# Homework Help: Groups whose order is a power of a prime

1. Sep 16, 2010

### R.P.F.

1. The problem statement, all variables and given/known data

Does every group whose order is a power of a prime p contain an element of order p?

2. Relevant equations

3. The attempt at a solution
I know it certainly can contain an element of order p. I also feel that
|G|=|H|[G] might be useful. Any help is appreciated!

2. Sep 16, 2010

### snipez90

Hmm, cool problem. I think the answer is yes, but I haven't thought it through completely, and more importantly I want to know your ideas. Have you tried using any of the corollaries of Lagrange's theorem that deal with the order (of a group or of an element in a group).

3. Sep 16, 2010

### Dick

It's called Cauchy's theorem.

4. Sep 16, 2010

### Office_Shredder

Staff Emeritus
Cauchy's theorem is overkill for this.

You can pick an arbitrary element of the group and construct an element of order p from it

5. Sep 16, 2010

### R.P.F.

True. But we haven't got to the part about the center of a group yet.

6. Sep 16, 2010

### R.P.F.

Yes I figured it out! That's what I did! Thanks!

7. Sep 16, 2010

### R.P.F.

I figured it out. I just picked an element g of the group and proved that there is an element in <g> that has order p.

8. Sep 16, 2010

### snipez90

Okay yeah I was slightly confused. At first I kept thinking that G had an order that was a multiple of p, and the tools I had did not solve it completely. When I realized that you actually stated ord(G) = p^n, I could do it (namely by using a^ord(G) = e and the fact that if a^m = e then ord(a)|m).

Then Dick mentioned Cauchy's Theorem, so I briefly glanced at a proof on wikipedia and noticed that the tools used were more advanced. Finally I realized that Cauchy's Theorem is what I originally had in mind. So I guess the only thing left to do now is learn Cauchy's Theorem.