If every element of $G/H$ has a square root and every element of $H$ has a square root, then every element of $G$ has a square root.

Question

I have tried to prove this statement but cannot. Does it require the group $G$ to be abelian? This assumption is not stated in the text I am reading, but all I can get is that if we consider that $gH = b^2H$ for some $b$, then $gb^{-2} \in H$ so $gb^{-2} = f^2$ for some $f$, but it requires the group to be abelian to say $g = (bf)^2$.

Answer 1

This is false if one leaves the realm of abelian groups or finite groups. Theorem 1.4 of Wehrfritz's The divisible radical of a group produces a non-divisible metabelian $2$-group that is the product of two of its divisible normal subgroups. It is a subgroup of the wreath product of two Pruefer groups $2^\infty$.