G is a finite group

• Sep 9th 2011, 06:44 PM
dwsmith
G is a finite group
G is a finite group and $\displaystyle x\in G$ and $\displaystyle |x|=m$. Prove that if m is odd, then $\displaystyle x=(x^2)^k$ for some $\displaystyle k\in\mathbb{Z}$, $\displaystyle k\geq 1$

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

I need k to be 0 though. What am I missing?
• Sep 9th 2011, 06:54 PM
Ackbeet
Re: G is a finite group
Try letting $\displaystyle m=2k-1,$ instead of $\displaystyle m=2k+1.$ Also, you have to use the order of $\displaystyle x$ to get another equation that will prove useful.
• Sep 9th 2011, 07:03 PM
dwsmith
Re: G is a finite group
Quote:

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

???????
• Sep 9th 2011, 07:38 PM
Ackbeet
Re: G is a finite group
What does |x| = m mean? Can you wrote down an equation that says the same thing?
• Sep 9th 2011, 07:52 PM
dwsmith
Re: G is a finite group
Quote:

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

Ok ok.