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:

\(\ds \left({ab^{-1} }\right)^k\)

\(=\)

\(\ds a^k \left({b^{-1} }\right)^k\)

Power of Product in Abelian Group

\(\ds \)

\(=\)

\(\ds a^k \left({b^k}\right)^{-1}\)

Powers of Group Elements

\(\ds \)

\(=\)

\(\ds ee^{-1}\)

as $a^k = b^k = e \in B$

\(\ds \)

\(=\)

\(\ds e\)

Identity is Self-Inverse

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

$\blacksquare$