Suppose that $\displaystyle |G| = p^n$ and further suppose that $\displaystyle G$ is simple. Prove that $\displaystyle |G| = p$.

- Feb 27th 2011, 10:33 AMjoestevensSuppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.
Suppose that $\displaystyle |G| = p^n$ and further suppose that $\displaystyle G$ is simple. Prove that $\displaystyle |G| = p$.

- Feb 27th 2011, 01:18 PMtonio
- Feb 27th 2011, 09:19 PMDrexel28
- Feb 28th 2011, 12:28 AMSwlabr
- Feb 28th 2011, 01:52 PMjoestevens
- Mar 1st 2011, 07:58 PMDrexel28
- Mar 2nd 2011, 12:36 AMSwlabr
Well, yes, but so is the fact that the centre is non-trivial. As the OP doesn't appear to know that the centre is non-trivial, we must assume they don't know that either! My point was, if one is going to prove one of `maximal subgroups have order p', `Z(G) is non-trivial', one would prove the latter.

Unless, of course, one doesn't know the class formula...

(I've used `one' as if I'd used `you' the last line is obviously false, because I'm pretty sure you know the class formula...so I was trying to not insult you, not trying to sound posh...)