$G$ is a group of odd order, show that $a^2=b^2 \Rightarrow a=b$

I've come across this question and I tried to prove it, but my solution seems a little stealthy to me, is it correct?

Let $|G|=2k+1$, then we have:

$a=ae=aa^{|G|}=a^{|G|+1}=a^{2k+2}=(a^2)^{k+1}=(b^2)^{k+1}=b^{2k+2}=b^{|G|+1}=eb=b$

I would like to know if I'm missing something

Solution 1:

Yep, that's a totally valid answer. Intuitively, the reason why this works is that in a group with an odd order, we can't have an element with an even order (by Lagrange's Theorem, trivially), so squaring something would never get us into a bind with the identity element that invalidates your $a = b \rightarrow a^2 = b^2$.

In fact, if you're interested in pursuing this a little further, you can think about why this approach could or couldn't prove that if $|G|$ has no common factors with some $n$, then $a^n = b^n \rightarrow a = b$. You can prove a more general result (though not necessarily one as clean as the one you have here!).