# Does regular representation of a finite group contain all irreducible representations?

I know that every irreducible representations of $S_n$ can be found in $\mathbb{C}S_n$. I wonder how can I prove that irreducible representations of a finite groups $G$ can be found in $\mathbb{C}G$. I thought to use the embedding of $G$ in $S_n$, but then I don't know how to go on. Could any of you help me, please?

Maybe I got it: I know that the character of the regular representation can be viewed as the sum of the characters of the irreducible representations with coefficients their dimension, so they are in the decomposition of the regular representation. Am I right?

To prove this statement using characters is kind of working backwards.

You can prove this statement without characters as follows: first, show that any simple $A=\mathbb{C}[G]$-module $M$ is a quotient of $A$. To do that, fix $0\neq m\in M$ and define $\mathbb{C}[G]\rightarrow M, \; \sum\alpha_gg \mapsto\sum\alpha_gg(m)$. Show that this is onto ($M$ is simple) and use the isomorphism theorem. Now, use Maschke's theorem to show that if $M$ is a quotient of $A$, then it is also a direct summand of $A$.

For details of all this, see e.g. Proposition 2.14 of my notes on representation theory.

$\mathbb{C}[G]$ is the free $\mathbb{C}[G]$-module on one generator. That means it satisfies $\text{Hom}_{G}(\mathbb{C}[G], M) \cong M$ for any $\mathbb{C}[G]$-module $M$ (since a $G$-morphism is uniquely determined by the image of $1$, which is arbitrary); in particular, this is nonzero for any simple $M$, so there exists a nonzero $G$-morphism whose image must be all of $M$ by simplicity. Then one appeals to Maschke's theorem.

In fact this argument tells you more: it immediately implies that the multiplicity of $M$ in $\mathbb{C}[G]$ is $\dim_{\mathbb{C}} M$ without character theory. Until the step involving Maschke's theorem, this argument is valid for an arbitrary ring $R$: any simple left $R$-module is a quotient of $R$ (as a left $R$-module). This implies that in fact it must be isomorphic to $R/m$ where $m$ is a maximal left ideal.

• I am struggling to follow your argument. Could you explain how exactly Maschke shows the multiplicity of $M$ in $\mathbb C[G]$ is $\dim_\mathbb C M$? This answer seems to claim a similar thing, but I don't understand why it holds. – Arrow Jun 26 '18 at 22:22
• @Arrow: if $M$ is simple, then the multiplicity of $M$ in another representation $V$ is $\dim \text{Hom}(M, V)$ by complete reducibility. This follows from writing $V$ as a direct sum of simples. If $M, V$ are both finite-dimensional then this multiplicity is also equal to $\dim \text{Hom}(V, M)$, which again follows from writing $V$ as a direct sum of simples. – Qiaochu Yuan Jun 26 '18 at 23:03
• But doesn't this also use Schur's lemma? – Arrow Jun 26 '18 at 23:04
• Sure, yes, you also need that. – Qiaochu Yuan Jun 26 '18 at 23:05
• Ah, okay :) I think that's missing from the answer I linked. Thanks! – Arrow Jun 26 '18 at 23:06