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**Statement:**

Let $G$ be a group. The set of all commutators $aba^{-1}b^{-1}$ for $a,b \in G$ generates a subgroup $C$ (the commutator subgroup) of $G$. This subgroup $C$ is a normal subgroup of $G$. Furthermore, if $N$ is a normal subgroup of $G$, then $G/N$ is abelian if and only if $C \leq N$.

**Proof:**

The commutators certainly generate a subgroup $C$; we must show that it is normal in $G$. Note that the inverse $(aba^{-1}b^{-1})^{-1}$ of a commutator is again a commutator, namely, $bab^{-1}a^{-1}$. Also $e=eee^{-1}e^{-1}$ is a commutator. Theorem 7.6 then shows that $C$ consists precisely of all finite products of commutators. For $x \in C$, we must show that $g^{-1}xg \in C$ for all $g \in G$, or that if $x$ is a product of commutators, so $g^{-1}xg$ for all $g \in G$. By inserting $e=gg^{-1}$ between each product of commutators occurring in $x$, we see that it is sufficient to show for each commutator $cdc^{-1}d^{-1}$ that $g^{-1}(cdc^{-1}d^{-1})g$ is in $C$. But

(1)which is in $C$. Thus $C$ is normal in $G$.

The rest of the theorem is obvious if we have acquired the proper feeling for the factor groups. One doesn't visualize in this way, but writing out that $G/C$ is abelian follows from

(2)Furthermore if $N$ is a normal subgroup of $G$ and $G/N$ is abelian, then $(a^{-1}N)(b^{-1}N)=(b^{-1}N)(a^{-1}N)$; that is, $aba^{-1}b^{-1}N=N$, so $aba^{-1}b^{-1} \in N$, and $C \leq N$. Finally, if $C \leq N$, then

(3)