# Homework Help: Prime p divides order of group

1. Aug 21, 2013

### mahler1

1. The problem statement, all variables and given/known data.
Let p be a prime number, m a natural number and G a group of order p^m. Prove that there exists an element a in G such that ord(a)=p.

3. The attempt at a solution.
I know of the existence of Lagrange theorem, so what I thought was: I pick an arbitrary element a (I exclude e, the identity element) of G and look at the group generated by that element, denoted as <a>. I also know that the order of <a> is equal to the order of the element a. Now I can apply Lagrange theorem: as the order of <a> divides the order of G, then the order of <a> must be of the form p^k for some natural number k, k≤m. I got stuck here, I can't deduce k=1 with only these statements, there is something missing.

2. Aug 21, 2013

### Axiomer

So if a has order p^k then which power of a has order p?

3. Aug 21, 2013

### verty

If the cyclic group generated by an element is always a subgroup, and if the order of that subgroup always divides the order of the group, then the claim will go through if there is always an element of order > 1. But on two sites I found, much more complicated proofs are given, induction on k with separate handling of the abelian/non-abelian cases. It could just be that they are written so as not to use Lagrange's Theorem.

4. Aug 21, 2013

### mahler1

I got it, but just in case I want to check: Is it p^(k-1)?, because then (a^p^(k-1))^p=a^(pp^(k-1))=a^p^k=1

Last edited: Aug 21, 2013
5. Aug 21, 2013

### mahler1

Maybe there is other way of proving this statement without using Lagrange's theorem, but, if you want to, check what Axiomer wrote and then the proof is completed using Lagrange's theorem. Anyway, I'll ask my professor to show me an alternative proof. Thanks.

6. Aug 21, 2013

Yep!