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
From Identity is Only Group Element of Order 1:
 * $\order e = 1$

and so from One Divides all Integers:
 * $\order e \divides n$

Thus $G_n \ne \O$.

Then:

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

Then from Product of Orders of Abelian Group Elements Divides LCM of Order of Product:
 * $\order {a b} = \divides \lcm \set {a, b}$

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.