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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.

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What is the easiest example of non-isomorphic finite groups $F$ and $G$ such that $F_g = F_h$? –  Matthew Kahle Jul 23 '10 at 0:17
    
Any examples arising from the Lazard or Baer correspondences or their generalizations would work. For instance, for odd primes $p$, there is a bijection between the multiplicative group $U(3,p)$ of upper triangular unipotent $3 \times 3$ matrices over the field of $p$ elements, and the elementary abelian group of order $p^3$. There are also some examples of order $2^4$, which don't strictly arise from the Lazard correspondence but arise from generalizations of it. –  Vipul Naik Jul 24 '10 at 19:22
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Is a 1-isomorphism any different from a bijection f satisfying $f(x^n)=f(x)^n$ for integers n? –  George Lowther Jul 24 '10 at 22:29
    
They look to be the same, yes. –  Harry Altman Jul 25 '10 at 0:53
    
George Lowther: Your definition of 1-isomorphism matches what I wrote. –  Vipul Naik Jul 27 '10 at 13:52

2 Answers 2

up vote 14 down vote accepted

Here is a counterexample of order $32$.

$G$ and $H$ will each have $3$ elements of order $2$ and $28$ elements of order $4$. In both cases all three elements of order $2$ will have square roots. That insures that $F_G=F_H$. But in $G$ one of them will have $4$ square roots while the others each have $12$, and in $H$ one of them will have $20$ square roots while the others each have $4$. That rules out a $1$-isomorphism.

Let $Q$ be a quaternion group of order $8$ and let $C\subset Q$ be a (cyclic) subgroup of order $4$. Inside $Q\times Q$ there are three subgroups of index $2$ that contain $C\times C$. Let $G$ be $Q\times C$ and let $H$ be the one that is neither $Q\times C$ nor $C\times Q$.

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it always seems to come back to the difference between the dihedral, quaternion and semi-dihedral groups! –  Sean Tilson Jul 28 '10 at 3:36
    
I don't think of this example as being about the difference between quaternion and dihedral. I didn't think about counterexamples of order $p^5$ for other primes. –  Tom Goodwillie Jul 28 '10 at 12:43
    
Tom Goodwillie: Thank you for the answer. I'm sorry I was away from Math Overflow for the last few days, so did not notice this answer from you. Your answer seems to have been auto-selected as the corrected answer. –  Vipul Naik Aug 5 '10 at 20:34
    
I didn't know that auto-selection was possible. –  Tom Goodwillie Aug 6 '10 at 1:09
    
If a question has a bounty, the highest-voted answer will be auto-accepted when the bounty period runs out if none has been accepted by then. –  Harry Altman Aug 7 '10 at 1:53

Not a complete answer, but a way of breaking the question down a bit.

We put a relation $R$ on a finite group $G$ by saying $x <_R y$ if $x$ is contained in the subgroup generated by $y$. This relation is preserved by 1-isomorphisms. Moreover, we can recover $F_G$ from $R$: the order of an element $x$ is the number of elements $y$ for which $y <_R x$, and $x$ is a $d$-th power if and only if there is a $y$ such that $x <_R y$ and $|y|=(d,|y|)|x|$.

  1. Can we determine the $R$-structure from $F_G$? In other words, if $F_G = F_H$, does this ensure there is an $R$-preserving bijection from $G$ to $H$?

  2. Suppose there is a bijection from $G$ to $H$ compatible with $R$. Are $G$ and $H$ 1-isomorphic? (It's easy to see that $R$-preserving maps are not always 1-isomorphisms: for instance $C_5$ has $24$ $R$-preserving permutations, but only $4$ (1-)automorphisms.)

There has also been some work on power graphs of finite groups which may be relevant here: the power graph has vertex set $G$ and an edge from $x$ to $y$ if $x <_R y$ or $y <_R x$. It was proved recently by Peter Cameron that this symmetric relation is enough to recover the $R$-structure. See: http://www.reference-global.com/doi/abs/10.1515/JGT.2010.023

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I think the answer to (2) is yes. Basically, we can use the relation R to determine a new symmetric relation where $x \sim y$ if and only if $x <_R y$ and $y <_R x$. This new symmetric relation just means that the two elements generate the same cyclic subgroup. This equivalent relation divides the group into orbits. We can then construct a directed acyclic graph on the set of these orbits and then get a bijection between the DAGs for the two groups, which then can be extended to a 1-isomorphism. So, the power graph seems to be a combinatorial object that stores the group up to 1-isomorphism. –  Vipul Naik Aug 6 '10 at 16:13
    
Yes: I suppose this is saying that although an R-isomorphism isn't necessarily a 1-isomorphism, we can make it into one by permuting elements within the equivalence classes. In the other direction, a 1-isomorphism from $G$ to some known group is certainly enough to determine the power graph of $G$ (not just up to isomorphism). The next natural question is: which finite graphs are power graphs of finite groups? This could well be an active area of research, but I don't know how far people have got with it. –  Colin Reid Aug 9 '10 at 10:29

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