|G|=p^k. Prove G has an element of order p.

[mathmajor2013]

Let p be a prime, k a pos. int., and G a group with |G|=p^k. Prove that G has an element of order p.

[disregardthat]

Consider x^{\frac{m}{p}} for some element x not equal to the identity, where m is the order of x.

[mathmajor2013]

This kind of seems like a contradiction because m/p is smaller than m, yet x^m/p=(x^m)^1/p=e^1/p=e.

[disregardthat]

This kind of seems like a contradiction because m/p is smaller than m, yet x^m/p=(x^m)^1/p=e^1/p=e.

What does (x^m)^(1/p) mean?

[mathmajor2013]

The pth root of x^m? I'm sorry I cannot see where this one is going

[disregardthat]

The pth root of x^m? I'm sorry I cannot see where this one is going

You have written it yourself, but it doesn't make any sense. x^{\frac{m}{p}} makes perfect sense, since m divides p^k.