If $k^n$ no simple $kG$ module, then we can find a basis in a special form.

Question

I'm reading Lam's book "A first course in noncommutative rings". In the proof of Burnside's First Theorem ((9.4), p151). The statement says:

Let $k$ be a field of characteristic $p \geq 0$ and let $G$ be a subgroup of $GL_n(k)$. If $G$ has exponent $N < \infty$ and $p \not| N$, then $|G| \leq N^{(n^3)}$.

In the proof, we can assume that $k$ is algebraically closed.

On a certain moment, the proof makes the following claim:

Assume $k^n$ is a non-simple $kG$-module. Then, after choosing a suitable basis for $k^n$ we may assume that the elements of $G$ have the form

$$\begin{pmatrix}g_1 \ h \\ 0 \ g_2\end{pmatrix}$$ where $g_1, g_2$ are square matrices of fixed size, say $n_1$ and $n_2$. Moreover, $g_i \in GL_{n_i}(k)$ for $i=1,2$.

Why is the above true? How do we choose this basis? Why can we assume our elements change?

Answer 1

I believe these matrices are intended to operate on the right of row vectors in $k^n$.

How do we choose this basis?

If $k^n$ is a nonsimple $kG$ module, it has a proper $kG$ submodule, call it $N$. After fixing a $k$-basis of $N$, we can go about figuring out just how $G$ acts on $N$, and those will become the matrices that go into the $g_2$ spot.

Now extend the $k$-basis for $N$ to all of $k^n$. Now we can work out representations of elements of $G$ that act on all of $k^n$. But look: this new matrix has to cause $N$ to be mapped into itself, since $N$ is a $kG$ submodule. Since that happens, it must be that the basis vectors not in $N$ have to be mapped to $0$, and this is what causes the lower left-hand block of zeros.

Why can we assume our elements change?

The elements don't change... just our representation of them changes. And after all, we are free to represent the vector space with whatever basis we prefer.

Why is $g_2\in GL_{n_2}(k)$?

Each element of $kG$ is acting as a $k$ linear transformation on the $n_2$ dimensional space $N$, and so it can be represented by a $n_2\times n_2$ matrix, i.e. $g_2$.

$g_1$ is an $n_1\times n_1$ matrix where $n_1=n-n_2$, just by your knowledge of the dimensions of the matrices and the size of that zero block. Using block-matrix arithemtic, you can easily show that the diagonal blocks are invertible, since the matrix is invertible (representing the invertible element $g\in G$.)