My textbook gives Schur's Lemma as follows:
Let $T_1: G \to GL(V_1)$ and $T_2: G \to GL(V_2)$ be two irreducible representations of some group $G$, and $A: V_1 \to V_2$ an intertwining operator between the two operators. Then either $A = 0$ and $A: V_1 \to V_2$ is an isomorphism between $V_1$ and $V_2$ (this latter case means, by definition, that the two representations are equivalent).
It then gives the following corollary, where I have bolded things I am unsure about/have added because I think they have been incorrectly omitted (and ask questions thereafter):
Let $T : G \to GL(V)$ be an irreducible representation of a finite group $G$ on a finite-dimensional complex vector space $V$ and $A : V \to V$ an operator that commutes with all $T (g)$; $AT(g) = T(g)A$ for all $g \in G$. Then $A = \alpha \textrm{id} \,V$ for some complex scalar $\alpha$.
I believe there are two problems with the corollary as stated, but wanted to check:
A) Surely we must state that $T$ is irreducible in the corollary's hypothesis (since Schur's Lemma is used in the proof)?
B) Why is the hypothesis that $G$ is finite presented? As far as I can tell, that hypothesis was not used in the proof. I wonder if instead one is meant to use that $V$ is finite-dimensional, since one uses that $A$ has an eigenvalue and this is only true if $V$ is finite-dimensional?
I then also have the following corollary, where again I have bolded things I am unsure of/have added because I think they have been incorrectly omitted :
If $G$ is a finite abelian group then all its irreducible representations on complex finite-dimensional vector spaces are one-dimensional.
C) This depends on the answer to the above, but I believe that since I am using the first corollary to Schur's Lemma I need the bolded hypotheses. Again, I am not sure why $G$ being finite is relevant.
Best Answer
Firstly, there is a general version of Schur's lemma that only relies on irreducibility, and not on any finiteness conditions or anything about the base field.
The proof is almost by definition of an irreducible representation. For example, by $G$-equivariance of $f$, the subspace $\ker f \subseteq V$ is a $G$-stable subspace, and then in (1) by irreducibility of $V$, we must either have $\ker f = V$ (hence $f$ is zero) or $\ker f = 0$ (hence $f$ is injective). The proofs of (2) and (3) are similar.
The next version of Schur's lemma holds for finite-dimensional irreducible representations over an algebraically closed field. $G$ can be infinite, but also note that if $G$ is finite, then any irreducible representation is automatically finite-dimensional.
The proof goes: since $k$ is algebraically closed and $V$ is finite-dimensional, $f$ has some eigenvalue $\lambda \in k$. The linear map $f - \lambda \operatorname{id}_V$ is then a $G$-equivariant map between irreducible representations, but it is not invertible, so it is zero by the general version of Schur's lemma. Hence $f = \lambda \operatorname{id}_V$.
The last result about abelian groups follows from the algebraically closed version of Schur's lemma. Again it needs no finiteness conditions on $G$.
The proof goes like this. For any $g \in G$ we can define the left-multiplication map $L_g: V \to V$ by $L_g(v) = gv$. Since $G$ is abelian, $L_g$ is a $G$-equivariant map: for any $h \in G$ we have $L_g(hv) = ghv = hgv = h L_g(v)$. Applying the algebraically closed version of Schur's lemma, we see that $L_g$ is a scalar, and therefore every element of $G$ just scales vectors. So for any nonzero $v \in V$, the subspace spanned by $v$ is a subrepresentation, and by irreducibility must be the whole space.