5
$\begingroup$

$\operatorname{GL}(n,F)$ 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?

$\endgroup$
3
  • 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. $\endgroup$ Feb 9, 2014 at 22:01
  • $\begingroup$ 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? $\endgroup$
    – user40491
    Feb 9, 2014 at 22:02
  • 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 $K\to\color{Red}{\rm Aut}(H)$. $\endgroup$
    – anon
    Feb 9, 2014 at 22:04

1 Answer 1

7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.