Let G be an Abelian group. Prove or disprove that
H={g^2|g is a member of G}
is a subgroup of G.
Indeed it's true, take $\displaystyle a,b\in G$ then $\displaystyle a^2*b^2=a*a*b*b=a*b*a*b=(a*b)*(a*b)=(a*b)^2$ (*)
Where we've used the fact that G is abelian and that the associativity holds.
So if $\displaystyle a'\in H$ and $\displaystyle b' \in H$ then by (*) we have $\displaystyle a'*b' \in H$
Also the identity belongs to H since $\displaystyle e^2=e$
And each element has an inverse, say we have $\displaystyle a' \in H$ then $\displaystyle a'=a^2$ for some $\displaystyle a \in G$
Now $\displaystyle a$ has an inverse, namely $\displaystyle a^{-1} \in G$
So note that $\displaystyle a^2*a^{-2}=a*a*a^{-1}*a^{-1}=e$ thus $\displaystyle a^{-2}=(a^{-1})^2\in H$ is the inverse of a'