First of all I am far from being an expert in representation theory, so it is possible (likely) that the following question is trivial (in fact a trivial reference question):

Let $\Gamma$ be a, let's say finite (abelian) group, and $R$ be an effective representation of $\Gamma$. Is it true, that every irreducible $\Gamma$-representation is a subrepresentation of the $k$-th symmetric power representation $S^kR$, for some $k=0,1,\dots$ ?
(Note that it is enough to show that any regular representations of $\Gamma$ is a subrepresentation of $S^kR$.)

I strongly suspect that this is a well know fact in representation theory, but I couldn't find any reference so far and would therefore appreciate any kind of literature reference.


1 Answer 1


Edit: The answer is yes for all finite groups $G$ (even, as far as this makes sense, independently of the ground field). A reference is Theorem 1 on page 45 of Alperin: "Local representation theory". While the claim of this theorem only refers to the tensor product $V\otimes \ldots \otimes V$ for some faithful $kG$-module $V$, the proof actually constructs a free submodule in $$ k[V] \cong \bigoplus_{i=0}^{\infty} S^k V $$ which gives you exactly what you need.

For a finite abelian group $G$ the answer is yes. Assume $V=\langle v_1\rangle\oplus \ldots \oplus \langle v_n \rangle$ is the decomposition of a faithful $\mathbb C G$-module $V$ into irreducibles. Denote the character associated to $v_i$ by $\chi_i$. Since $V$ is faithful the $\chi_i$ must generate the character group $\textrm{Hom}(G,\mathbb C^\times)$. So we can write any irreducible character $\psi$ of $G$ as $$ \psi = \chi_1^{k_1}\cdots \chi_n^{k_n} $$ and then the vector $v_1^{k_1}\cdots v_n^{k_n} \in S^{k_1+\ldots+k_n} V$ spans a one-dimensional submodule of this symmetric power with character $\psi$.

  • $\begingroup$ Actually, symmetric powers of the natural permutation module of S_n do contain every irreducible. This is one of the few cases of which I'm sure! $\endgroup$ Aug 18, 2012 at 17:09
  • $\begingroup$ @John: Yes, you are right. I misread the paper (they just show that the characters of the symmetric powers do not span $\mathbb C \textrm{Irr}_{\mathbb C}(S_n)$). $\endgroup$ Aug 18, 2012 at 17:13
  • $\begingroup$ A remark for those who don't own a copy of Alperin's book: The relevant pages can be found at books.google.be/… $\endgroup$ Aug 18, 2012 at 18:51

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