Exercise 4.5.35 (If $P \in Syl_p(G)$ and $H \leq G$, then $gPg^{-1} \cap H \in Syl_p(H)$ for some $g \in G$)

Question

Let $P \in Syl_p(G)$ and let $H \leq G$. Prove that $gPg^{-1} \cap H$ is a Sylow $p$-subgroup of $H$ for some $g \in G$. Give an explicit example showing that $h P h^{-1} \cap H$ is not necessarily a Sylow $p$-subgroup of $H$ for any $h \in H$ (in particular, we cannot always take $g = 1$ in the first part of this problem, as we could when $H$ was normal in $G$).

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $P \in Syl_p(G)$ and let $H \leq G$.

Let $Q$ be a Sylow $p$-Sylow subgroup of $H$. Since $Q$ is a $p$-subgroup of $G$,  by the conjugacy part of Sylow's Theorem, there is some $g \in G$ such that $Q \subseteq gPg^{-1}$. Then $Q \subseteq gPg^{-1} \cap H$, and $gPg^{-1} \cap H$ is a $p$-subgroup of $H$, which contains the  $p$-Sylow subgroup $Q$ of $H$, hence 
$$gPg^{-1} \cap H = Q,$$
so $gPg^{-1} \cap H$ is a  $p$-Sylow subgroup of $H$.

\bigskip
To give the required explicit example, consider the group $G = A_4$, and the subgroups $P = \langle (1\ 2\ 3) \rangle$, $H = \langle (1\ 2\ 4) \rangle$. Then $P$ is a $3$-Sylow subgroup of $G$, and for $g = 1 = ()$,
$$gPg^{-1} \cap H = P \cap H= \{1\}$$
is not a $3$-Sylow subgroup of $H$.
\end{proof}

Exercise 4.5.35 (If $P \in Syl_p(G)$ and $H \leq G$, then $gPg^{-1} \cap H \in Syl_p(H)$ for some $g \in G$)

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Exercise 4.5.35 (If $P \in Syl_p(G)$ and $H \leq G$, then $gPg^{-1} \cap H \in Syl_p(H)$ for some $g \in G$)

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