# P-groups and orders of elements

1. Apr 11, 2010

### kimberu

1. The problem statement, all variables and given/known data
Let p be a prime and let H be a group of order p^n, some n > 0. Prove that for any x not equal to 1 in H, some power of x has order p.

2. Relevant equations
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3. The attempt at a solution
I know that by lagrange, for any x in G, if x is not the identity, then x has an order p^r for some r>0. Also I know that there is some element of order p. But I don't know how to show that some power of every p^r = p, even though it seems almost intuitive.

Thanks very much!

2. Apr 11, 2010

### Petek

Suppose that x is not the identity and has order $p^2$. Can you find a power of x that has order p? Now try to solve the general case where x has order $p^r$.

Petek

3. Apr 11, 2010

### kimberu

Sorry, I don't even know which power of x would yield order p in the case of $p^2$! Is it $x^2$?

4. Apr 11, 2010

### Dick

If you know x has order p^r for some r, then x^(p^r)=1. What's (x^(p^(r-1)))^p?

5. Apr 11, 2010

### kimberu

Using the example of the cyclic group of order 9, I found that (x^(p^(r-1)))^p is 1 as well. Is that correct?

6. Apr 11, 2010

### Dick

Probably. But I meant it more generally. In any group, how does (x^(p^(r-1)))^p compare with x^(p^r)?

7. Apr 11, 2010

### kimberu

(x^(p^(r-1)))^p is equal to x^(p^r)?

So, is (x^(p^(r-1))) always of order p?

Last edited: Apr 11, 2010
8. Apr 11, 2010

### Dick

Yes, just use the rules of exponents. (x^n)^m=x^(nm).

9. Apr 11, 2010

### kimberu

Okay, so I have:

For any element x, |x|=p^r, x^p^r = 1 = (((x^p)^(r-1))^p) by the properties of exponents. Then x^p^(r-1) equals an element of order p.

Thank you so much!