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Is it true that any finite group determined by representation over closed field? In other words, are there exists two different groups with the same representations? For example, any non-abelian group in order 8 is $D_4$ or $Q = \{ \pm 1, \pm i. \pm j, \pm k \}$, because: $$|G| = 2^2+1^1+1^1+1^1+1^1=2^2+2^2.$$

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What does it mean for the representations to be the "same"? – Qiaochu Yuan Jul 21 '11 at 15:19
What do you mean by "the same representations"? It matters. For instance, certainly $\mathbb{C}[G_1] \cong \mathbb{C}[G_2]$ does not imply $G_1 \cong G_2$ (simplest example: take any two abelian groups of the same order). On the other hand, a finite abelian group can be recovered up to isomorphism by its character table, but this is not true for arbitrary finite groups... – Pete L. Clark Jul 21 '11 at 15:19
@Alyushin: but to does not make much sense that two representations, each over a different group, be isomorphic... – Mariano Suárez-Alvarez Jul 21 '11 at 15:33
Probably not what the questioner wanted, but the question could make sense if asked about equivalence of categories of finite-dimensional complex repns of finite two groups...? with direct sums and tensor products, say? – paul garrett Jul 21 '11 at 15:52
It looks like you think that one of $Q$ and $D_4$ has irreducible representations of dimension $(2,2)$. This is wrong. Both of these groups have irreps of dimensions $(1,1,1,1,2)$. – David Speyer Jul 21 '11 at 17:15
up vote 3 down vote accepted

Capsulizing some comments and adding a bit: the question as (originally) posed does not formally make sense, because we lack an elementary sense of what it would mean for representations of _different_groups_ to be "the same". While there is some justification in dismissing the issue, the numerical example given (even if slightly garbled) already illustrates one possible meaning given to the question, namely, if the list of dimensions of irreducibles is the same, are the groups isomorphic? Or, what senses can this be given?

We can start with some "numerology", as mentioned in the question and comments: f or abelian groups of the same order, the list is all 1's, so we fail to distinguish them. For non-abelian groups of order 8, there are 4 1's and a 2, failing to distinguish the quaternion group from the dihedral group.

How to forget less about the repns?

One way that fails is to look at the isomorphism class of the rings $\mathbb C[G]$ (with convolution, naturally). The reason this fails is that these rings are products of simple rings (matrix rings, over $\mathbb C$), of sizes equal to the dimensions. So for abelian groups we get a product of copies of $\C$, no good, and for the two groups of order $8$, we get 4 copies of $\mathbb C$ and one 2-by-2 matrix ring, in both cases. No good.

As Pete C. mentioned, we could look at "character tables", or, equivalently, (here a segue could start) isomorphism classes of repns _with_tensor_product_. Irreducibles are distinguishable in this category, even without tensor products. Characters for abelian groups form a group themselves, as do irreducible (one-dimensional) repns under tensor product, so we recover the dual group to the original abelian group, and, hence, the group itself.

(Not every category of modules over a ring admits a tensor product. Group algebras, universal enveloping algebras, and others with appropriate involution do. Edit: see Brad's comment below for further...)

However, for the order-8 non-abelian groups, the character table and/or tensor-product-of-irreducibles relations are insufficient to distinguish the two. Maybe this can be made visible in an idiosyncratic but potentially convincing fashion: first, since both groups modulo the central $\mathbb Z/2$ are $2,2$ groups, the $4$ 1-D repns have the same tensor-product relations in both cases, that is, those parts of the character table are identical, so do not distinguish. This only leaves room for a single isomorphism class of 2-D repn $\rho$. For any 1-D repn $\alpha$, necessarily $\alpha\otimes \rho\approx \rho$. The tensor product could not be a sum of two 1-D irreducibles, because $\alpha^\vee\otimes(\alpha\otimes\rho)\approx (\alpha^\vee\otimes \alpha)\otimes \rho \approx 1\otimes \rho \approx \rho$, where $\vee$ is contragredient. $\rho$ is its own contragredient, since there's no other irred 2-D repn. Thus, our last hope of distinguishing the two groups is to have $\rho\otimes\rho$ be different in the two cases. But the fact that $\alpha\otimes\rho\approx \rho$ gives an equality of characters, $\alpha\cdot \chi_\rho=\chi_\rho$ (identifying $\alpha$ with its trace). That is, $\alpha$ is $=1$ on the support of $\chi_\rho$, for all 1-D repns $\alpha$, so the support of $\chi_\rho$ must be the commutator subgroup $[G,G]$ in both cases. Since $\langle \chi_\rho, \alpha\rangle=0$, the two non-zero values of $\chi_\rho$ must be equal but of opposite sign on $[G,G]$. Thus, $\langle \chi_{\rho\otimes \rho},\chi_\rho\rangle=0$, and no copy of $\rho$ occurs in $\rho\otimes\rho$. And the inner product with all 1-D characters must be the same, sooo, "sadly", in both cases $\rho\otimes\rho$ is the sum of all $4$ 1-D repns. That is, the tensor structure fails to distinguish, already in this little example, and for reasons that will tend to be manifest more broadly.

For any beginners who've not thought about categories of repns of (e.g.,) finite groups before, it might be interesting to attempt to capture their essential features axiomatically. And, e.g., see whether the above discussion still "forgets" something whose recollection would allow distinguishing the two groups.

In contrast to this story, in various categories of repns of Lie groups, the subcategories of repns of distinct groups can (to various degrees) be distinguished by their categorical structure...

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You can distinguish the category of representations of $Q_8$ from $D_4$ if you add in exterior powers as extra data, but even this is not enough in general. There is a nice discussion of this issue at… . – Qiaochu Yuan Jul 21 '11 at 23:21
Thanks for the link, Qiaochu Y! – paul garrett Jul 21 '11 at 23:23
The extra structure you need on a ring A to equip the category of representations of A with an (associative) tensor product (with unit) is that of a bialgebra, i.e. compatible comultiplication (and counit). The additional structure of the antipodal involution on a Hopf algebra is what's needed to make sense of taking the dual of a representation. – Brad Jul 22 '11 at 0:02
Thx @Brad! .... – paul garrett Jul 22 '11 at 0:16

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