# Elements of Abelian Group whose Order Divides n is Subgroup

## Theorem

Let $G$ be an abelian group whose identity element is $e$.

Let $n \in \Z_{>0}$ be a (strictly) positive integer .

Let $G_n$ be the subset of $G$ defined as:

$G_n = \set {x \in G: \order x \divides n}$

where:

$\order x$ denotes the order of $x$
$\divides$ denotes divisibility.

Then $G_n$ is a subgroup of $G$.

## Proof

$\order e = 1$

and so from One Divides all Integers:

$\order e \divides n$

Thus $G_n \ne \O$.

Then:

 $\displaystyle x$ $\in$ $\displaystyle G_n$ $\displaystyle \leadsto \ \$ $\displaystyle \order x$ $\divides$ $\displaystyle n$ $\displaystyle \leadsto \ \$ $\displaystyle \order {x^{-1} }$ $\divides$ $\displaystyle n$ Order of Group Element equals Order of Inverse $\displaystyle \leadsto \ \$ $\displaystyle x^{-1}$ $\in$ $\displaystyle G_n$

Let $a, b \in G_n$ such that $\order a = r, \order b = s$.

$\order {a b} = \divides \lcm \set {a, b}$

But $r \divides n$ and $s \divides n$ by definition of $G_n$.

Therefore, by definition of lowest common multiple:

$\order {a b} \divides n$

Thus we have:

$G_n \ne \O$
$x \in G_n \implies x^{-1} \in G_n$
$a, b \in G_n \implies a b \in G_n$

and the result follows by the Two-Step Subgroup Test.

$\blacksquare$