# Why all irreducible representations appear in the regular representation?

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?

• 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 $R \to V$. – Derek Holt May 30 '16 at 15:39

• 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? – user1620696 May 30 '16 at 20:47
• @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 $R$. – arctic tern May 30 '16 at 22:52
• 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 $R=\bigoplus_{i\in I} a_i V_i$. – Colin McLarty May 31 '16 at 0:58