Showing a set is a subgroup of abelian group Let $H$ be a subgroup of abelian group $G$, and let $K = \{x \in G: \text { for some integer } n > 0, x^n \in H\}$. Prove that $K$ is a subgroup of $G$.

Let $x, y \in K.$ Then by definition of $K$, $x^n, y^n \in H.$ Since $H$ is closed under multiplication, $x^ny^n \in H.$ Since $G$ is abelian, $x^ny^n =(xy)^n.$ Thus, $xy \in K.$
Let $x \in K.$ By definition of $K$, $x^n \in H.$ Since $H$ is closed under inverses, $(x^n)^{-1} \in H.$  Since $(x^n)^{-1} = (x^{-1})^n, x^{-1} \in K$ by definition of $K$.
Since $e \in H$ and $e^n = e, e^n \in H.$ So, $e \in K$ by definition of $K$.
Please, check my work.
edit: if this question looks like a duplicate of my previous question, is it possible to delete it?
 A: The main issue with your argument is that you assume the same exponent $n$ works for all your elements. 
This is how you can argue. Since $e = e^1 \in H$, $K$ is nonempty (and contains $e$). Given $x,y\in K$, $x^n \in H$ and $y^m \in H$ for some integers $m,n > 0$ (not necessarily the same, mind you). Let $k$ be the least common multiple of $m$ and $n$. The $m$ and $n$ divide $k$. So since $H$ is closed under multiplication, $x^n \in H$ implies $x^k = (x^n)^{k/n} \in H$; $y^m \in H$ implies $y^k = (y^m)^{k/m} \in H$. Again, by closure under multiplication in $H$,
$$(xy)^k = x^k y^k \in H,$$
showing that $xy\in K$. So $K$ is closed under multiplication. To see that $K$ is closed under inverses, let $x\in K$, and let $n$ be a positive integer such that $x^n \in H$. Then $(x^{-1})^n = (x^n)^{-1} \in H$ as $x^n \in H$ and $H$ is closed under inverses. Therefore $x^{-1}\in K$. Consequently, $K$ is a subgroup of $G$.
A: $K$ is the pullback of the torsion subgroup of $\frac{G}{H}$. Pullbacks of subgroups are subgroups.
