I will be making reference to this answer by David E. Speyer. Representation theory, as taught by Artin's introductory text, has interested me greatly. Unfortunately he does not prove the result that the degree of an irreducible representation divides the order of the group, or even the more surprising results such as the degree divides the index of the centre. TlDr; my question is how can elements of $\Bbb C[G]$ be reconciled with elements of a vector space $V$, if $G$ is irreducibly represented over $V$?
I was looking for as elementary a proof as possible, as my knowledge of abstract algebra is not good enough to parse proofs involving e.g. integral extensions and the abstract Cayley-Hamilton theorem. David's answer is great, but I have two concerns.
The notation that they use:
$G$ is a finite group, $\rho_V:G\to GL(V)$ an irreducible representation of $G$ over a finite dimensional complex vector space $V$, $\chi_V$ the character of this representation; the $g_i$ are representative elements for the conjugacy classes of $G$, which are denoted by $C(\cdot)$.
The main first concern is the notation $\Bbb Z[G]$ or $\Bbb C[G]$ for a finite group $G$.
Let's define: $$P(g)=\sum_{h\in C(g)}h$$This is an element in $\Bbb Z[G]$.
It acts on $V$ through $\rho_V$ via $P_V(g)=\sum_{h\in C(g)}\rho_V(g)$, and this expression is now a map $G\to GL(V)$.
Continuing:
Consider $Q_V=\sum_i P_V(g_i)\chi_V(g_i^{-1})$. [...] On the other hand if you expand out $Q_V$ in $\Bbb C[G]$, you'll see that the coefficient of every group element is an algebraic integer.
This last sentence confuses me. To my mind, $\Bbb C[G]$ is a ring of polynomials in the elements of $g$ with complex coefficients, e.g. $g_1^3-(1+i)g_2$. $Q_V$ is an element of $GL(V)$, so how are we able to talk about "the coefficients of $g$"? My best guess is that, because $Q_V(G)\subseteq\rho(G)$, we are identifying the linear operators in $\rho(G)$ with elements of $g$ - although we aren't given that the representation is faithful, so this seems weird.
If David had instead written:
[...] $Q_V=\frac{|G|}{\dim V}\mathrm{Id}_V$. Since $Q_V$ is a linear combination of linear operators with algebraic-integer coefficients (the $\chi_V(g_i^{-1})$) it follows by a basis-span argument that $\frac{|G|}{\dim V}$ is an algebraic integer. [...]
I'd have accepted it entirely without batting an eye. However he passed to a different formalism, and that makes me wonder - is my "rephrasing" incorrect?
Another concern is that someone commented:
"What do you mean by 'if you expand out $Q_V$ in $\Bbb C[G]$, you'll see that the coefficient of every group element is an algebraic integer'? I don't think there is a 'sum of algebraic integers is an algebraic integer' result for noncommutative rings"
This is compounding my confusion. Does that comment really invalidate David's argument? I think not, since we are dealing with coefficients in the complex numbers and in that number system we definitely do have that the sum of algebraic integers is an algebraic integer. Besides, polynomial rings are obviously commutative, so I really have no idea what was meant.
I apologise if this is too many questions, but this last one is a quick one: David provides a nice variant proof, but they argue that the characteristic polynomial of a matrix with integer entries has rational coefficients, when I believe integer matrices have integer monic characteristic polynomials. Furthermore, $V^{\oplus\dim V}\subseteq\Bbb C[G]$ is claimed - how are polynomials in $G$ being interpreted as elements of a product vector space here? I think I have a strong misunderstanding of the meaning of $\Bbb C[G]$.
