Elements of Abelian Group whose Order Divides n is Subgroup

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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:

\(\displaystyle x\) \(\in\) \(\displaystyle G_n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \order x\) \(\divides\) \(\displaystyle n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \order {x^{-1} }\) \(\divides\) \(\displaystyle n\) 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 $\order a = r, \order b = s$.

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.

$\blacksquare$


Sources