There are continuum 2-generated groups where all proper subgroups are cyclic of order $p$ (for the same prime $p\sim 10^{70}$), the Tarski monsters. All these groups have the same lists of proper subgroups, and the same lattice of subgroups. Also all cyclic groups of prime order have the same proper subgroups: the trivial group. So in general the answer is "no". But every finite non-cyclic abelian group is determined by the list of its proper subgroups, which follows easily from the description of finite abelian groups (note that I and, I think, the question, are not talking about the lattice of subgroups, but about the list of all proper subgroups). I suspect that at least for a large class of finite solvable groups, the list of proper subgroups determines the group. I would look first at A-groups, finite groups with all Sylow subgroups abelian (see the book of Huppert, "Endliche Gruppen") because the structure of subgroups of A-groups is more known, and one can use induction on the order of the group.

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