Does there exist a group of infinite order with the property of containing precisely one nontrivial (proper) subgroup. Is it possible to disprove the statement?
My intuition tells me that no such group exists becausr we can thin out the group in multiple ways, however I'm not sure how to show it.
Suppose there exist a group $G$ which satisfies those conditions., and let $G'$ be its only nontrivial proper subgroup. Since this subgroup is proper, there is a non-identity element $g$ which is $G$ but isn't in $G'$. Consider the subgroup $\langle g \rangle$. This subgroup cannot be equal to $G'$, as $g \notin G'$, and as it is a nontrivial subgroup it has to equal the whole $G$.
However, $G= \langle g \rangle$ is easily seen to have many subgroups. As it is infinite then $g^n \neq g^k$ for any distinct $n$ and $k$. For example $\langle g^p \rangle$ is a distinct subgroup for any prime $p$ (consisting of all elements of the form $g^{np}$ where $n \in \mathbb{Z}$). This is a contradiction with $G'$ being the only nontrivial proper subgroup.