A set $M$ on which a composition law $*$ cannot be defined so that $(M,*)$ it is a cyclic group is $\mathbb{R,N,Z,Q}$?
I received this question from a colleague and I think it is $\mathbb{Q}$ but I am not 100% sure I do not want to tell him something wrong.
If a group $G$ is cyclic, i.e. $G=\langle g\rangle$ for some $g\in G$, then $G$ is either finite or countable. That's because there's a group epimorphism $\mathbb{Z}\to G$, $n\mapsto g^n$. This automatically tells us that $\mathbb{R}$ cannot be a cyclic group.
On the other hand if $X$ is a countable set, meaning infinite countable or finite, then indeed it can be turned into a cyclic group. First of all, if $G$ is any group, $X$ a set and $f:G\to X$ a bijection, then $X$ can be turned into a group such that $f$ becomes an isomorphism. This group structure on $X$ is given by $x\cdot y:=f(f^{-1}(x)f^{-1}(y))$. With that you only need to realize that for a countable set $X$ there exists a bijection $\mathbb{Z}/n\mathbb{Z}\to X$ for some $n\in\mathbb{Z}$, turning $X$ into a cyclic group. In particular $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ can be turned into cyclic groups.
