Order of Element in Quotient Group

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Theorem

Let $G$ be a group, and let $H$ be a normal subgroup of $G$.

Let $G / H$ be the quotient group of $G$ by $H$.


The order of $a H \in G / H$ divides the order of $a \in G$.


Proof

Let $G$ be a group with normal subgroup $H$.

Let $G / H$ be the quotient of $G$ by $H$.

From Quotient Group Epimorphism is Epimorphism, $G / H$ is a homomorphic image of $G$.

Let $q_H: G \to G / H$ given by $\map f a = a H$ be that quotient epimorphism.

Let $a \in G$ such that $a^n = e$ for some integer $n$.

Then, by the morphism property of $q_H$:

\(\ds \map {q_H} {a^n}\) \(=\) \(\ds \paren {\map {q_H} a}^n\)
\(\ds \) \(=\) \(\ds \paren {a H}^n\)
\(\ds \) \(=\) \(\ds a^n H\) Definition of Coset Product
\(\ds \) \(=\) \(\ds e H\) by hypothesis
\(\ds \) \(=\) \(\ds H\)

Hence $\order H$ divides $n$.

$\blacksquare$


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