Alternative ways to determine the centre of $GL_n(K)$ Does anyone know a non calculative way the determine the centre of $GL_n(K)$?
Could we for example classify all normal subgroups of $GL_n(K)$ and unite them? (I know this will not work [look at $SL_n(K)$ for example, this normal subgroup is allready to big] it's just an example of the kind of proof I'm looking for).
I thought about this problem for some hours and I guess there just is no way to proof that the centre is $\{\lambda E , \lambda \in K^*\}$ without "opening the group" and doing some matrix calculations.
 A: First, you can show that the center of $M_n(k)$ is $k$ (embedded as the diagonal matrices) using the double centralizer theorem. This part is very conceptual. Next, you can show that any element in the center of $GL_n(k)$ must lie in the center of $M_n(k)$ by showing that every element of $M_n(k)$ is a linear combination of elements of $GL_n(k)$. 
If $k$ is an infinite field this is easy: then if $X \in M_n(k)$ is any matrix the polynomial $\det (X + t I)$ is nonzero and hence takes nonzero values for all but finitely many nonzero values $t \in k$. If $t_0 \in k$ is nonzero and such that $\det (X + tI) \neq 0$, then $X$ is the difference of the invertible matrices $t I$ and $X + t I$.
I don't have quite so clean a way to handle the finite case. It suffices to write down a basis of $M_n(k)$ consisting of invertible matrices. For starters we can consider the identity matrix $I$ together with matrices of the form $I + E_{ij}$ where $E_{i,j}$ is the matrix with a $1$ in the $i,j$ entry and $0$ otherwise, and also $i \neq j$. Next, to handle the diagonal, if the characteristic of $k$ isn't equal to $2$ we can consider the diagonal matrices with entries $(1, -1, 1, \dots), (1, 1, -1, \dots)$, etc. If the characteristic is $2$ we can consider matrices of the form $I - E_{i,i} + E_{(i-1),i} + E_{i,i-1}$ for $i = 2, \dots n$. 
