Let $G$ be a group that every element have infinite order. If $G$ has a cyclic subgroup with finite index then $G$ is cyclic.

First I show that $G$ is finitely generated. If $\langle x\rangle$ be the cyclic subgroup that is mentioned above then assuming that $|G:\langle x\rangle|=n$, $G$ has elements like $g_2,\ldots,g_n$ such that $G=\langle x,g_2,\ldots,g_n\rangle$.

Now I want to show that $G=\langle x\rangle$ but I don't know how to continue.

  • 2
    $\begingroup$ (1) As a nit-picky note, the problem statement probably says "every element except for the identity". (2) It's not necessarily true that $G=\langle x\rangle$ (following your notation). So you can't prove your last claim because it's not true. I think you have a good start, but a different approach is needed to continue. $\endgroup$ – zipirovich May 25 '18 at 5:09
  • $\begingroup$ @zipirovich Thank you. $\endgroup$ – Yasmin May 25 '18 at 5:12
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    $\begingroup$ You could prove that $|G:Z(G)|$ is finite. There is then a theorem of Schur you could use that $|G:Z(G)|$ finite implies $[G,G]$ finite, and hence $[G,G]=1$ and $G$ is abelian. I don't whether you can avoid using Schur's theorem. $\endgroup$ – Derek Holt May 25 '18 at 8:32
  • $\begingroup$ @DerekHolt Thanks, but how should I show $|G:Z(G)|$ is finite. I tried before but I couldnt. $\endgroup$ – Yasmin May 25 '18 at 8:36
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    $\begingroup$ Each $g_i$ has infinite order, so some power of $g_i$ is equal to $x^{n_i}$ for some $n_i \ne 0$ and hence $g_i$ centralizes $x^{n_i}$. The intersection of the subgroups $\langle x^{n_i} \rangle$ has finite index in $G$. $\endgroup$ – Derek Holt May 25 '18 at 9:00

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