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 $$G$$ over an algebraically closed field of characteristic $$0$$ 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 $R \to V$. Commented May 30, 2016 at 15:39
• I wanted to make sure I understand the conclusion of this argument. For each irreducible $V_i$ appearing in the decomposition of $R$, we can compose $R \to V$ with an embedding $V_i \to R$ to obtain a homomorphism $V_i \to V$ of irreducible $FG$-modules. By Schur's lemma, this must either be the zero map or an isomorphism. We cannot have that every such embedding is the zero map, for then $R \to V$ would be the zero map, so $V$ must be isomorphic to one of the $V_i$'s.
– nahp
Commented Sep 5, 2020 at 18:58

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.

• 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?
– Gold
Commented May 30, 2016 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 $R$.
– anon
Commented May 30, 2016 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 $R=\bigoplus_{i\in I} a_i V_i$. Commented May 31, 2016 at 0:58

I will answer the part of the question where you ask, But what guarantees that any irreducible representation of $$G$$ appears in that decomposition of $$R$$'.

To answer this, choose some irreducible representation $$V$$. Choose nonzero $$v\in V$$. We get a homomorphism of representations $$R\to V$$ by sending $$\sum a_ge_g\mapsto \sum a_ggv$$. This map is surjective because $$V$$ is simple. As $$R$$ is semisimple (Maschke's theorem) the kernel of the map $$R\to V$$ is a direct summand $$R=\ker \oplus \tilde{V}$$ where $$\tilde{V}$$ is a copy of $$V$$ inside $$R$$.

This is good question which haunted me for a while as a first learner. I think the book "David S. Dummit, Richard M. Foote - Abstract Algebra 3rd ed" answers it well, in Sec.18.2 Example(3).

The concept one needs to be clear is that the regular representation is afforded by CG ring itself when seen as a CG-module. To see what irreducible representation appear in it amounts to decomposing this CG-module. According to Wedderburn's Theorem, this CG seen as a ring is isomorphic to a decomposition of matrix rings (ideals): $$\mathbb{C} G \cong M_{n_{1}}(\mathbb{C}) \times M_{n_{2}}(\mathbb{C}) \times \cdots \times M_{n_{r}}(\mathbb{C}).$$ Furthermore, each $$M_{n_{i}}(\mathbb{C})$$ decomposes further as direct sum of $$n_i$$ isomorphic simple left ideals. These left ideals seen as CG-module give a complete set of isomorphism classes of irreducible CG-modules, which again affords the regular representation.

Thus we reach the conclusion: the regular representation (over C) of G decomposes as the direct sum of all irreducible representations of G, each appearing with multiplicity equal to the degree of the irreducible representation.

This is just an expansion of Derek Holt's comment, but some extra detail may be useful. Throughout I assume that $$G$$ is a finite group, and that representations are taken over a field $$F$$ (of any characteristic).

The "regular representation" is $$\mathcal F(G)$$, the $$F$$-linear span of a set $$\{e_g: g\in G\}$$, which is assumed to be linearly independent (this is sometimes expressed by saying $$\mathcal F(G)$$ is the vector space of "formal linear combinations of the group elements"). This technique of associating an $$F$$-vector space to a finite set can be viewed as a functor from the category of $$\textbf{FinSet}$$ of finite sets to the category $$\textbf{FDVect}_F$$ of finite-dimensional $$F$$-vector spaces, where if $$X$$ is a finite set, $$\mathcal F(X)$$ is a vector space with basis $$\{e_x:x \in X\}$$. More precisely, instead of saying "formal linear combinations" one can define $$\mathcal F(X) = \text{Map}(X,F)$$ the space of $$F$$-valued functions on $$X$$. This a vector space because $$F$$ is, i.e. we can define addition and scalar multiplication pointwise, and it has a basis $$\{e_x: x\in X\}$$ where, for $$x,x' \in X$$ we set $$e_x(x')=1$$ if $$x=x'$$ and $$e_x(x')=0$$ otherwise. If $$f\colon X\to Y$$, then we can define $$\mathcal F(f)\colon \mathcal F(X)\to \mathcal F(Y)$$ by extending the definition of $$f$$ to all of $$\mathcal F(X)$$ by linearity, that is $$\mathcal F(f)(\sum_{x \in X} \lambda_x e_x) = \sum_{x \in X} \lambda_x.e_{f(x)} \forall x\in X.$$

If $$G$$ is a finite group, and we can consider the category $$\mathbf{FinSet}_G$$ of finite sets equipped with an action of $$G$$, that is, a homomorphism $$\alpha\colon G\to \text{Sym}(X)$$ from $$G$$ to the group of bijections from $$X$$ to itself. Our $$\mathcal F$$ then induces a functor $$\mathcal F_G$$ from the category of $$\mathbf{FinSet}_G$$ to the category of finite-dimensional representations of $$G$$, where if $$(X,\alpha)$$ is a $$G$$-set, $$\mathcal F_G(X)= (\mathcal F(X),\mathcal F(\alpha))$$, that is, $$\mathcal F_G(X)$$ is the vector space $$\mathcal F(X)$$ equipped with the homomorphism $$G\to \text{GL}(\mathcal F(X))$$ given by $$g \mapsto\mathcal F(\alpha(g))$$.

If $$X$$ and $$Y$$ are finite sets, then we have a natural isomorphism $$\mathcal F(X\times Y) \cong \mathcal F(X)\otimes \mathcal F(Y)$$ given by $$e_{(x,y)} \mapsto e_x\otimes e_y$$, and so we can use the group multiplication map $$m\colon G\times G \to G$$ to obtain a linear map $$\mathcal F(G)\otimes \mathcal F(G)\to \mathcal F(G)$$, or equivalently a bilinear map $$*\colon \mathcal F(G)\times \mathcal F(G)\to \mathcal F(G)$$, which is just given by $$e_g*e_{g'} = e_{gg'}$$.

In this way $$\mathcal F(G)$$ becomes an algebra over $$F$$, called the `group algebra''. Note that $$\{e_g: g \in G\}\subseteq \mathcal F(G)$$ is a subgroup of the group of units of the algebra $$\mathcal F(G)$$ which is isomorphic to $$G$$ and generates $$\mathcal F(G)$$ as an $$F$$-algebra. A consequence of this is that any representation $$(V,\rho)$$ of $$G$$ on an $$F$$-vector space is naturally an $$\mathcal F(G)$$-module, and conversely any $$\mathcal F(G)$$-module is a $$G$$-representation. Indeed an $$\mathcal F(G)$$-module is an $$F$$-vector space by restriction of the action of $$\mathcal F(G)$$ to $$F.e_{1_G}\cong F$$, and this $$F$$-vector space has a linear action of $$G$$ because $$G \cong \{e_g: g \in G\}\subseteq \mathcal F(G)^*$$ is isomorphic to a subgroup of the group of units of $$\mathcal F(G)$$. Conversely, if $$(V,\rho)$$ is a $$G$$-representation and $$x = \sum_{g \in G} \lambda_g.e_g \in \mathcal F(G)$$ for some $$\lambda_g \in F$$, $$g \in G$$, then we define $$\tilde{\rho}\colon \mathcal F(G)\to \text{End}(V)$$ by setting $$\tilde{\rho}(x) = \sum_{g \in G}\lambda_g\rho(g)$$.

If $$M$$ is any $$\mathcal F(G)$$-module and $$u \in M$$ then $$\mathcal F(G).u = \{a.u: a \in\mathcal F(G)\}$$ is a submodule of $$M$$ (the "cyclic submodule generated by $$u$$). It is the image of the module homomorphism $$ev_u\colon \mathcal F(G) \to M$$ given by $$ev_u(x) = x.u$$, and so a homomorphic image of $$\mathcal F(G)$$. Since $$u = 1_G.u \in \mathcal F(G).u$$, the submodule $$\mathcal F(G).u$$ is nonzero whenever $$u \neq 0$$. Now if we assume that $$M$$ is irreducible, then for any $$u \in M$$ nonzero we must have $$\mathcal F(G).u = M$$, and hence $$M$$ is cyclic and hence a quotient of $$\mathcal F(G)$$ as claimed.

This shows that any irreducible representation of $$G$$ occurs as composition factor of $$\mathcal F(G)$$, but that does not immediately imply that any irreducible occurs as a subrepresentation of $$\mathcal F(G)$$. If $$\text{char}(F)$$ is coprime to $$|G|$$, then Maschke's theorem shows that any quotient representation is isomorphic to a subrepresentation (if $$q\colon V\to W$$ is a surjective homomorphism where $$V$$ is semisimple, then $$\ker(q)$$ has a complementary subrepresentation $$W'$$, and $$q_{|W'}\colon W' \to W$$ is an isomorphism). In fact all irreducibles occur in $$\mathcal F(G)$$ even when Maschke's theorem does not hold, because it is always true that $$\mathcal F(X) \cong \mathcal F(X)^*$$, i.e. if $$X$$ is a $$G$$-set, the $$G$$-representation $$\mathcal F(X)$$ is self-dual. This can be seen using the map $$\phi \mapsto \phi^*$$, where $$\phi^*(\psi) = \sum_{x \in X} \phi(x)\psi(x)$$, which is an isomorphism because it maps $$e_x$$ to the basis of $$\mathcal F(X)^*$$ dual to $$\{e_x: x\in X\}$$.

But if $$q\colon \mathcal F(G)\to V$$ is a surjective map onto an irreducible representation $$V$$, then $$V^*\cong (\text{ker}(q))^0\leq (\mathcal F(G))^*$$, so that $$V^*$$ is a subrepresentation of $$\mathcal F(G)^*$$. Since the dual of an irreducible representation is irreducible, it follows that every irreducible representation of $$G$$ is a subrepresentation of $$\mathcal F(G)$$ in any characteristic.

The fact that the multiplicity with which $$V$$ occurs is equal to its dimension is a more delicate fact, which is only true if $$\mathcal F(G)$$ is semisimple. In particular, it can fail if the characteristic of $$F$$ divides the order of $$G$$: for example if $$G = \{1_G,\sigma\}$$ is the group with two elements, then if $$\text{char}(F)=2$$, then $$\{0\}\subseteq F.(1_G+\sigma)\subseteq F(G)$$ are all the subrepresentations of $$F(G)$$, and both $$F.(1_G+\sigma)$$ and $$F(G)/F.(1_G +\sigma)$$ are copies of the trivial representation, thus its multiplicity is 2 not 1.

There is a theorem that states:

Let $$V$$ be a linear representation of $$G$$ s.t. $$V = W_1 \oplus \dots\oplus W_k$$ (all of these are irreducible representation of $$G$$) with $$\phi$$ being the character. If $$W$$ is the irreducible representation with character $$\chi$$, then the number of $$W_i$$ which is isomorphic to $$W$$ is the scalar product of $$(\phi|\chi)$$.

[Serre-Linear representation of finite groups-Ch-2-theorem 4]

So, when it says that "Any irreducible representation $$V$$ of G appears in the regular representation", it takes into consideration the isomorphs, as we normally do in algebra.