subgroup of a group for non abelian cases

Question

Let $G$ be any group and $H =\{ x^2: x \in G \}$ then $H$ is subset of $G$. Now if $G$ is an abelian group then clearly $H$ is normal subgroup of $G$. But I have doubt on the case when $G$ is non abelian. Is $H$ is subgroup of $G$ if $G$ is non abelian? And further if yes , then is $H$ is normal subgroup also of non abelian $G$ ?

Answer 1

Take G = A_{4}, the alternating group on 4 letters, then H is subset of cardinality 9, certainly it can not be a subgroup since 9 does not divide 12. Also if H is a subgroup then it would be normal because gx^{2}g^{-1} = (gxg^{-1})^{2}

Answer 2

In general, $\{x^2: x \in G\}$ is a normal set, since $g^{-1}x^2g=(g^{-1}xg)^2$ for any $g \in G$. Hence, $\langle x^2: x \in G \rangle$ is always a normal subgroup, say $N$. And in $G/N$ squaring any element gives the identity, so this group must be abelian. Hence $G' \subseteq N$.