Abelian Quotient Group

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: $\S 47 \zeta$