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Let $G$ be a group and $C$ the normal closure of the subgroup of $G$ generated by elements of the form $$aba^{-1} b^{-1}$$ for $a,b \in G$.

Show that $G/C$ is Abelian. $C$ is called the commutator subgroup of $G$.

Thanks in advance!

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If $x, y \in G$, then $xyx^{-1}y^{-1} \in C$. Hence, $$Cxyx^{-1}y^{-1} = C,$$ or $$Cxy = Cyx,$$ which means that $G/C$ is abelian.

Edit: corrected a typo pointed out by Loki Clock.

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    $\begingroup$ $xyC=xyCxyx^{-1}y^{-1}, Cxy=Cyx$ $\endgroup$
    – Loki Clock
    Mar 2 '13 at 6:36
  • $\begingroup$ @LokiClock: Thanks, Loki. I corrected the typo. $\endgroup$ Mar 2 '13 at 6:41
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Perhaps it is instructive to see how you get to construct such a subgroup.

Given a group $G$ and a normal subgroup $N$, you may inquire when is the quotient group $G/N$ abelian.

You see that this happens if and only if $$ aN \cdot bN = bN \cdot aN $$ for all $a, b \in G$, so that, multiplying on the right first by $(aN)^{-1} = a^{-1}N$ and then by $(bN)^{-1}$ $$ a b a^{-1} b ^{-1} N = N, $$ that is, if and only if $$ [a, b] = a b a^{-1} b ^{-1} \in N $$ for all $a, b \in G$.

Now it is natural to consider the group generated by all these commutators $[a,b]$: $$ G' = \{ a b a^{-1} b ^{-1} : a, b \in G \}. $$ First of all it is normal, because if you conjugate a commutator, you get another commutator $[a, b]^x = [a^x, b^x]$. Alternatively, if $b \in G'$, then $$ b^{a^{-1}} = a b a^{-1} = a b a^{-1} b ^{-1} b = [a, b] b \in G'. $$ And then, because the steps above are reversible, $G/G'$ is abelian.

We have thus seen that $G'$ is the smallest normal subgroup $N$ such that $G/N$ is abelian.

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The commutator subgroup is always normal in $G$ so you don't need to take the normal closure. Now the quotient is saying you are forcing the cosets of $aba^{-1}b^{-1}$ to be the identity coset. What does this say about $(aC) (bC)$ and $(bC) (aC)$?

See this link to see a proof for why $[G,G]$ is normal http://planetmath.org/DerivedSubgroupIsNormal.html

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