Proving Abelian p-Group: gnhn = (gh)n

[*FaerieLight*]

Homework Statement

How would I prove that gⁿhⁿ = (gh)ⁿ, for g,h $\in$G where G is a p-group, and all its elements have order p?

The Attempt at a Solution

My aim is to prove this in order to prove that G is Abelian, but I don't want to prove it using centres. I've supposed that gⁿhⁿ = (gh)^m for some m, and now I'm trying to prove that m must be congruent to n modulo p, as this is the only way that (gh)^m = (gh)ⁿ since gh has order p. And this is where I strike a wall, so to speak. Is there a better way to prove that G is Abelian, if this isn't good enough?

Thanks!

 

[Robert1986]

Think: generator.

 

[*FaerieLight*]

As in...gh generates a cyclic group?

 

[Robert1986]

Well, yes, but don't think about gh. You know that if g is in G, then g has order p, correct?

 

[Robert1986]

That is, can you think of a way to write any element of G in terms of g?

 

[micromass]

Yeah, what you're trying to prove is false. A counterexample is given by

$(\mathbb{Z}_3\rtimes \mathbb{Z}_3)\times \mathbb{Z}_3$

All (non-identity) elements have order 3 and the group is non-abelian.