Finite group with three proper subgroups The Klein-$4$ group is a finite group with exactly three subgroups $H$ such that $1<H<G$. 
Conversely, if $G$ is a finite group with exactly three subgroups $H$ such that $1<H<G$, then what can be said about $G$? 

It seems to me that $G$ must be either the Klein four, or a cyclic group of order $p^4$ for some prime $p$.


*

*A cyclic group of order $\prod_{i=1}^kp_i^{a_i}$ has $\prod_i(a_i+1)$
subgroups (including the trivial ones), so $k=1, a_1=4$ is the only
possibility. 

*A non-cyclic abelian group $G$ has a subgroup $H$ isomorphic to $C_p\times
   C_p$. That group has $p+1$ proper subgroups, so we must have $p=2$ and $H=G$. This leaves the Klein four as the only alternative. 

*A non-abelian $p$-group $G$ has $G/[G,G]$ as a non-cyclic abelian head, so the argument of the previous bullet works.

*A finite non-$p$-group with a non-normal Sylow subsgroup has at least three Sylow subgroups of that order, so that is ruled out, because it has at least one more Sylow subgroup.

*That leaves the case of a non-abelian group $G$ such that all its Sylow subgroups are normal. Then $G$ thus a direct product of its Sylow subgroups, and one of those has order $p^3$ at least, because otherwise $G$ is abelian. We are done.

