## The commutator subgroup is normal

Let $G$ be a group. Prove that $N = \langle [x,y] \ |\ x,y \in G \rangle$ is a normal subgroup of $G$ and that $G/N$ is abelian. (Recall that $[x,y] = x^{-1}y^{-1}xy$.) This $N$ is called the commutator subgroup of $G$.

Lemma: Let $G$ be a group and $x,y,g \in G$. Then $g[x,y]g^{-1} = [gxg^{-1},gyg^{-1}]$. Proof: We have $g[x,y]g^{-1} = gx^{-1}y^{-1}xyg^{-1} = gx^{-1}g^{-1}gy^{-1}g^{-1}gxg^{-1}gyg^{-1}$ $= (gxg^{-1})^{-1}(gyg^{-1})^{-1}(gxg^{-1})(gyg^{-1}) = [gxg^{-1}, gyg^{-1}]$. $\square$

Now if $g \in G$, then $gNg^{-1} = g \langle [x,y]\ |\ x,y \in G \rangle g^{-1}$ $= \langle g[x,y]g^{-1} \ |\ x,y \in G \rangle$ $\langle [gxg^{-1}, gyg^{-1} \ |\ x,y \in G \rangle$ $= \langle [x,y] \ |\ x,y \in G \rangle$ $= N$. So $N$ is normal in $G$.

Moreover, $G/N$ is abelian by this previous exercise.

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### Comments

• Gobi Ree  On November 22, 2011 at 2:36 am

A simplication: Since $g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}] \in N$ for all $g \in G$, we have that $g \langle [x,y] | x,y \in G \rangle g^{-1}=gNg^{-1} \subseteq N$ for all $g \in G$.

• nbloomf  On November 22, 2011 at 10:18 am

This is a vast improvement. Thanks!

• Gobi Ree  On November 22, 2011 at 2:50 am

Also, $G/N$ is abelian by the previous problem.