This is a very slight variant on the question order information enough to guarantee 1-isomorphism? that I asked a while back, with an answer in the negative.

Background repeated:

I define a 1-isomorphism between two groups as a bijection that restricts to an isomorphism on every cyclic subgroup on either side. There are plenty of examples of 1-isomorphisms that are not isomorphisms. For instance, the exponential map from the additive group of strictly upper triangular matrices to the multiplicative group of unipotent upper triangular matrices is a 1-isomorphism. Many generalizations of this, such as the Baer and Lazard correspondences, also involve 1-isomorphisms between a group and the additive group of a Lie algebra/Lie ring.

Consider the following function *F* associated to a finite group *G*. For divisors $d_1$, $d_2$ of *G*, define $F_G(d_1,d_2)$ as the number of elements of *G* that have order equal to $d_1$ and that can be expressed in the form $x^{d_2}$ for some $x \in G$.

New question: If *G* is a finite *abelian* group and *H* is a finite (not necessarily abelian) group such that $F_G = F_H$, is it necessary that there is a 1-isomorphism between *G* and *H*.

For the original question, I had not insisted that one of the groups be abelian, and Tom Goodwillie provided a counterexample with both groups non-abelian of order 32.

The reason for my interest is as follows: I want to determine which groups are 1-isomorphic to abelian groups. This will help me with exploring some generalizations of the Lazard correspondence. To do this properly, I would need to construct a combinatorial structure (such as the directed power graph) that stores all the information of the group up to 1-isomorphism. However, constructing this structure and then verifying whether the graphs thus constructed for two groups are isomorphic is computationally somewhat harder. On the other hand, $F_G$ can be stored easily and we can quickly check for two groups whether their $F$s coincide.

Apart from this computational perspective, the question is also of academic interest to me.