Let $G$ be a finite group and $R$ the regular representation. That is, as a vector space $R = F(G)$ is the free vector space with basis $G$. If the basis is $\{e_g : g \in G\}$ the action is defined by

$$g \cdot e_{g'}=e_{gg'}$$

and extended by linearity.

Now, in the book I'm studying the author states the following corolary:

Corollary 2.18: Any irreducible representation $V$ of $G$ appears in the regular representation $\dim V$ times.

The "proof" for this is a little argument before the statement:

We know tha the character of $R$ is simply

$$\chi_R(g)=\begin{cases}0, & g\neq e, \\ |G|, & g= e\end{cases}$$

Thus, we see first of all that $R$ is not irreducible if $G\neq \{e\}$. In fact, if we set $R = \bigoplus V_i^{\oplus a_i}$, with $V_i$ distinct irreducibles, then:

$$a_i = (\chi_{V_i},\chi_R)=\dfrac{1}{|G|}\chi_{V_i}(e)|G|=\dim V_i.$$

All I get from that is: if we decompose $R$ into a direct sum of irreducible representations, the multiplicites are the dimensions.

But what guarantees that any irreducible representation of $G$ appears in that decomposition of $R$? Why all irreducible representations of $G$ appear in the direct sum decomposition of the regular representation?

  • $\begingroup$ Clearly any irreducible $G$-module $V$ is generated as a module by any nonzero element of $V$. Since the module $R$ of the regular representation of $G$ is a free $FG$-module with a single generator, there is an $FG$-module epimorphism $R \to V$. $\endgroup$ – Derek Holt May 30 '16 at 15:39

By definition, a simple module is nonzero. Hence its dimension, which equals its multiplicity in the decomposition, is positive. This guarantees that any irreducible representation does appear in it.

  • 1
    $\begingroup$ But again, if we have that $R = a_1V_1\oplus \cdots \oplus a_k V_k$, we indeed have that $a_i > 0$. That is fine. But what guarantees that there is not another irreducible representation of $G$, say $V_{k+1}$ such that $R = a_1V_1\oplus \cdots \oplus a_kV_k \oplus a_{k+1}V_{k+1}$? My question is: when we decompose $R$ as a direct sum of irreducible representations, what guarantees that all irreducible representations of $G$ will appear in the decomposition? $\endgroup$ – user1620696 May 30 '16 at 20:47
  • 1
    $\begingroup$ @user1620696 Start with $V_i$ being a list of all irreducible representations. Then write $R=\bigoplus_{i\in I} a_i V_i$. Then find all $a_i$ are positive. So all irreps are in $R$. $\endgroup$ – arctic tern May 30 '16 at 22:52
  • $\begingroup$ OP's question is not whether there is a sum $\bigoplus_{i\in I} a_i V_i$ of all irreducible representations of $G$, but how we know the right $a_i$ give the regular representation $R=\bigoplus_{i\in I} a_i V_i$. $\endgroup$ – Colin McLarty May 31 '16 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.