Semi-direct product in general linear groups

Question

$\operatorname{GL}(n,F)$ can be written as a semidirect product : $\operatorname{GL}(n,F) = \operatorname{SL}(n,F) ⋊ F^\times$ where $F^\times$ is multiplicative group of the field $F$. According to the definition of semi direct product we must have a homomorphism between $\operatorname{SL}(n,F)$ and $F^\times$. How can we define this homomorphism?

Answer 1

This is not the correct piece of data for defining this semidirect product. The correct piece of data is a homomorphism $F^{\times} \to \text{Aut}(\text{SL}_n(F))$. This homomorphism can be written down as follows.

Theorem: Let $H$ be a normal subgroup of a group $G$, so that there is a short exact sequence

$$1 \to H \to G \to G/H \to 1.$$

Then $G$ can be written as a semidirect product $H \rtimes G/H$ iff this short exact sequence splits on the right in the sense that there is a map $r : G/H \to G$ which, after projecting back down to $G/H$, is the identity. In this case the action of $G/H$ on $H$ is the restriction of the action of $G$ on $H$ via conjugation to the image of $r$.

We of course have a short exact sequence

$$1 \to \text{SL}_n(F) \to \text{GL}_n(F) \xrightarrow{\text{det}} F^{\times} \to 1.$$

An example of a splitting of this short exact sequence is

$$F^{\times} \ni a \mapsto \left[ \begin{array}{ccc} a & 0 & \cdots \\ 0 & 1 & \cdots \\ \vdots & \vdots & \ddots \end{array} \right]$$

so the action is given by conjugation by this matrix.