\section{Elements of Abelian Group whose Order Divides n is Subgroup}
Tags: Order of Group Elements, Subgroups, Abelian Groups

\begin{theorem}
Let $G$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] .
Let $G_n$ be the [[Definition:Subset|subset]] of $G$ defined as:
:$G_n = \set {x \in G: \order x \divides n}$
where:
:$\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$
:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then $G_n$ is a [[Definition:Subgroup|subgroup]] of $G$.
\end{theorem}

\begin{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:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = G_n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order x
      | o = \divides
      | r = n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {x^{-1} }
      | o = \divides
      | r = n
      | c = [[Order of Group Element equals Order of Inverse]]
}}
{{eqn | ll= \leadsto
      | l = x^{-1}
      | o = \in
      | r = G_n
      | c = 
}}
{{end-eqn}}
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 [[Definition:Lowest Common Multiple of Integers|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]].
{{qed}}
\end{proof}


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