# 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

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41 \epsilon$