Quotient Group is Abelian iff All Commutators in Divisor/Proof 2

Theorem

Let $G$ be a group.

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

Let $G / N$ be the quotient group of $G$ by $N$.

Then the quotient group $G / N$ is abelian if and only if:

$\forall x, y \in G: \sqbrk {x, y} \in N$

where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.

That is, if and only if $x y x^{-1} y^{-1} \in N$ for all $x, y \in G$.

Proof

Let $G / N$ be abelian.

Then:

\(\ds \forall a N, b N \in G / N: \, \)

\(\ds a N b N\)

\(=\)

\(\ds b N a N\)

\(\ds \leadsto \ \ \)

\(\ds a b N\)

\(=\)

\(\ds b a N\)

\(\ds \leadsto \ \ \)

\(\ds a b \paren {b a}^{-1}\)

\(\in\)

\(\ds N\)

\(\ds \leadsto \ \ \)

\(\ds a b a^{-1} b^{-1}\)

\(\in\)

\(\ds N\)

The argument reverses.

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

\(\ds \forall a, b \in G: \, \)

\(\ds a b a^{-1} b^{-1}\)

\(\in\)

\(\ds N\)

\(\ds \leadsto \ \ \)

\(\ds a b \paren {b a}^{-1}\)

\(\in\)

\(\ds N\)

\(\ds \leadsto \ \ \)

\(\ds a b N\)

\(=\)

\(\ds b a N\)

\(\ds \leadsto \ \ \)

\(\ds a N b N\)

\(=\)

\(\ds b N a N\)

$\blacksquare$