Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Statement: Let G be a group and p a prime that divides $|G|$. Prove that if $K\le G$ such that $|K|$ is a power of p, K is contained in at least one Sylow p-group.

I just started studying Sylow p-groups, so although I'm familiar with Sylow theorems and a couple of corollaries, I don't know how to get started with this problem. Any hint is more than welcome.

PS: I looked for something related here at Math.SE but didn't find anything. Sorry if it's a duplicate.

share|improve this question
1  
Hint: Let $P$ be a Slow $p$-subgroup of $G.$ Consider the action of $K$ in the permutation action of $G$ on the right cosets of $P.$ –  Geoff Robinson May 27 '12 at 22:56

3 Answers 3

Assume $\,|G|=p^nm\,,\,(p,m)=1$:

I assume you already know the following Sylow theorems: Sylow $p$-sbgps. exist in $\,G\,$ , they all are conjugate in $\,G\,$ and their number equals $\,1\pmod p:$ let $\,K\,$ be a $p$-sbgp. and let $\,\mathcal{P}:=\{P\leq G \mid P\text{ is a Sylow }p\text{-sbgp. of }\, G\}\,$. As noted above, $\,|\mathcal{P}|\equiv 1\pmod p$.

Let now $K$ act on $\,\mathcal{P}\,$ by conjugation: $\,k\in K\,,\, P\in\mathcal{P}\Longrightarrow k\cdot P\to k^{-1}Pk=:P^k\,$ . If there is an orbit with one single element , say $\,\mathcal{O}_Q=\{Q\}\,\,,\,\,\text{for some}\,Q\in\mathcal{P}$, then $\,Q^k=Q\,\,\forall\,k\in K\Longrightarrow QK=KQ\Longrightarrow QK\,$ is a $p$-subgroup of $\,G\,$ , so if $\,K\,$ is not contained in any Sylow $p$-sbgp then $\,|QK|>|Q|=p^n\,$, which is absurd, so that all the orbit have size a power of $p$, but this means $\,|\mathcal{P}|\equiv 0\pmod p$, which of course is also absurd since, as we mentioned above, $\,|\mathcal{P}|\equiv1\pmod p\Longrightarrow \,$ it must be that $\,K\,$ is contained in some Sylow $p$-sbgp.

share|improve this answer
1  
After 10 month, someone made me to read this nice approach. + –  B. S. Apr 13 '13 at 19:28

Suppose that $|G|=p^nm$, where $p\nmid m$. Let $P$ be a Sylow $p$-subgroup of $G$, and let $\mathscr{C}$ be the set of left cosets of $P$; clearly $|\mathscr{C}|=m$. Let $G$ act on $\mathscr{C}$ in the obvious way: $g\cdot aP=(ga)P$ for each $g\in G$ and $aP\in\mathscr{C}$.

Suppose that $g\cdot aP=bP$; then $h\in\operatorname{Stab}_G(bP)$ iff $hg\cdot aP=g\cdot aP$ iff $g^{-1}hg\cdot aP=aP$ iff $g^{-1}hg\in\operatorname{Stab}_G(aP)$. In particular, $h\in\operatorname{Stab}_G(aP)$ iff $a^{-1}ha\in\operatorname{Stab}_G(P)=P$. Thus, $\operatorname{Stab}_G(aP)=aPa^{-1}$.

Now consider the restriction of this action to $K$. For each $aP\in\mathscr{C}$ we have $$\operatorname{Stab}_K(aP)=K\cap\operatorname{Stab}_G(aP)=K\cap aPa^{-1}\;.\tag{1}$$ $|K|$ is a power of $p$, as is $|aPa^{-1}|=p^n$, so

$$|\operatorname{Orb}_K(aP)|=\frac{|K|}{|\operatorname{Stab}_K(aP)|}=p^{k(aP)}$$ for some $k(aP)\in\Bbb N$. The orbits of the action of $K$ on $\mathscr{C}$ partition $\mathscr{C}$, so $$m=|\mathscr{C}|=\sum_{C\in\mathscr{C}}|\operatorname{Orb}_K(C)|\;.$$

Here are some questions to direct you towards finishing the argument. If you get completely stuck, I’ve finished it below the questions but left it spoiler-protected; mouse-over to see it.

  1. Why is it impossible that $k(aP)\ge 1$ for every $aP\in\mathscr{C}$?
  2. If $k(aP)=0$, what does $(1)$ tell you about $K$ in relation to $aPa^{-1}$?
  3. What kind of subset of $G$ is $aPa^{-1}$?

If $k(C)$ were positive for every $C\in\mathscr{C}$, $m$ would be a multiple of $p$. It’s not, so there must be at least one $aP\in\mathscr{C}$ such that $|\operatorname{Orb}_K(aP)|=p^0=1$, and hence $K=\operatorname{Stab}_K(aP)=K\cap aPa^{-1}$ by $(1)$. But then $K\subseteq aPa^{-1}$, where $aPa^{-1}$ is a $p$-Sylow subgroup of $G$.

share|improve this answer

Here's another approach. You want to show for any $p$-subgroup $K$, $K\subseteq aPa^{-1}$ where $P$ is a Sylow $p$-subgroup and $a\in G$ . Let $X=\{aP \mid a\in G\}$ be the set of left cosets of $P$ in $G$. Now let $K$ act on $X$ in the following manner $k \cdot(aP)=(ka)P$ .

Let $|G|=p^n m$ where $p \nmid m$ , we know that $|P|=p^n$ since $P$ is a Sylow $p$-subgroup. Note that $[G:P]=|X|= \displaystyle\frac {|G|}{|P|}=m$, then $p \nmid |X|$. We also know that if a $p$-group acts on a set $X$ then $p\mid |X|-|X_f|$ where $X_f$ is the fixed set under the action (I can supply the proof if needed). But $p \nmid |X|$ then $p \nmid |X_f|$ (otherwise $p \mid |X|-|X_f|+|X_f|=|X|$ which is a contradiction) then $|X_f|\not= 0$ then there is an element $aP\in X$ such that $k(aP)=P \ \forall\ k\in K$. $kaP=aP \implies a^{-1}(ka) \in P \implies k\in aPa^{-1}$ for all $k\in K$ hence $K \subseteq aPa^{-1}$ .

As a side note you can use this as a lemma to prove the second Sylow theorem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.