# Finite groups with elements of the same order

Given a finite group $G$, let $\{(1,1),(m_1,n_1),\ldots,(m_r,n_r)\}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n_1+\cdots+n_r$, and the pair $(1,1)$ accounts for the neutral element.

Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet not) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple.

Edit. Nick's answer gives the correct terminology, of conformal groups. Ben's answer speaks of the refined notion of almost conjugate subgroups. Is there any other related notion ?

• For what it's worth, it's true for finite abelian groups. – Gerry Myerson Sep 24 '10 at 12:40
• In response to your edit: the survey of Mazurov and Shi that I mention in my answer also mentions that the notion of conformality is sometimes known as the Grassmann equivalence. I don't know the origins of either piece of terminology (conformality/ Grassmann equivalence), but in any case I really don't like either of them! I much prefer order portrait which a couple of people have used in their answers below. – Nick Gill Nov 18 '15 at 16:17
• @NickGill - Are you sure that you mean 'Grassmann equivalence' and not 'Gassmann equivalence'? I ask because 'Gassmann equivalence' is the same thing as the almost conjugate condition I mentioned in my answer. In fact, I'll edit the answer right now to explain the history of Gassmann equivalence. – Ben Linowitz Nov 18 '15 at 21:25
• @BenLinowitz, I checked my source and it definitely says 'Grassmann'. However, after reading your answer below, it seems quite possible that this is a misprint. One query though: the source I quote asserts that "conformality" and "G(r)assmann equivalence" are the same thing, where as you saying that the latter implies the former, right? Is there an obvious counterexample for the other implication? – Nick Gill Nov 19 '15 at 10:02
• @NickGill - I think we finally have everything sorted out! On pages 352 and 353 of Perlis paper (link below) he shows that Gassmann equivalent subgroups have the same order portraits and conversely that non-isomorphic groups with the same order portraits are Gassmann equivalent. sciencedirect.com/science/article/pii/0022314X77900701 – Ben Linowitz Nov 19 '15 at 15:14

## 5 Answers

There are easy examples that are $p$-groups. For instance, the mod 3 Heisenberg group is the nilpotent group with presentation $\left < a,b,c \;\bigg |\, [a,b] = c, [a,c] = [b,c] = a^3 = b^3 = c^3 = 1 \right >$ has order 27, and all but the trivial element of order 3. This has the same order portrait as $C_3^3$ where $C_3 = \mathbb Z / 3\mathbb Z$ is the cyclic group of order 3.

This question was answered more than five years ago, but I am just now noticing it and wanted to point out that non-isomorphic groups with the same order portraits arise very naturally in spectral geometry and underlie Sunada's Method of constructing isospectral Riemannian manifolds.

Let $G$ be a finite group and $H_1, H_2$ be subgroups of $G$. We say that $H_1$ and $H_2$ are almost conjugate if every $G$-conjugacy class intersects both subgroups in the same number of elements.

An easy argument shows that almost conjugate subgroups have the same order portraits.

The connection of these groups to geometry stems from a theorem of Sunada that says that if $G$ acts by isometries on a Riemannian manifold $M$ and $H_1$ and $H_2$ are almost conjugate subgroups of $G$ then the quotient orbifolds $M/H_1$ and $M/H_2$ are isospectral with respect to the Laplace operator.

This paper by Brooks, Gornet and Gustafson gives explicit examples of arbitrarily large families of almost conjugate subgroups of Heisenberg groups defined over various commutative rings. The examples are then applied so as to construct families of isospectral non-isometric genus $g$ Riemann surfaces having cardinality $g^{c\log(g)}$ (here $c>0$ is a positive constant).

EDIT: In light of Nick's comment above I wanted to say a few words about Gassmann equivalence. To start with, by definition, two subgroups $H_1$ and $H_2$ of a finite group $G$ are Gassmann equivalent if and only if they are almost conjugate. The history of the term arises from a connection of these groups with number theory.

To an algebraic number field $k$ one can associate a Dedekind zeta function $\zeta_k(s)$. (If $k$ were the field of rational numbers then $\zeta_k(s)$ would be the Riemann zeta function.) This function determines all sorts of arithmetic properties of $k$: the discriminant, signature, splitting of primes, product of class number times regulator, etc. It is therefore natural to ask whether $k$ is determined up to isomorphism by $\zeta_k(s)$. It turns out that this is not in general true. We will therefore say that two number fields with the same Dedekind zeta function are arithmetically equivalent. The first examples of arithmetically equivalent number fields were discovered by Fritz Gassmann in 1925 and made use of the almost conjugacy condition defined above.

The term 'Gassmann equivalent' is due, I believe, to Robert Perlis. On page 344 of this paper Perlis defines subgroups $H_1$ and $H_2$ of $G$ to be Gassmann equivalent if every $G$-conjugacy class intersects $H_1$ the same number of times as $H_2$, which of course is precisely the same thing as saying that $H_1$ and $H_2$ are almost conjugate. Perlis then proves the following remarkable theorem.

Theorem (Perlis): Let $k$ be a Galois number field and $k_1, k_2$ be subfields of $k$. Then $\zeta_{k_1}(s)=\zeta_{k_2}(s)$ if and only if $\text{Gal}(k/k_1)$ and $\text{Gal}(k/k_2)$ are Gassmann equivalent subgroups of $\text{Gal}(k/\mathbb Q)$.

• A very nice comment indeed ! – Denis Serre Nov 17 '15 at 14:05
• Gasmann equivalence, being the same as Sunada's almost conjugacy, confirms that zeta functiosn have something to do with spectral analysis of Laplacians. This encourage us to look for a spectral argument in the way of a proof of Riemann's hypothesis. – Denis Serre Nov 19 '15 at 9:39

The smallest counterexamples have order $16$. Up to isomorphism, there are $14$ groups of order $16$; these fall into $9$ distinct equivalence classes w.r.t. order portrait. The $3$ equivalence classes containing more than one group can be found with GAP as follows:

gap> OrderPortrait := G -> Collected(List(AsList(G),Order));;
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->List(cl,IdGroup)); # in terms of catalog id numbers of groups
[ [ [ 16, 5 ], [ 16, 6 ] ], [ [ 16, 2 ], [ 16, 4 ], [ 16, 12 ] ],
[ [ 16, 3 ], [ 16, 10 ], [ 16, 13 ] ] ]
gap> List(last,Length); # two classes have length 3, and one has length 2
[ 2, 3, 3 ]
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->List(cl,StructureDescription)); # the structure of the groups
[ [ "C8 x C2", "C8 : C2" ], [ "C4 x C4", "C4 : C4", "C2 x Q8" ],
[ "(C4 x C2) : C2", "C4 x C2 x C2", "(C4 x C2) : C2" ] ]
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->OrderPortrait(cl)); # the actual order portraits
[ [ [ 1, 1 ], [ 2, 3 ], [ 4, 4 ], [ 8, 8 ] ],
[ [ 1, 1 ], [ 2, 3 ], [ 4, 12 ] ], [ [ 1, 1 ], [ 2, 7 ], [ 4, 8 ] ] ]


Let me join Stefan in resurrecting this question. I want to mention a fascinating observation of Thompson.

Apparently two groups $G$ and $G'$ that, in Denis' terms "have the same list" are, in the literature called conformal (see Mazurov-Shi in Groups St Andrews 1999).

J.G.~Thompson famously gave an example of two non-isomorphic conformal groups: $2^4:A_7$ and $L_3(4):2_2$. These two groups both appear as maximal subgroups of $M_{23}$, one of the sporadic Mathieu groups.

Along with this observation, Thompson posed the following question:

Suppose that $G$ and $G'$ are conformal, and that $G'$ is solvable. Is $G$ solvable?

As far as I know, this question remains open. However there has been a lot of progress in the study of conformality amongst non-solvable groups. I'm thinking of results of the kind: Suppose that $G$ and $G'$ are conformal, and that $G'$ is [insert name of some simple group], then $G\cong G'$.

I refer those interested to the above-mentioned survey of Mazurov--Shi for several results of this kind (and many other interesting results).

• See also this question that I asked quite some time ago: mathoverflow.net/questions/55071/… – Tom De Medts Nov 17 '15 at 13:52
• @TomDeMedts, Thanks for the link - it's a very nice question, and very much in the spirit of Thompson's question. Great minds think alike, eh? :-) – Nick Gill Nov 17 '15 at 15:07

See this question and the first answer:

Order information enough to guarantee 1-isomorphism?

A fortiori, any counterexample given to that question will work for your question as well.