$\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\rank{rank}$Let $G$ be a finite group, and $C=\Rep(G)$ be the monoidal category of complex finite-dimensional representations of $G$. As $C$ is finite and semisimple, one can get all representations from $\oplus$ and a finite set $I$ of irreducible representations. By classical character theory, there is a (noncanonical) bijection between $I$ and $\mathrm{Conj}(G)$. In this thread, I hope to understand a bijection, if any, between both sides with the consideration of $\otimes$.

To be more precise, let $V$ be an irreducible faithful representation of $G$. Then every representation occurs as a submodule of $V^{\otimes n}$ for some $n$ (cf this and this), and vice versa! We then say that $V$ itself generates $C$ under $\otimes$ and Cauchy completion. However, not every group has an irreducible faithful representation. In the same post, we can see that this largely deals with the "rank" of the socle of $G$.

To summarize, define the rank, $\rank(G)$, to be the minimal number of elements needed to generate $\mathrm{socle}(G)$ under conjugation. Define the rank, $\rank(C)$, to be the minimal number of irreducible elements needed to generate $C$ under $\otimes$ and Cauchy completion. Then

$$ \rank(G) = 1 \Leftrightarrow \rank(\Rep(G)) = 1 $$


Does this equivalence generalize to

$$ \rank(G) = n \Leftrightarrow \rank(\Rep(G)) = n, $$

for each natural number $n$?

(EDIT As Qiaochu pointed out in the comment, this is true for finite abelian groups by Pontrjagin duality.)

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    $\begingroup$ Small comment: it is at least reasonably straightforward to see that this is true for $G$ finite abelian. $\endgroup$ Dec 26, 2020 at 1:02

1 Answer 1


The answer to your question is yes and is the main theorem of the paper Žmudʹ, È. M. On isomorphic linear representations of finite groups. Mat. Sb. N.S. 38(80) (1956), 417–430.

It can be found in Theorem 5 on page 245 of Characters of Finite Groups. Part 1. by Berkovich and Žmudʹ. The theorem is phrased in a different, but equivalent way, and is proved in a very similar way to Gaschutz's theorem.

The theorem of Žmudʹ says that $G$ has a faithful representation with $k$ irreducible constituents if and only if the socle of $G$ can be generated as a normal subgroup by at most $k$ elements. In particular, the least number of normal generators of $\mathrm{socle}(G)$ coincides with the least number of irreducible constituents in some faithful representation of $G$.

It now suffices to observe $\mathrm{rank}(C)$ is exactly the minimal number of irreducible constituents in a faithful representation of $G$. Indeed, if $V$ is any faithful representation, then the Burnside theorem (or R. Steinberg's generalization) shows that every irreducible module is a direct summand in a tensor power of $V$ and so the irreducible constituents of $V$ generate $C$ under tensor product, direct sums and taking direct summands. On the other hand, if $\rho_1,\ldots, \rho_k$ are irreducible representations whose direct sum is not faithful, then $\ker \rho_1\cap\dots\cap \ker \rho_k$ acts as the identity on all modules in the subcategory generated by the corresponding simple modules under the operations of direct sum, tensor product and taking direct summands and so these irreducible representations cannot generate $C$.

Thus $\mathrm{rank}(G)=\mathrm{rank}(C)$


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