Since $R$ is a Sylow $p$-subgroup of $H$,  $P \cap H$ is a Sylow $p$-subgroup of $H$.

\bigskip
Moreover, since $P \unlhd G$, then $P \cap H \unlhd H$. This proves that $P \cap H$ is the unique Sylow $p$-subgroup of $H$.
\end{proof}

Exercise 4.5.33 (If $P \in Syl_p(G)$ and $H \leq G$, then $P \cap H \in Syl_p(H)$)

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Exercise 4.5.33 (If $P \in Syl_p(G)$ and $H \leq G$, then $P \cap H \in Syl_p(H)$)

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