Elements of Abelian Group whose Order Divides n is Subgroup
Jump to navigation
Jump to search
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:
\(\ds x\) | \(\in\) | \(\ds G_n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \order x\) | \(\divides\) | \(\ds n\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \order {x^{-1} }\) | \(\divides\) | \(\ds n\) | Order of Group Element equals Order of Inverse | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^{-1}\) | \(\in\) | \(\ds G_n\) |
Let $a, b \in G_n$ such that $\order a = r, \order b = s$.
Then from Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:
- $\order {a b} = \divides \lcm \set {a, b}$
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: $\S 41 \epsilon$