Commutator subgroup $G'$ is the unique smallest normal subgroup $N$ such that $G/N$ is abelian.

If $G$ is a group, then $G'$ is a normal subgroup of $G$ and $G/G'$ is abelian. If $N$ is a normal subgroup of $G$ , then $G/N$ is abelian iff $N$ contains $G'$ .

Proof

Let $f:G\to G$ be any automorphism. Then $\displaystyle f(aba^{-1}b^{-1})=f(a)f(b)f(a)^{-1}f(b)^{-1}\in G'.$

It follows that $f(G')\leq G'$ . In particular, if $f$ is the automorphism given by conjugation by $a\in G$ , then $aG'a^{-1}=f(G')\leq G'$ , so $G'\unlhd G$ .

Since $(ab)(ba)^{-1}=aba^{-1}b^{-1}\in G'$ , $abG'=baG'$ and hence $G/G'$ is abelian.

( $\implies$ ) If $G/N$ is abelian, then $abN=baN$ for all $a,b\in G$ . Hence $ab(ba)^{-1}=aba^{-1}b^{-1}\in N$ . Therefore, $N$ contains all commutators and $G'\leq N$ .

( $\impliedby$ ) If $G'\subseteq N$ , then $ab(ba)^{-1}=aba^{-1}b^{-1}\in G'\subseteq N$ . Thus $abN=baN$ for all $a,b\in G$ . Hence $G/N$ is abelian.