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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.

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A comment about the abelian case: A finite abelian group $G$ is specified up to isomorphism by the size of $G^n$ as $n$ runs through all prime powers (which is part of the information given by $F_G$). Consider the invariant factor decomposition $A_1 \times \dots \times A_n$, where each $A_i$ is cyclic and divides the order of the subsequent $A_i$. Then $log_p(|G|/|G^p|)$ is the number of $A_i$ divisible by $p$, $log_p(|G^p|/|G^{p^2}|)$ the number divisible by $p^2$, and so on. Given that the order of $A_i$ divides the order of $A_{i+1}$, we can obtain all the numbers $|A_i|$. –  Colin Reid Aug 12 '10 at 11:38
    
cdr: Yes, I am aware of that. If both <em>G</em> and <em>H</em> were abelian groups, we'd be able to say that they are isomorphic. However, in this situation, only one of the groups is given to be abelian, and there are many examples where the other one could be non-abelian. –  Vipul Naik Aug 12 '10 at 17:54
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