# Find Sylow $p$-subgroup within subgroup

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 $$H$$?

• 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... Sep 8 '20 at 23:44
• @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 $H$? Sep 8 '20 at 23:52
• exactly! (have to insert more characters to satisfy character limit of comment...) Sep 8 '20 at 23:56
• @AmiteshDatta: Nice! So in fact, $H\cap T$ already equals the found Sylow $p$-subgroup of $H$. Sep 8 '20 at 23:59
• @AmiteshDatta: Is it possible to use $G=S_3$, $H=\langle (1\, 2)\rangle$ and $S = \langle (1\, 3)\rangle$ as an example for $g\neq e_G$? Sep 9 '20 at 0:07

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 $$1 \bmod p$$.

• Nice answer! Could you explain how to let $H$ act on the left cosets $G/S$ in order to use orbit-stabilizer theorem? Sep 9 '20 at 0:34
• @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 $G$. Sep 9 '20 at 0:41
• 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 $H$!) Sep 9 '20 at 10:35
• @Mor: you can get from one to the other by conjugating by $g$. Sep 9 '20 at 18:00