Burnside's Theorem

Lemma

Density Theorem

Statement

Let $E$ be a finite-dimensional vector space over an algebraically closed field $k$, and $R$ be a subalgebra of $En{d}_k(E)$, i.e. $k\subset R\subset M_n(k)$. If $k^n(=E)$ is a simple $R$-module, then $R=En{d}_{R^{\prime }}(E)$.

Proof

Let $E=k^n$.

Step 1 : Show $k=En{d}_R(E)=R^{\prime }$.

We know that $En{d}_R(E)$ is a division ring by Schur’s lemma, and as a commutant $k$ is a central subalgebra. So $k\subset R^{\prime }$. However, since $k$ is an algebraically closed, for $\alpha \in R^{\prime }$, $k(\alpha )=k$. Hence $\alpha \in k$, which implies $R^{\prime }\subset k$.

Step2: $R=En{d}_k(E)=R^{\prime \prime }$

By the hypothesis, we have $R^{\prime \prime }\subset R$. So, we only need to show that $R\subset R^{\prime \prime }$. Let $A\in R^{\prime \prime }$ and choose a $k$-basis $\left\{e_1,\cdots ,e_n\right\}$ for $E=k^n$. By the density theorem, there exists $r\in R$ such that $A(e_i)=r{e}_i$ for $i=1,\cdots ,n$. So that $A$ is just a multiplying $r$ to each basis. Since $A$ is only determined by the choice the basis, therefore we can conclue $En{d}_k(E)=R$.

Remark

If $k$ is not algebraically closed, it does not holds. For example, let $k=\mathbb{R}$ and $R=\mathbb{C}\subset M_2\left(\mathbb{R}\right)$, then

$$i\mapsto \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\notin En{d}_{R^{\prime }}(E).$$

Example

Burnside Theorem

Meaning(?)

I think this theorem is the key point to go to the linear representation. We are builiding the representation space $kG$-module $M$, so we need to clarify the $kG$ action on $M$.

There is a comment, “the vector space is our hometwon in heart” in the certain book. If we can write $kG$ action as a matrix, then, in my opinion, I can handle it much more easily.