element of odd finite group has unique square root let G be an odd finite group show that each element of G has a unique square root (if $x \in G$ then there exists $y \in G$ such that $x=y^2$)
Is this problem right?
If there is another condition such as G is commutative, I can solve this problem
But with only this, My logic fails.
My question is that "do this problem without condition: abelian,  make sense?? Is there any counterexamples? (odd finite group but there exists an element which doesn't have square root)"

First we prove that the square root (if it exists) is unique:  If $\#G=2k-1$ and $g_1^2=g_2^2$ then we write $$g_1=g_1^{2k}=(g_1^2)^{k}=(g_2^2)^k=g_2^{2k}=g_2$$
Now we remark that this suffices.  Indeed, if $f:S\to S$ is a function from a finite set to itself, then injectivity is equivalent to surjectivity.
