Subgroup of Elements whose Order Divides Integer

Theorem
Let $A$ be an abelian group.

Let $k \in \Z$ and $B$ be a set of the form:
 * $\left\{{x \in A : x^k = e}\right\}$

Then $B$ is a subgroup of $A$.

Proof
First note that the identity $e$ satisfies $e^k = e$ and so $B$ is non-empty.

Now assume that $a, b \in B$.

Then:

Hence, by the One-Step Subgroup Test, $B$ is a subgroup of $A$.