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Let $p$ be a prime number, $G$ a group with subgroup $H$ and $S$ a Sylow $p$-subgroup of $G$. Show that there exists $g\in G$ such that $H\cap gSg^{-1}$ is a Sylow $p$-subgroup of $H$. Moreover, come up with an example that shows that $g\neq e_G$ holds in general.

My attempt: By the first Sylow theorem applied to $H$, we find a Sylow $p$-subgroup of $H$. This should be a $p$-group of $G$ and thus, by the second Sylow theorem, contained in a Sylow $p$-subgroup, say $T$ of $G$. As all Sylow $p$-subgroups are conjugate, we find $g\in G$ such that $gSg^{-1}=T$. Is it possible to conclude that $H\cap T$ is a Sylow $p$-subgroup of $H$?

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  • $\begingroup$ Your reasoning is almost complete! You just need to observe that $H\cap T$ is a $p$-subgroup of $H$ containing a Sylow $p$-subgroup of $H$, which means... $\endgroup$ Sep 8 '20 at 23:44
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    $\begingroup$ @AmiteshDatta: ...that $H\cap T$ already has to be a Sylow $p$-subgroup of $H$ because otherwise, it would contradict the maximality of this Sylow $p$-subgroup of $H$? $\endgroup$ Sep 8 '20 at 23:52
  • $\begingroup$ exactly! (have to insert more characters to satisfy character limit of comment...) $\endgroup$ Sep 8 '20 at 23:56
  • $\begingroup$ @AmiteshDatta: Nice! So in fact, $H\cap T$ already equals the found Sylow $p$-subgroup of $H$. $\endgroup$ Sep 8 '20 at 23:59
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    $\begingroup$ @AmiteshDatta: Is it possible to use $G=S_3$, $H=\langle (1\, 2)\rangle$ and $S = \langle (1\, 3)\rangle$ as an example for $g\neq e_G$? $\endgroup$ Sep 9 '20 at 0:07
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You don't need Sylow I to do this, and in fact it can be used to prove Sylow I! Consider the action of $H$ on the left cosets $G/S$. The stabilizer of the coset $gS$ consists of all $h \in H$ such that

$$hgS = gS \Leftrightarrow g^{-1}hg \in S$$

and hence $\text{Stab}(gS) = g^{-1}Hg \cap S$; in particular it must have order a power of $p$. On the other hand, dividing up $G/S$ into its $H$-orbits and applying orbit-stabilizer gives

$$|G/S| = \sum_{|H\backslash G/S|} \frac{|H|}{|\text{Stab}(gS)|} = \sum_{|H \backslash G / S|} \frac{|H|}{|g^{-1}Hg \cap S|}.$$

Since $S$ is Sylow $|G/S|$ is not divisible by $p$ so some term on the RHS is not divisible by $p$. This says precisely that there is some $g$ such that $g^{-1} Hg \cap S$ has index in $H$ coprime to $p$, and hence $g^{-1} Hg \cap S$ is Sylow!

An example where we need $g \neq e$ can be obtained by finding any $G$ such that $S$ is not normal and setting $H$ to be a nontrivial conjugate of $S$; your example in the comments is minimal with this property. Note that setting $H$ to be another $p$-subgroup of $G$ now immediately proves Sylow II for any $G$ containing a Sylow.

This lemma, which I hear is due to Frobenius, can be used to prove Sylow I by explicitly constructing Sylow $p$-subgroups for any family of groups into which all finite groups embed. Historically this was first done for $G = S_n$ the symmetric groups; it's a tiny bit annoying to explicitly write down the Sylows but it can be done (I hear it was first done by Cayley) and it's a bit easier if $n = p^k$ is a prime power. It's easier for $G = GL_n(\mathbb{F}_p)$; here the upper triangular matrices with $1$s on the diagonal (the unipotent subgroup) give a Sylow $p$-subgroup, and there's even an easy proof, again without the Sylow theorems (or the above argument), that every $p$-subgroup of $G$ is conjugate to a subgroup of this unipotent subgroup, and also an easy proof that the index of the normalizer of the Sylow is $1 \bmod p$.

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  • $\begingroup$ Nice answer! Could you explain how to let $H$ act on the left cosets $G/S$ in order to use orbit-stabilizer theorem? $\endgroup$ Sep 9 '20 at 0:34
  • $\begingroup$ @physicist23: it's by left multiplication: $h \in H$ sends the coset $gS$ to the coset $hgS$. It's the restriction of the natural action of $G$. $\endgroup$ Sep 9 '20 at 0:41
  • $\begingroup$ I might be missing something but you've shown that $\operatorname{Stab}(gS)=H\cap gSg^{-1}$, namely all elements of $H$ that are conjugate to elements of $S$ by $g$. (rather than $\operatorname{Stab}(gS)=g^{-1}Hg\cap S$, which might not even be a subset of $H$!) $\endgroup$
    – Mor A.
    Sep 9 '20 at 10:35
  • $\begingroup$ @Mor: you can get from one to the other by conjugating by $g$. $\endgroup$ Sep 9 '20 at 18:00

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