Can someone give me an example of a non-Abelian group $G$ with a normal subgroup $H$ such that $G/H$ is Abelian?
[Math] Example of nonAbelian group with a normal subgroup such that the quotient group is Abelian
abstract-algebra
abstract-algebra
Can someone give me an example of a non-Abelian group $G$ with a normal subgroup $H$ such that $G/H$ is Abelian?
Best Answer
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)