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I wonder whether it is possible that the only proper subgroups of a finite group are p-groups (of course, except the case when the whole group is a p-group).

I know that I can't find it in solvable groups.

After investigating some small simple groups, I think it is possible to prove that such a group doesn't exist. How to do it?

Edit: Except $|G|=pq$ and $|G|$=$p^2q$, $p,q$-primes. I forgot about it.

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    What should we make of, say, the dihedral group of order $10$? Its only subgroups have orders $2$ and $5$, but they don't have much of a choice! – 2017-02-04
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    I didn't get it. The least non p-group $S_3$ actually has only $p$-groups as a proper subgroups. – 2017-02-04
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    @pjs36: Such a consideration led me to add "proper" as a qualification to subgroups, as obviously the OP knows the whole group being a $p$-group easily admits this condition. – 2017-02-04
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    @hardmath Sure, but I still think these are counterexamples to what OP is asking about. I'm suggesting that *any* group of order $pq$ with $p, q$ prime must be a counterexample, and wondering if such solutions count -- or if OP is also trying to exclude these cases. – 2017-02-04
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    @pjs36: In my convoluted way I'm saying you are right (unless I misunderstood what the OP wants). – 2017-02-04
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    Not all groups of order $p^2q$ have this property. For example all abelian groups of this order also have subgroups of order $pq$. It does work for some, though. For example $A_4$ has no subgroup of order six, so it has this property. – 2017-02-04
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    I know it, but it is ok. I'm looking for some more complicated groups. I'm going to write something about subgroups in a finite groups and I'm looking for something between Sylow theorem and Hall theorem (which holds for solvable groups). If it is possible to find some "big" groups which doesn't have any additional subgroups (except p-subgroups), then it is not possible to find some general rule (like Sylow theorem) for existence of subgroups of some order. I hope I described it clearly. – 2017-02-04
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    Thompson proved in his thesis that a finite group is solvable, if it has a maximal subgroup that is nilpotent of odd order. So you won't find non-solvable examples. For solvable groups try to prove that it is the semidirect product of an elementary abelian group $N$ (look at a minimal normal subgroup!) and a cyclic group whose operation on $N$ is irreducible. – 2017-02-05
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    That's exactly what I'm looking for. In solvable case it is also enough to take a group generated by a minimal normal subgroup (p-group) and some other group (of order q). It works except two cases I wrote about. So the conclusion is that groups usually have many subgroups. Is there any theorem which is stronger than Sylow theorem (just existence)? I mean is there any rule that for my group of order n (I know nothing more except the order) I can say that subgroup of some order exists? – 2017-02-05
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    A nilpotent finite group is a direct product of its Sylow subgroups---so examples are not nilpotent. – 2017-02-05
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    The product of a normal subgroup with any subgroup is a subgroup. So if you have a cyclic group $Z$ acting on an elementary abelian group $E$ of coprime order, then a subgroups $N\le E$ normal in the whole group gives you a subgroup $NZ$ of mixed order. As $E$ is elementary abelian you can view it as vector space on which $Z$ acts. Having no proper $Z$-invariant subspace means being irreducible. – 2017-02-06

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