TITLE: Finite group with three proper subgroups
QUESTION [5 upvotes]: 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 $G$?

REPLY [3 votes]: 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.