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 iff $H$ 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:

	\(\displaystyle \forall aH, bH \in G / H:\)	\(\displaystyle a H b H\)	\(=\)	\(\displaystyle b H a H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a b H\)	\(=\)	\(\displaystyle b a H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a b \left({b a}\right)^{-1}\)	\(\in\)	\(\displaystyle H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a b a^{-1} b^{-1}\)	\(\in\)	\(\displaystyle H\)	\(\displaystyle \)

• The argument reverses.

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

	\(\displaystyle \forall a, b \in G:\)	\(\displaystyle a b a^{-1} b^{-1}\)	\(\in\)	\(\displaystyle H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a b \left({b a}\right)^{-1}\)	\(\in\)	\(\displaystyle H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a b H\)	\(=\)	\(\displaystyle b a H\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a H b H\)	\(=\)	\(\displaystyle b H a H\)	\(\displaystyle \)

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 47 \zeta$