Sufficient conditions on the isomorphism of two groups 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.

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 https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups 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.)
