$\operatorname{GL}(n,F)$ 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?
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1$\begingroup$ @anon: the copy of $F^\times$ here is not (typically) central. It is the copy consisting of diagonal matrices in which all but the top-left entry agrees with the identity matrix. so { [t,0,0;0,1,0;0,0,1] : t in F, t≠0 } is an example with n=3. $\endgroup$– Jack SchmidtFeb 9, 2014 at 22:01
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$\begingroup$ I am a bit confused about the multiplicative group of a finite field for example GF(4)? Would you please more details in this case? and another question : if we assume the identity automorphism semidirect is direct product, can we conclude that F× is a normal subgroup of GL(n,F) as well? $\endgroup$– NilFeb 9, 2014 at 22:02
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2$\begingroup$ @Jack You are right. OP: A semidirect product $H\rtimes K$ presupposes no homomorphism at all between $H$ and $K$. Rather, there is a map $K\to\color{Red}{\rm Aut}(H)$. $\endgroup$– anonFeb 9, 2014 at 22:04
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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.
answered Feb 9, 2014 at 22:34