Exercise 4.5.2 (If $H\leq G$ and $Q \in Syl_p(H)$ then ${gQg^{-1} \in Syl_p(gHg^{-1})}$ )

Question

Prove that if $H$ is a subgroup of $G$ and $Q \in Syl_p(H)$ then ${gQg^{-1} \in Syl_p(gHg^{-1})}$ for all $g \in G$.

Richard Ganaye · Accepted Answer

\begin{proof}

 We suppose that $Q \leq H \leq G$, where $Q \in Syl_p(H)$, so that there are positive integers $\alpha, l$ such that
$$|Q| = p^\alpha,\qquad |H| = p^\alpha\,  l, \qquad p 
mid l.$$
For any $g \in G$, since $Q \leq H$, $g Qg^{-1} \leq gHg^{-1}$. Since the map $\gamma_g : G 	o G$ defined by $\gamma_g(x) = gxg^{-1}$ is bijective, 
$$|g Qg^{-1} | = |G| = p^\alpha,\qquad |gHg^{-1}| = |H| = p^\alpha l,\qquad p 
mid l,$$
thus $gQg^{-1} \in Syl_p(gHg^{-1})$.
\end{proof}

Exercise 4.5.2 (If $H\leq G$ and $Q \in Syl_p(H)$ then ${gQg^{-1} \in Syl_p(gHg^{-1})}$ )

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Exercise 4.5.2 (If $H\leq G$ and $Q \in Syl_p(H)$ then ${gQg^{-1} \in Syl_p(gHg^{-1})}$ )

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