Example of a specific type of infinite group

Question

Give example (if exists) of an infinite non-cyclic group such that every non-trivial subgroup of it has finite index ; off-course the group must be non-abelian ; motivated form $G/H$ is a finite group so $G\cong\mathbb Z$

Answer 1

I think there is no such group. First, $G$ is a torsion-free and virtually cyclic. For, if $g$ is a non-trivial element of $G$, then $\langle g\rangle$ has finite index in $G$, so $G$ is virtually cyclic. And, since $G$ is infinite, the element $g$ must have infinite order. As $g$ was arbitrary, $G$ is torsion-free. Also, note that $G$ is finitely generated. (See below.)

Now, if $G = \langle x_1,\ldots, x_n\rangle$, then $C_G(x_i)$ is non-trivial, so it has finite index in $G$. Therefore, $Z(G) = \bigcap_{i=1}^n C_G(x_i)$ has finite index in $G$, so the derived subgroup $[G,G]$ of $G$ is finite, by Schur's Theorem. Since $G$ is torsion-free, it follows that $G$ is abelian. Now $G$ must be infinite cyclic.

ADDED

Here is the requested proof that $G$ is finitely generated. We have that $\langle g\rangle$ is a subgroup of finite index in $G$, where $g$ is any non-trivial element of $G$. Therefore, we can write $G$ as the disjoint union of cosets: $$G = y_1\langle g\rangle\cup y_2\langle g\rangle\cup\cdots\cup y_m\langle g\rangle,$$ for suitable elements $y_1, y_2,\ldots, y_m\in G$, where $m$ is the index of $\langle g\rangle$ in $G$. (We could take, for example, $y_1 = 1\in\langle g\rangle$, but it's not important here.) This means that an arbitrary element $u$ in $G$ belongs to (exactly) one of those cosets, say, $u\in y_i\langle g\rangle$. Therefore, $u = y_ig^s$, for some integer $s$. This shows that every element of $G$ can be written as a product of powers of elements of the set $\{ y_1, y_2,\ldots, y_m, g\}$, so $G$ is generated by this finite set.

(Exercise. Show that a group is finitely generated if it has a finitely generated subgroup of finite index.)