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.
I think it is possible. If G has prime-power order $\displaystyle p^a$, then the subgroup of order $\displaystyle p^{a-1}$ works (assuming $\displaystyle a\ge{2}$).
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 $\displaystyle \frac{|G|}{p}$ and $\displaystyle \frac{|G|}{q}$ 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.