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If you have a non-abelian group $G$ with some normal subgroup $K$, is it possible to have a non-abelian quotient group $G/K$?

Besides actually sitting down and trying to generate quotient groups through exhaustion, I have been thinking about using the fundamental theory of homomorphisms to pick a small non-abelian group like $D_6$ and find the quotient group it is isomorphic to. Does this seem like a good tactic?

I'm not looking for answers, just confirmation that this is a useful way to be thinking about it.

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    $\begingroup$ Why don't you simply take $K = \{e\}$? $\endgroup$ – Najib Idrissi Mar 17 '14 at 19:37
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Yes, it is a good tactic.

Think of a product

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Of $A$, a non-abelian group, and $B$, an abelian group

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$G=A\times B$ and $K=\{e_A\}\times B$

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  • $\begingroup$ great! thank you. i took $G = D_6$x$C_2$ and then $K = {e_A}$x$C_2$, so then $G/K$ has ${(F, e_B)}$ in it, which is not going to commute with all the other elements in the quotient. $\endgroup$ – epsilonics Mar 17 '14 at 19:29
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In general, there is the following fact:

If $N$ is a normal subgroup of $G$, then $G/N$ is abelian if and only if $[G,G]$ (the commutator subgroup) is a subgroup of $N$.

So the quotient will not be commutative precisely when your normal subgroup does not contain the commutators.

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