If K is a finite field, proof that $Gl_n(K)$ is not commutative

Question

The following property was stated during a lecture in Algebra:

If K is a finite field and $n \ge 2$ then $Gl_n(K)$ is a non-abelian finite group.

I know how to proof that $Gl_n(K)$ is finite but, is it true that it is not commutative? how could I proof it?

What do you mean you known how to prove it is commutative. That is precisely what you need to show it is not. And showing this is just a matter of picking two matrices suitably. — Tobias Kildetoft, Mar 04 '16 at 11:11
@Tobias Kildetoft I'm sorry I meant "I know how to proof it is finite" — user1868607, Mar 04 '16 at 11:18

score 3 · Answer 1 · answered Mar 04 '16 at 11:28

I could give you two matrices, that do not commute over any field, but I think the following hint gives you the chance to find such matrices by yourself.

Hint: $X\begin{pmatrix}0&1\\1&0\end{pmatrix}$ swaps the columns of $X$, while $\begin{pmatrix}0&1\\1&0\end{pmatrix}X$ swaps the rows of $X$. Hence you only have to find some $X$, where swapping columns and rows respectively produces different matrices.

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

The easiest way to see that the group $GL(n,K)$ is commutative if and only if $n=1$ is to produce matrices $A$ and $B$ with $AB\neq BA$ for $n\ge 2$. Another way is to use that we know that the center $Z$ of $GL(n,K)$ is equal to $K^{\times}\simeq \{ \lambda I_n\mid \lambda\neq 0 \}$ for all $n\ge 1$. Several proofs on MSE can be found, e.g., see here. However, a group $G$ is abelian if and only if $Z(G)=G$. This is for $GL(n,K)$ only the case when $n=1$.

If K is a finite field, proof that $Gl_n(K)$ is not commutative

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