# Subgroup of Elements whose Order Divides Integer

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## 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:

\(\displaystyle \left({ab^{-1} }\right)^k\) | \(=\) | \(\displaystyle a^k \left({b^{-1} }\right)^k\) | Power of Product in Abelian Group | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^k \left({b^k}\right)^{-1}\) | Powers of Group Elements | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle ee^{-1}\) | as $a^k = b^k = e \in B$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) | Identity is Self-Inverse |

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

$\blacksquare$