Thread: One biggest subgroup

Let G be a group and H a proper subgroup which contains every proper subgroup of G. What can be said about G?

I can't seem to figure out what the question is referring to.

my guess would be that G is cyclic. let's see if we can prove this.

suppose H is such a group. since H is proper, there must be some x in G, with x not in H.

now <x> is a subgroup of G. if it were proper, then by our assumption on H, <x> would be contained in H (since H contains every proper subgroup).

in particular, x would have to be in H, since x is in <x>. but this contradicts our choice of x.

hence <x> must not be proper, in other words, <x> = G, so G is cyclic.

Yes, G is cyclic, but is it possible to have a proper nontrivial H with such property?

I think it is possible. If G has prime-power order , then the subgroup of order works (assuming ).

In fact, I think if G is finite, it has to have prime-power order. If p and q are distinct prime divisors of |G|, then the subgroups of orders and are contained in H - meaning H cannot be a proper subgroup.

Also it can't be infinite, since then it would be isomorphic to the integers and the integers don't have a single largest subgroup.