TITLE: Are these two statements about a finite group equivalent?
QUESTION [7 upvotes]: Suppose $G$ is a finite group. Are these two statements about it equivalent?

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

REPLY [4 votes]: 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.