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Can someone give me an example of a non-Abelian group $G$ with a normal subgroup $H$ such that $G/H$ is Abelian?

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Pick any abelian group $A$ and any nonabelian group $N$. Then the direct product $G=A\times N$ provides an example with $H=N$. – Mariano Suárez-Alvarez May 28 '12 at 2:48

Smallest example: take $G=S_3$, $H=\{1,(123),(132)\}$. Then $G/H$ has order $2$, so it is abelian.

(Of course, you can always just take $H=G$; then $G/H$ is trivial, hence abelian; more generally, if $G$ is a group, then there is a smallest normal subgroup $N$ such that $G/N$ is abelian; it's called the "commutator subgroup of $G$", denoted $[G,G]$ or $G'$, and is the subgroup generated by all elements of the form $[x,y] = x^{-1}y^{-1}xy$; $G$ is abelian if and only if $[G,G]=\{1\}$. It is possible that $[G,G]=G$, of course, for example, with $G=A_5$, but since you did not specify that you wanted $H$ to be nontrivial, this subgroup always works and is the smallest one that does)

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