Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

Question

Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

I don't understand the quotient group $G/H$ being cyclic. What does it mean? From what I understand, $G/H$ is a group that includes groups and not elements, right?

I know that the reverse is right (If $G$ is cyclic then $G/N$ is cyclic.)

How do I disprove it?

Answer 1

This is a fun one!

To address the question in the title, note that:

• $D_4$ is not cyclic,
• the group $C_4 = \{\rho_0, \rho_{90}, \rho_{180}, \rho_{270} \}$ (the rotation subgroup) is cyclic, and a normal subgroup of $D_4$
• $D_4/C_4$ is of order $2$ and is thus cyclic.

Answer 2

Of the small finite non-cyclic groups $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ of order $8$ is the smallest which doesn't have a cyclic normal subgroup with a cyclic quotient.

Note that a subgroup of index $2$ is always normal. Do you know any groups which have a cyclic subgroup of index $2$? If they are not themselves cyclic, they will be counterexamples.

You should be able to find both abelian and non-abelian examples of small order, and also some general cases. Also try to find a small example of odd order (the smallest non-abelian example of odd order has order $21$).

This is a test and suggestion to develop a repertoire of examples of groups with/without interesting properties. Working hands-on will teach you a great deal about what is possible and what isn't.

Answer 3

The smallest counterexample is Klein's $4$-group, $G=\{e,a,b,ab\}$. Take e.g. $H=\{e,a\}$; then, $H\unlhd G$, $H\cong C_2$ and $G/H\cong C_2$ (it's worthwile to explicitly write the two elements of the quotient, and squaring the nonidentity one), but $G$ is not cyclic.