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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  • 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.

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