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 $x \in G$.

Question: Suppose *G* and *H* are finite groups of the same order such that $F_G = F_H$. Does there necessarily exist a 1-isomorphism between *G* and *H*?

Note that the converse is obviously true: if there exists a 1-isomorphism between *G* and *H*, then $F_G = F_H$.

Incidentally, *just* knowing the orders of elements does not determine the group up to 1-isomorphism. There are many counterexamples of order 16, with two non-abelian groups (one being the direct product of the quaternion group and the cyclic group of order two, and the other a semidirect product of cyclic groups of order four) having the same statistics on orders of elements as $\mathbb{Z}_4 \times \mathbb{Z}_4$, but neither being 1-isomorphic to it because they don't have the same number of squares.

Similarly, just knowing how many elements are there of the form $x^d$ for each divisor *d* of the order is not sufficient to determine the group up to 1-isomorphism. Again, there are counterexamples of order 16.

fsatisfying $f(x^n)=f(x)^n$ for integersn? $\endgroup$ – George Lowther Jul 24 '10 at 22:29