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

**joestevens** · February 27th 2011, 10:33 AM

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

**tonio** · February 27th 2011, 01:18 PM

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. $N_k\leq G$

3) Solve your problem now.

Tonio

**Drexel28** · February 27th 2011, 09:19 PM

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.

**Swlabr** · February 28th 2011, 12:28 AM

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).

**joestevens** · February 28th 2011, 01:52 PM

Originally Posted by tonio

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

done.

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.

Originally Posted by tonio

3) Solve your problem now.

Tonio

n must be 1. then the only normal subgroups have order p and 1, which are $G$ and $\{e\}$ .

**Drexel28** · March 1st 2011, 07:58 PM

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?

**Swlabr** · March 2nd 2011, 12:36 AM

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...)

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