Let $G$ be a group and $H=\{g^2 : g\in G\}$ then which of the following is/are true?
$(1)H $ is always a subgroup of $G$
$(2)H$ may not be a subgroup of $G$
$(3)$ If $H$ is a subgroup of $G$, then it must be normal in $G$
$(4)H$ is a normal subgroup of $G$ only if $G$ ia abelian.
My attempt
$(3)$ is true
Let $H$ be a subgroup and $x^2 \in H$.
Then $gx^2g^{-1}=(gxg^{-1})^2 \in H$
So yes if $H$ is a subgroup , then it is normal.
$(4)$ Let $G=Q_8$ , the group of Quaternions.
Then $H=\{1,-1\}$ is a normal subgroup but $G$ is non-abelian.
I guess $H$ is not necessarily a subgroup since $H$ may not be closed under multiplication but I can't find one example.
I think I have to look for non-abelian groups (may be of odd order, but not sure..)
Can you give an example?
Thanks for your time.
For abelian groups, $g^2h^2 = ghgh = (gh)^2 \in G^2$ so $H$ is always a (normal) subgroup of $G$ when $G$ is abelian. As you correctly pointed out, $Q_8$ is a counterexample to the "only if" direction of (4).
(3) is fine.
(1) and (2) are mutually exclusive, so you need to prove the statement or find a counterexample. I think that the smallest counterexample is $A_4$: trivially, every $3$-cycle is a square (as if $x$ has order $3$ then $x=(x^2)^2$), and there are eight $3$-cycles in $A_4$. Every other nontrivial element is the product of two disjoint $2$-cycles, and as such has order $2$, so it cannot be the square of anything as $A_4$ doesn't have elements of order $4$.
So, $H$ is the set of all $3$-cycles in $A_4$, and it has $9$ elements (eight plus the identity). But $|A_4|=12$, and $9$ does not divide $12$, so $H$ cannot be a subgroup. And, indeed: $$(123)(124) = (13)(24) \not \in H$$