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21 votes
2 answers
679 views

Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?

Let $p$ be a prime; $\mathbb{F}_{p}$ is the field with $p$ elements and $\mathbb{F}_{p}[t]$ the ring of polynomials in $t$ over $\mathbb{F}_{p}$. Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the ...
A Stasinski's user avatar
  • 3,813
1 vote
0 answers
67 views

Relation Among Conjugacy Classes

This is more a request to find out if there is any work in the literature discussing certain things. Is there a naturally defined partial ordering on the set of conjugacy classes of a finite group G? ...
P Vanchinathan's user avatar
9 votes
2 answers
860 views

What is natural about the well-known bijection between conjugacy classes and irreps of a symmetric group?

Symmetric groups possess a well-known bijection between conjugacy classes and irreducible representations. More precisely, both sets are indexed by Young diagrams. Question: To what extent is this ...
Alexander Chervov's user avatar
35 votes
2 answers
3k views

Examples of finite groups with "good" bijection(s) between conjugacy classes and irreducible representations?

For symmetric group conjugacy classes and irreducible representation both are parametrized by Young diagramms, so there is a kind of "good" bijection between the two sets. For general finite groups ...
Alexander Chervov's user avatar
8 votes
4 answers
2k views

Product of conjugacy classes - is there an analog of Tanaka-Krein reconstruction ?

Consider a finite group G. The product of conjugacy classes can be defined in natural way just by multiplying the representatives and counting multiplicities (see e.g. MO 62088). So we get ring with ...
Alexander Chervov's user avatar
12 votes
1 answer
602 views

Why would dim primitive irrep divide size of some conjugacy class ?

From Isaacs et.al. 2005 Conjecture C. Let χ be a primitive irreducible character of an arbitrary finite group G. Then χ(1) divides | clG(g)| for some element g ∈ G. Here, of course, we ...
Alexander Chervov's user avatar