Prove if 0(a) is odd, show a is a square. Let $G$ be a group and a is an element of $G$. If $o(a)$ is odd, show $a$ is a square. 
I started by supposing that $o(a) = 2n+1$ which implies $a^{2n+1}=1$. I am not sure if I am on the right track or what I should do next. 

If $o(a) = 2m + 1$ then this implies that $$a^{2m+1} = 1.$$ 
If you multiply both sides by $a$ you get $$a^{2m+2} = a$$
Which is to say that $$(a^{m+1})^2 = a$$ Thus $a$ is a square.
