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Consider two sets M, N such that C[M] is isomorphic to C[N] as representations of G, somewhat surprisingly it does not imply M, N are isomorphic as G sets. (Everything is finite.)

However it does imply that 1) |M| = |N| 2) number of orbits in M = number of orbits in N, because orbits correspond to invariant functions.

Question: Are there some other implications of C[M]=C[N] for the actions structure on M, N and corresponding module structure C[M]=C[N]? E.g. some numerical inequalities on sizes of orbits (there sizes can be different but may be not too much ?), stabilizers of points (can be non-conjugated, but may be somehow related), .... ?

As Benjamin Steinberg suggests number of orbits in MxMxM...xM (any times) will be the same as number of orbits in NxNxN...N, since these are dims of invariant functions in $C[M]^{\otimes L}$. Can it be the characterization ?

As far as I understand sets of conjugacy classes and irreducible representations of G provide such sets M,N for Aut(G). (hope I am correct ? See first sentence here). Can something more specific be said in this situation ? (e.g. Geoff Robinson mentions Glauberman correspondence here).

PS

Question What are natural examples/constructions/classifications of such sets ?

PSPS

The very related MO question by Vipul Naik is Brauer's permutation lemma -- extending to some other finite groups? and Ben Webster's answer with reference to his paper on the subj... But somehow I do not see the answer there...

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(It seems my brain is off at the moment.. Could you provide an example of isomorphic C[M]=C[N] with nonisomorphic G-sets M,N please? Thank you! ) –  Qfwfq Sep 11 '12 at 19:45
    
@QfwfqTake vector space V over F_q, take M=V, N=V^* take any G\in GL(V), people say that would be an example... Fourtier transform will give isomorphism... I've heard this today, so I am not big expert... Another example I added in the post - Irr(G) and Conj(G) under Aut(G) - if I understand correctly –  Alexander Chervov Sep 11 '12 at 19:53
    
@Qfwfq: there is an example with $G = \text{SL}_3(\mathbb{F}_2)$ at en.wikipedia.org/wiki/Gassmann_triple . –  Qiaochu Yuan Sep 11 '12 at 20:39
    
Could you please retag to use "finite-groups" instead of "finite-group". This is the only question tagged finite-group while for the other there are many; it is standard to use pluralized forms. Thanks in advance. –  quid Feb 10 '13 at 14:13
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4 Answers

up vote 5 down vote accepted

An old paper of mine [On exponentiation of G-sets, Discrete Math. 135 (1994) 69-79] contains, among other things, the result that the free complex-linear actions generated by two $G$-sets $X$ and $Y$ are isomorphic if and only if the power sets, $P(X)$ and $P(Y)$, with the obvious induced actions of $G$, are isomorphic as $G$-sets. Curiously, the "intermediate" property, that the power sets are isomorphic as $(\mathbb Z/2)$-linear representations of $G$, is not equivalent to these.

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This seems to me to really answer the question. –  Benjamin Steinberg Sep 12 '12 at 2:59
    
This is of course the same as P(X) and P(Y) being isomorphic as modules of the semiring P(G). What other idempotent semirings R have the property that $\mathbb CX\cong \mathbb CY$ iff $RX\cong RY$. –  Benjamin Steinberg Sep 12 '12 at 3:01
    
@Andreas Blass Thank you very much ! Very nice result ! –  Alexander Chervov Sep 12 '12 at 6:02
    
@Andreas Blass May I ask you ? Vaguely: is the proof constructive ? More precisely may be there is something like "correspondence" between X,Y which is G-ivariant ? Can the isomrphism between P(x) P(Y) be chosen uniquely ? or "almost" uniequely up to something... –  Alexander Chervov Sep 18 '12 at 11:59
    
@Alexander Chernov: The proof is not directly constructive. It involved comparing the characters of the linear representations over $\mathbb C$ with the marks (in Burnside's sense) of the power-set permutation actions. I haven't really considered the question whether one can convert it into something more constructive, setting up a specific, canonical correspondence between the two sorts of representations. –  Andreas Blass Sep 18 '12 at 13:20
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To answer the question about natural examples/classification, Tim Dokchitser and I have completely classified all such sets in the following sense.

If $\tilde{G}\leq G$, and $M$, $N$ are two $\tilde{G}$-sets giving rise to isomorphic permutation representations, then their inductions to $G$ are $G$-sets with isomorphic permutation representations. Similarly, if $\bar{G}$ is a quotient of $G$ and $M$, $N$ are $\bar{G}$-sets that give rise to isomorphic permutation representations, then their lifts to $G$ give rise to isomorphic permutation representations. We call such a relation between two sets imprimitive if it comes from a proper subquotient. The whole space of such relations can be given the structure of a group (it is a subgroup of the Burnside ring of $G$), and the imprimitive relations generate a subgroup. We determine the structure of the whole group of relations modulo the imprimitive ones, and give explicit generators for this quotient. It turns out that the groups that have primitive relations at all fall into a finite number of families, a bit like in the classification of simple groups. In other words, we give a morally finite list, such that all relations between $G$-sets are obtained from this list using induction from subgroups, lifts from quotients, and disjoint union of sets (more precisely linear combinations in the Burnside ring).

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Thank you very much ! It is very interesting. –  Alexander Chervov Sep 12 '12 at 6:02
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In this context representations are completely determined by their character, so two $G$-sets $M, N$ have this property if and only if the number of fixed points of any $g \in G$ acting on $M, N$ are always the same. This implies various generalizations of Benjamin Sternberg's answer. For example, the number of orbits of $G$ acting on the subsets ${M \choose k}, {N \choose k}$ of $M, N$ of size $k$ are the same.

Triples $(M, N, G)$ with this property are called Gassmann triples. They were famously used by Sunada to construct pairs of isospectral but not isometric Riemannian manifolds, and can also be used to construct pairs of isospectral (have the same zeta function) but not isomorphic number fields. Ben Webster has a nice pair of blog posts on the subject here and here, and the Wikipedia article has references. Apparently they are not easy to construct, but hopefully these keywords will provide a good starting point from which to search in the literature.

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Thank you very much ! Reference/explaining Sunada is especially nice ! –  Alexander Chervov Sep 12 '12 at 5:56
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The number of orbits of $G$ on $M\times M$ is the same as the number of orbits of $G$ on $N\times N$ since both are the dimension of the centralizer algebra.

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Wow, so fast :) Agree, thank you ! By the way what about MxMxM..xM ? more than two times ? –  Alexander Chervov Sep 11 '12 at 19:28
    
Might add that \sum d_i^2 = number of orbits in MxM, where d_i are dims(irreps) in C[M]. –  Alexander Chervov Sep 11 '12 at 19:30
    
Ahh... about MxMxM...xM it is also obviously yes. May be it is characterization ? –  Alexander Chervov Sep 11 '12 at 20:02
    
@Alexander: no. $M^k$ and $N^k$ have the same number of orbits if and only if $\sum_{g \in G} \text{Fix}_M(g)^k = \sum_{g \in G} \text{Fix}_N(g)^k$ for all $k$ if and only if the multiset $\{ \text{Fix}_M(g) \}$ is the same as the multiset $\{ \text{Fix}_N(g) \}$, but this is strictly weaker than the claim that $\text{Fix}_N(g) = \text{Fix}_M(g)$ for all $g$ since, for example, $M$ and $N$ may be related by an outer automorphism of $G$ which exchanges two elements $g_1, g_2$ such that $\text{Fix}_N(g_1) \neq \text{Fix}_N(g_2)$. –  Qiaochu Yuan Sep 11 '12 at 20:24
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The number of orbits of $M \times M \times M \times ... \times M$ for all lengths of the product do not characterize the representation. For instance consider to actions of $G \times G$ on a set, one factoring through the first $G$ and one being the same action factoring through the second $G$. –  Will Sawin Sep 11 '12 at 20:27
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