The subset of elements of order dividing $k$ in an abelian group is a subgroup

Question

Suppose $G$ is an abelian group and $k$ is a natural number.

Prove $H = \{ g \in G : g^k = 1 \}$ is a subgroup of G.

I know I need to show that $1_G \in H$, existence of inverse element in group, and closure, but how?

Answer 1

$1 \in H$ since $1^k = 1$.

Suppose $g \in H$. That is, $g^k = 1$. Then $(g^{-1})^k = (g^{k})^{-1} = 1$. So $g^{-1} \in H$.

If $x,y \in H$, this means $x^k = 1$ and $y^k = 1$. Then $(xy)^k = x^ky^k = 1$ since $G$ is an abelian group. So $xy \in H$.

$H$ is a subgroup.

Answer 2

It's probably overkill but $x \mapsto x^k$ is a homomorphism $G\to G$ when $G$ is abelian. The set $H$ is the kernel of that homomorphism and hence is a subgroup.