# 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 $q_H$:

\(\displaystyle \map {q_H} {a^n}\) | \(=\) | \(\displaystyle \paren {\map {q_H} a}^n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a H}^n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^n H\) | Definition of Coset Product | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e H\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle H\) |

Hence $\order H$ divides $n$.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \delta$