In the following problems, let $G$ be an Abelian group.
1) Let $H = \{ x \in G: x=y^{2} \text{ for some } y \in G \}$; that is, let $H$ be the set of all the elements of $G$ which have a square root. Prove that $H$ is a subgroup of $G$.
(i). Let $a, b \in H$, then $a = c^{2}$ and $b = d^{2}$ for some $c$ and $d \in G$. The product $ab = c^{2}d^{2}$ shows that $ab$ has a square root because $c^{2}d^{2} = ccdd =(cd)(cd) = (cd)^{2}$. Because $G$ is a group, it is closed under multiplication so $cd \in G$.
(ii). Let $a \in H$, then $a = c^{2}$ for some $c \in G$. Since $c \in G, c^{-1} \in G$ because $G$ is a group.
This is where I am confused:
Am I allowed to make this conclusion: $a^{-1} = (c^{2})^{-1} = (c^{-1})^{2}$ which shows that $a^{-1} \in H$?
Comments on my next proof would be appreciated.
2) Let $H$ be a subgroup of $G$ and let $K = \{ x \in G: x^{2} \in H \}$. Prove that $K$ is a subgroup of $G$.
(i). Let $a, b \in K$, then $a, b \in G$ and $a^{2}, b^{2} \in H$. We need to show that $ab \in G$ and $(ab)^{2} \in H$. Since $G$ is a group, it must be closed with respect to group multiplication. Since $a, b \in G$, the product $ab \in G$ as well. Since $H$ is a subgroup, it must also be closed with respect to multiplication. Since $a^{2}, b^{2} \in H$, the product $a^{2}b^{2} = aabb = (ab)^{2} \in H$ making use of the fact that $G$ is Abelian.
(ii.) In order to show that $K$ is a subgroup of $G$, it is necessary to show that for every element $a$, there is an inverse element $a^{-1}$ such that $a^{-1} \in G$ and $(a^{-1})^{2} \in H$. Because $G$ is a group, it is closed with respect to inverses. So for every element $a$ there is an inverse element $a^{-1} \in G$. Also since $a^{2} \in H$ and $H$ is a subgroup, it is closed to inverses. So $(a^{2})^{-1} = (a^{-1})^{2} \in H$.