Do there exist groups $X,Y$ such that $X/Y,Y$ are commutative/cyclic but $X$ is not?

Question

Let $X$ be a group and $Y$ a normal subgroup of $X$. I am interested in the following two statements:

1. If $Y$ and $X/Y$ are commutative, then so is $X$.

and

2. If $Y$ and $X/Y$ are cyclic, then so is $X$.

For 1, I am able to conclude $b^{-1}a^{-1}ba \in Y$ for all $a,b \in X$. Similarly $a^{-1}b^{-1}ab \in Y$. I think the quotient of the dihedral group $D_6$ divided by the group of rotations provide a counterexample to 1.

For 2, I am not aware of any potential counterexamples. If 2 is false, then it immediately follows that 1 is false.

The converse to both of these statements are true.

Answer 1

A good way to approach such questions is to try this: Let $X$ be the smallest non-Abelian (respectively, non-cyclic) group. If it has a proper normal subgroup, then that must be a solution to your question, because both $Y$ and $X/Y$ will be smaller than $X$, and nothing smaller than $X$ is non-Abelian (respectively, non-cyclic).

Your example, of $D_6$, modding out the group of rotations, works as a counterexample to both, even though it is not the smallest non-cyclic group there is.

(By $D_6$, I’m assuming you mean the dihedral group of order $6$, or the symmetries of an equilateral triangle, sometimes called $D_3$. If you mean the symmetries of a regular hexagon, which is a group of order $12$, sometimes called $D_{12}$, then it still works as a counterexample to both claims.)

Answer 2

Your suggestion for the first problem (the dihedral group $D_6$) is correct. And it works for the other problem too. The subgroup $C_3$ of $D_6$ whose elements are the rotations form a normal subgroup and $D_6/C_3\simeq\mathbb{Z}_2$, which is cyclic too. Besides, $C_3$ is cyclic. However, $D_6$ is not cyclic.