Let $G_1, G_2$ be two groups with at least one nontrivial proper subgroup each.
Let $S_1, S_2$ be the sets of proper subgroups of, respectively $G_1, G_2$.
Suppose there exists a bijective function $f: S_1 \rightarrow S_2$ such that $\forall A\in S_1, f(A)$ is isomorphic to $A$.
When can I conclude that $G_1, G_2$ are isomorphic?
I think that, if $G_1$ and $G_2$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It. Moreover, I haven't found any counterexample for nonabelian finite groups.
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$\begingroup$There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4\times C_4$ and $C_4\rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2\times C_2$
3 subgroups isomorphic to $C_4\times C_2$
(See for the subgroups of $C_4\rtimes C_4$.)
Another easy pair of examples is $C_9\times C_3$ and $C_9\rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
$\endgroup$ $\begingroup$Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $\phi(n)$ elements of order $n$.
Lemma 2Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
ProofThe p-Sylow, P, of G is of the form $\mathbb{Z}_{p^{a_1}} \times ... \times \mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(\leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(\leq p^k)}=\Pi_{i\leq n}{\min (p^{a_i}, p^k)}$$Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
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