2

Let $ G $ be a finite simple group which acts on finite set $ X $ non-trivially. The goal is that the largest prime number $p$ dividing $ \left|G\right| $ is also divide $ \left|X\right| $.

I know $G$ is embedded in $S_{X}$. But, I can't get any other things.

jawlang
  • 685

1 Answers1

0

Hmm, I'm not sure to understand exactly the question : do you want to prove this for every non-trivial action or do you want a condition for it to be true. Because I think in general it's false (I even think that for almost every group $G$ you can find a non-trivial action for which this is false).

Take $G$ of order $n=p^km$, with $m \wedge p=1$ and $p$ larger than all the prime factors of $m$. Consider a $p$-Sylow $P$ of $G$ and the quotient $X=G/P$. $G$ act transitivly and non-trivially on $X$ by translation, however $|X|=m$ so $p$ does not divide $|X|$.

Is there a mistake there ?

Sylvain L.
  • 1,265
  • Well, wether you consider the left quotient or the right quotient, either the left translation or the right translation will map $X$ to itself, no ? – Sylvain L. Jun 21 '15 at 17:05
  • I'm not sure to understand. If you consider $X=\lbrace gP, g\in G \rbrace$. Can't you consider the left action : $h.gP=(hg)P$ ? – Sylvain L. Jun 21 '15 at 17:09
  • Actually, I think the only case where this counter-example does not work is for is for $p$-groups, as $X$ will be of size one, so the action will be trivial. But as OP asks for $G$ simple, it forbids $p$-groups. – Sylvain L. Jun 21 '15 at 17:19
  • Why would $P$ be a normal subgroup ? – Sylvain L. Jun 21 '15 at 17:25
  • Hmm, I don't think that's the case : for example take $G=A_5$, we have $n=60$, $p=5$ and $m=12$, so $m$ is not equal to 1 mod $p$ and yet, there are 6 non-normal 5-Sylow. I think it's enough that there exists $q$ dividing $m$ such that $q=1 ; \text{mod} ; p$ for the possibility of a non normal $p$-Sylow (and if this is no the case, $G$ is not simple). – Sylvain L. Jun 21 '15 at 17:42
  • Actually, there is not that much to wonder here I think. Take any simple group $G$. Then there will exist a $p$-Sylow and it is not normal, as $G$ is simple (well, except if $G=\mathbb{Z}/p\mathbb{Z}$). – Sylvain L. Jun 21 '15 at 17:46
  • Oh, $A_{5}$ is a counter-example. Thank you. This problem was a test problem. This problem ask me to prove this, and show that index of every non-trivial subgroup of $ G $ is at least $ p $ using this. Showing the second part is easy, more over the proof doesn't need this. – jawlang Jun 22 '15 at 14:22