# Thread: G is a finite group

1. ## G is a finite group

G is a finite group and $x\in G$ and $|x|=m$. Prove that if m is odd, then $x=(x^2)^k$ for some $k\in\mathbb{Z}$, $k\geq 1$

$x^m=x^{2k+1}=(x^{2})^kx$

I need k to be 0 though. What am I missing?

2. ## Re: G is a finite group

Try letting $m=2k-1,$ instead of $m=2k+1.$ Also, you have to use the order of $x$ to get another equation that will prove useful.

3. ## Re: G is a finite group

Originally Posted by Ackbeet
Also, you have to use the order of $x$ to get another equation that will prove useful.
???????

4. ## Re: G is a finite group

What does |x| = m mean? Can you wrote down an equation that says the same thing?

5. ## Re: G is a finite group

Originally Posted by Ackbeet
What does |x| = m mean? Can you wrote down an equation that says the same thing?
Ok ok.