Can $G/H$ be cyclic if $G$ is nonabelian?

Question

I've learned in class that if $G/Z(G)$ is cyclic then $G$ is abelian.

I was wondering if $H\lhd G$, where $G$ is nonabelian if $G/H$ is cyclic, when $H$ is not $G$ itself. If so, could someone give me an example?

Answer 1

How about $G=S_3$ (the smallest non-abelian group) and $H=\langle(123)\rangle$?

$G/H$ has order $2,$ so it must be cyclic.

Answer 2

The center is very special in this regard. And to find examples demonstrating this, we can look to the smallest non-abelian groups we have, like $S_3$ and $D_4$. Those both have proper, non-trivial normal groups, and the resulting quotients are cyclic, as they have order $2$.

Answer 3

Take for $G$ a group of permutations on $n$ elements and for $H$ the subgroup of even permutations. Then $G/H$ is cyclic of order $2$. Now, if $n \geqslant 3$, $G$ is non-abelian (and, as Galois would have told you, not even solvable for $n \geqslant 5$).

Answer 4

Sure. For instance, consider $S_n$ and $A_n$. The quotient is $C_2$. And for $n\ge3$, $S_n$ is not abelian.

Answer 5

The first answer that occurred to me was $S_3$, the smallest nonabelian group, but that's already included in all four of the answers already given. So here's a different example: Take any non-abelian group $H$ and any nontrivial cyclic group $C$; let $G$ be the direct product $H\times C$.