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


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

$\blacksquare$


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