If a finite group has no element of order $2,$ does every element have a unique square root?

Question

Suppose $G$ is a finite group. Are these two statements about it equivalent?

1. $\forall g \in G \exists ! h \in G$ such that $h^2 = g$

2. There are no elements of order $2$ in $G$

It is quite obvious that the first statement implies the second one: if $g = e$, then $h = e$ is the only possible $h$.

However, I do not know whether the second statement implies the first one or not.

Answer 1

Suppose $(2)$. By Cauchy’s theorem, the order $n$ of the group is an odd number, $n=2m+1$. Now given $g\in G$, take $h=g^{m+1}$ it satisfies $h^2=g$ and it is unique because if $h^2=h’^2$ then $h^{2(m+1)}=h’^{2(m+1)}$ or $h=h’$ becaue $h^{2m+1}=e=h’^{2m+1}$, and $(1)$ follows.

Answer 2

Suppose there are no elements of order $2$ in $G$, then every element is of odd order.

Suppose $g^{2r-1}=1$ then $(g^r)^2=g$ - so there is at least one element - $g^r$ - whose square is $g$.

Suppose now $h^2=g$ so that $h^{4r-2}=1$ and $(h^{2r-1})^2=1$, then $h^{2r-1}$ is not of order $2$ so $h^{2r-1}=1$ and $g^r=h^{2r}=h$. So the element whose square is $g$ is unique.