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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 $G\cong\mathbb Z$

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Souvik Dey
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  • duplicate: http://math.stackexchange.com/questions/513700/infinite-group-must-have-infinite-subgroups – Denis Oct 08 '14 at 13:33
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    @Denis: How is it a duplicate ?? then provide me with the group – Souvik Dey Oct 08 '14 at 13:36
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    It doesn't @Denis: that answer's examples are of infinite groups with subgroups of obvious infinite index. the OP here wants example(s) of infinite non-cyclic groups all of which proper non-trivial subgroups have finite index. – Timbuc Oct 08 '14 at 13:56
  • @Denis: Sorry , but there cannot be any abelian examples because if it is abelian , then it is cyclic , but I want non-cyclic – Souvik Dey Oct 08 '14 at 13:56
  • @SouvikDey. I didn't get your last remark: how do you know that if an infinite abelian group has all its proper non-trivial subgroups of finite index then it must be cyclic? Writing "...because if it is abelian then it is cyclic" sounds pretty dangerous. – Timbuc Oct 08 '14 at 14:02
  • I don't get why the answer does not fit your requirements. For instance dyadics in $\mathbb Q/{\mathbb Z}$. – Denis Oct 08 '14 at 15:36
  • @ Timbuc : The very question I linked shows that "if an infinite abelian group has all its proper non-trivial subgroups of finite index then it must be cyclic" – Souvik Dey Oct 09 '14 at 12:49

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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.)

James
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