# Abelian Quotient Group

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## Theorem

Let $G$ be a group.

Let $H$ be a normal subgroup of $G$.

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

Then $G / H$ is abelian if and only if $H$ contains every element of $G$ of the form $a b a^{-1} b^{-1}$ where $a, b \in G$.

## Proof

Let $G / H$ be abelian.

Then:

\(\, \displaystyle \forall a H, b H \in G / H: \, \) | \(\displaystyle a H b H\) | \(=\) | \(\displaystyle b H a H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a b H\) | \(=\) | \(\displaystyle b a H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a b \paren {b a}^{-1}\) | \(\in\) | \(\displaystyle H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a b a^{-1} b^{-1}\) | \(\in\) | \(\displaystyle H\) |

The argument reverses.

Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in H$.

\(\, \displaystyle \forall a, b \in G: \, \) | \(\displaystyle a b a^{-1} b^{-1}\) | \(\in\) | \(\displaystyle H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a b \paren {b a}^{-1}\) | \(\in\) | \(\displaystyle H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a b H\) | \(=\) | \(\displaystyle b a H\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a H b H\) | \(=\) | \(\displaystyle b H a H\) |

$\blacksquare$

## Sources

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