# Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

• February 27th 2011, 10:33 AM
joestevens
Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.
Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.
• February 27th 2011, 01:18 PM
tonio
Quote:

Originally Posted by joestevens
Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.

1) Using the Class Equation, show that any finite p-group has non-trivial center,

2) By induction on n, show that for any $0\leq k\leq n$ there exists a normal

sbgp. $N_k\leq G$

Tonio
• February 27th 2011, 09:19 PM
Drexel28
Quote:

Originally Posted by joestevens
Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.

Or, use the fact that if a $p$-group has order $p^n$ then any subgroup of order $p^{n-1}$ must be normal.
• February 28th 2011, 12:28 AM
Swlabr
Quote:

Originally Posted by Drexel28
Or, use the fact that if a $p$-group has order $p^n$ then any subgroup of order $p^{n-1}$ must be normal.

You would also need to prove subgroups of order $p^{n-1}$ exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).
• February 28th 2011, 01:52 PM
joestevens
Quote:

Originally Posted by tonio
1) Using the Class Equation, show that any finite p-group has non-trivial center,

done.

Quote:

Originally Posted by tonio
2) By induction on n, show that for any $0\leq k\leq n$ there exists a normal sbgp. $N_k\leq G$

done.

Quote:

Originally Posted by tonio

Tonio

n must be 1. then the only normal subgroups have order p and 1, which are $G$ and $\{e\}$.
• March 1st 2011, 07:58 PM
Drexel28
Quote:

Originally Posted by Swlabr
You would also need to prove subgroups of order $p^{n-1}$ exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).

I'm sorry, isn't the fact that if $p^k\mid |G|$ where $|G|<\infty$ then $\exists H\leqslant G$ such that $|H|=p^k$ commonly known?
• March 2nd 2011, 12:36 AM
Swlabr
Quote:

Originally Posted by Drexel28
I'm sorry, isn't the fact that if $p^k\mid |G|$ where $|G|<\infty$ then $\exists H\leqslant G$ such that $|H|=p^k$ commonly known?

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...)