If $G$ has only 2 non-trivial proper subgroups H, N

, then H, N are cyclic subgroup of $G$.

I searched essentially same problem at

If $G$ has only 2 proper, non-trivial subgroups then $G$ is cyclic

So, I already know the solving method of my question.

  1. $G$ is cyclic.
  2. subgroup of cyclic subgroup is cyclic.
  3. Therefore H, N are cyclic.

$$ $$ In this time,

I'd like to know other method solving my question.


Can anybody show my question, not through "$G$ is cyclic."

$$ $$

Umm, maybe this is useless question.... :-(

But I spent lots of time for solving this question.

Thank you for your attention to this matter.


Any group with at most one non-trivial proper subgroup is cyclic: any element not in the non-trivial proper subgroup (or not the identity if there is none) must generate the whole group. Neither of the subgroups $H,N$ of the question can have more than one non-trivial proper subgroup (as with the subgroup itself that would give more than two non-trivial proper subgroups of $G$), so by the above they must be cyclic.

  • $\begingroup$ Thanks a lot. That's exactly what I need. $\endgroup$ – user143993 Oct 23 '14 at 14:45

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