Abstract Algebra groups and subgroups Let $G$ be an abelian group. Let $H$ be a subgroup of $G$. Define the set 
$S(H)=\{x\in G:x^2 \in H\}$. Prove that $S(H)$ is a subgroup of $G$.
The notation is confusing to me, any type of help would be appreciated. 
 A: $S(H)=\{x\in G :x^2\in H\}$


*

*Prove that $S(H)$ is non empty.

*Take $x,y\in S(H)$ and prove that $x\cdot y\in S(H)$

*Take $x\in S(H)$ and prove $x^{-1}\in S(H)$


You are supposed to use that $G$  is abelian. 
A: You should read $S(H)$ is the set of all elements (say $x$) in $G$ such that its square is in $H$ (i.e. $x^2\in S(H)$.
If you want to check for example if $e\in S(H)$ you do this:

$e\in G$ and $e^2=e\in H$

These are  the  exact  conditions that any element (so in particular $e$) should have to be in $S(H)$. Hence we can safely say: $e\in S(H)$. 
If we want $S(H)$ to be a subgroup, we should show this:

$a,b\in S(H)\implies ab^{-1}\in S(H)$

A: I suppose you now that if $\pi : G_1 \rightarrow G_2$ is a group homomorphism and $H\subset G_2$ is a subgroup then $\pi^{-1}(H)$ is a subgroup of $G_1$
Define $\pi(x):=x^2$, since $G$ is abelian this is a group homomorphism. Now $\pi^{-1}(H)=S(H)$.
A: Here is a generalization.
Proposition Let $G$ be a group, $H$ a subgroup of $G$ and let $S(H)=\{x \in G : x^2 \in H\}$. If the commutator subgroup $G' \subseteq H$, then $S(H)$ is a (in fact normal) subgroup of $G$.
Proof Since $1 \in S(H)$ and $S(H)$ is closed under taking inverses, we have to show that if $x, y \in S(H)$, also $xy \in S(H)$. Note that $G' \subseteq H$ implies that $H$ is normal. But then $$(xy)^2=xyxy= xyyxx^{-1}y^{-1}xy=xy^2x[x,y]=xy^2x^{-1}\cdot x^2 \cdot[x,y] \in H,$$ since $x^2$ and $y^2$ are both elements of $H$.
A: Maybe some examples are helpful. 
Example 1: Consider $G =\mathbb{Z}$. Let $H = 2\mathbb{Z}$. Then $S(H)$ consists of all elements of $G$ with square in $H$. That is, you want all integers $n$ such that $2n$ (remember that $\mathbb{Z}$ is an additive group) is even. So $S(H) = G = \mathbb{Z}$ (so it is a group).
Example 2: Consider again $G = \mathbb{Z}$. Let $H = 3\mathbb{Z}$. Then $S(H)$ consists of all integers $n$ such that $2n$ is a multiple of $3$. Now $3$ is prime, so that means $n$ must be a multiple of $3$. So $S(H) = 3\mathbb{Z}$.
Exercise 1: Consider again $G = \mathbb{Z}$. Let $H = 4\mathbb{Z}$. What is $S(H)$?
Exercise 2: Consdier again $G = \mathbb{Z}$. Let $H = \{0\}$. What is $S(H)$?
Example 3: Let $G = \mathbb{Z} / 8\mathbb{Z}$ (additive finite group). Let $H = \{\bar{0}, \bar{2}, \bar{4}, \bar{6}\}$. Now $S(H)$ consists of all elements $\bar{n}$ such that $2\bar{n}$. Now $\bar{1} + \bar{1} = \bar{2}$, so $\bar{1}\in S(H)$. It is not hard to see that $S(H) = G$.
Example 4: Let $G = (\mathbb{Z} / 7\mathbb{Z})^\times$ (multiplicative group). Let $H = \{\bar{1}\}$. Now we are looking for all $\bar{n}$ such that $\bar{n}^2$ is $\bar{1}$. But $$
\bar{2}^2 = 4\not\in H\\
\bar{3}^2 = \bar{9} = \bar{2} \not\in H \\
\bar{4}^2 = \bar{16} = \bar{2}\not\in H \\
\bar{5}^2 = \bar{25} = \bar{4}\not\int H \\
\bar{6}^2 = \bar{36} = \bar{1}\in H\\
\vdots
$$ 
You get $S(H) = \{\bar{1},\bar{6}\}$.
Example 5: Let $G = \mathbb{Z}$. Let $H = 4\mathbb{Z}$. Then you are looking for the integers $n$ such that $2n$ is a multiple of $4$. That is $n$ is a multiple of $2$. So $S(H) = 2\mathbb{Z}$.
Example/Exercise: Let $G = \mathbb{R}^\times$. Let $H = \mathbb{Q}^\times$. Then you are looking for the real numbers $n$ such that $n^2$ is rational. You see, for example, that $\sqrt{2}^2 \in \mathbb{Q}^\times$ but that $(\sqrt[4]{2})^2 \not\in \mathbb{Q}^\times$. 
It seems that 
$$
S(H) = \{\frac{\sqrt{n}}{\sqrt{m}} : n,m\in \mathbb{Z}, m\neq 0\}.
$$
Is that right? (Think about it). Is this a group?
