Let $H \leq G$: prove that $K=\{x\in G:\exists n \in\Bbb{N},\,x^n\in H\}$ is a subgroup of $G$? $H$ is a subgroup of an abelian group $G$. We define
$$K = \{ x \in G : \exists n \in \Bbb{N}, x^n \in H\}.$$
How to prove that $K$ is a subgroup of $G$ ?
 A: To show $K$ is a subgroup of $G$, we need to show $(i)$ $K$ is nonempty (which we will do my showing $e$, the identity in $G, H \in K$; $(ii)$ that  $K$ is closed under taking inverses; and $(iii)$ that $K$ is closed under the group operation. 
I'll work through $(i), \;(ii)\;\;$ and help set up what you need to show $(iii)$.

(i) Identity: 


*

*We know that since $H$ is a subgroup of $G$, it contains $e$, the identity element of $G$. And $e\in G, e^n = e \in H$ for $n \in \mathbb{Z}$


(ii) Closure under Inverses: 


*

*Now we need to show that for every element in $x \in K$, $x^{-1} \in
   K$.

*Use the fact that if $x \in K,$ then $x^n \in H$ for some $n \in \mathbb{N}$. Let $x \in K,$ so x^{n} \in H$. Now  $x^n \in H \implies (x^{n})^{-1} \in H$, since $H$ is a subgroup, and therefore is closed under inverses. But $(x^n)^{-1} = x^{-n} = (x^{-1})^n \in H \implies x^{-1} \in K$.


(iii) Closure under the group operation: 


*

*Then we need to show for any $x, y \in K, (x\circ y)^k \in K,$ for
some $k\in \mathbb{N}$.

*Use the facts that for $x, y \in K$, $x^n, y^m \in H$ for some $ n, m \in \mathbb{N}$. So we need to show that $(x \circ y) \in K$ by showing $(x\circ y)^k \in H$, for some $k \in \mathbb N$. Note that since $H$ is a subgroup, it is closed under the group operation, and since $x^n, y^m \in H$, so is $x^n\circ y^m \in H$. 

*What can you say about $x^n\circ y^m \in H,$ knowing that $G$ is abelian? How can we use these facts to show $x \circ y \in K$?

A: First do the case $H=0$, then $K$ is known as the torsion subgroup of $G$. In general, $K$ is the torsion subgroup of $G/H$ pulled back along the projection $G \twoheadrightarrow G/H$.
A: It is the set of all elements $x\in G$ such that
$$
\overline{x}=xH
$$
has finite order in $G/H$.
In other words, if we denote
$$
\pi:G\longrightarrow G/H
$$
the canonical surjection, and if we set $L$ to be the set of all finite order elements in $G/H$, then
$$
K=\pi^{-1}(L).
$$
So it suffices to check that $\pi$ is a group homomorphism and that $L$ is a subgroup of $G/H$. Both are easy.
This is far from being the fastest way, but that's how I see it.
A: HINT: Just check the requirements for being a subgroup: show that $K$ is non-empty, closed under the group operation, and closed under taking inverses.


*

*Closure under the operation: If $x,y\in K$, then there are $m,n\in\Bbb N$ such that $x^m,y^n\in H$. Can you find from $m$ and $n$ a positive integer $k$ such that $(xy)^k\in H$? Use the fact that $G$ is Abelian.

*Closure under taking inverses: If $x\in K$, then there is an $n\in\Bbb N$ such that $x^n\in H$. Can you find a $k\in\Bbb N$ such that $(x^{-1})^k\in H$?
