# Prove $\dim N = \dim \rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$ for irreducible N

I am an undergrad who is self studying representation theory by working through Martin Isaac's Character Theory. I am trying to prove 1.17(c) just for the case of group algebras, but I need some help. Please do not give a full proof, but answer (or hint at) my specific question at the bottom. Here is the problem statement: Prove for an irreducible $$\mathbb{C}[G]$$-module $$N$$ that $$\dim N$$ equals the number of distinct copies of $$N$$ in the decomposition of $$\mathbb{C}[G]$$.

I learned from this post that it is possible to prove this by showing $$\dim N = \dim \rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$$, which follows from an isomorphism $$\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)\cong N$$ defined by sending all $$\psi\in \rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$$ to $$\psi(1)$$. This isomorphism makes sense to me. Since all $$\psi\in\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$$ are $$\mathbb{C}[G]$$-homomorphisms, then for all $$w\in\mathbb{C}[G]$$ and some $$\psi(1)=n\in N$$, we have $$\phi(w) = w\cdot \psi(1) = w\cdot n$$ which spans $$N$$. Each $$\psi$$ is therefore determined by where it sends $$1$$, so two homomorphisms are equal if $$1$$ is sent to the same place.

The next step is to construct a basis for $$\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$$, but I am confused though about what the basis of $$\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$$ looks like. I thought that we could construct a basis element $$\phi_i$$ by pointing out the decomposition of $$\mathbb{C}[G]$$ into $$\bigoplus_i V_i$$. Then for a set of irreducible submodules $$N_1, N_2,..., N_\alpha\cong N$$ represented in the decomposition $$\bigoplus_i V_i$$ we just define $$\phi_i$$ by sending the $$N_i$$ component of all $$w\in\mathbb{C}[G]$$ to its identity in $$N$$. Adding $$\phi_i$$ and $$\phi_j$$ together defines a homomorphism into a strict submodule of $$N_i\bigoplus N_j$$, so it is also a homomorphism into $$N$$, and this gives us the structure of a $$\mathbb{C}[G]$$-module. My problem is, if this is how each $$\phi_i$$ is defined, then $$\phi_i(1)$$ sends $$1$$ to the $$1$$ component of $$N_i$$, and $$\phi_j(1)$$ sends $$1$$ to the $$1$$ component of $$N_j$$, but shouldn't these be the same element in $$N$$?

• @MarianoSuárez-Álvarez I want to find the dimension of $\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$. The only way I know to do this is by explicitly forming a basis for $\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$. I know that $\dim N$ should equal the number of copies of $N$ in $\mathbb{C}[G]$, and I know that an isomorphism exists between $N$ and $\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$. So I expect to make a basis for $\rm{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], N)$, where each basis element is a function from a copy of $N$ in $\mathbb{C}[G]$ to $N$. Commented Dec 30, 2023 at 17:11
• @MarianoSuárez-Álvarez Yes. Obviously they have the same dimension. As I said, the statement is "Prove for an irreducible $\mathbb{C}[G]$ module $N$ that $\dim N$ equals the number of copies of $N$ in $\mathbb{C}[G]$." So, I would like to show that the decomposition of $\mathbb{C}[G]$ has $\dim N$ copies of $N$. Commented Dec 30, 2023 at 17:24
• @MarianoSuárez-Álvarez Okay, thank you for the help. I will try and complete a proof using those tools. Commented Dec 30, 2023 at 17:51