General semilinear group of degree one: Difference between revisions

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Let <math>K</math> be a [[field]]. The '''general semilinear group of degree one''' over <math>K</math>, denoted <math>\Gamma L(1,K)</math>, is defined as the [[general semilinear group]] of degree one over <math>K</math>. Explicitly, it is the [[external semidirect product]]:
Let <math>K</math> be a [[field]]. The '''general semilinear group of degree one''' over <math>K</math>, denoted <math>\Gamma L(1,K)</math>, is defined as the [[general semilinear group]] of degree one over <math>K</math>. Explicitly, it is the [[external semidirect product]]:
<math>\Gamma L (1,K) = GL(1,K) \rtimes \operatorname{Aut}(K) = K^\ast \rtimes \operatorname{Aut}(K) </math>
where <math>GL(1,K) = K^\ast</math> is the [[multiplicative group of a field|multiplicative group]] of <math>K</math>, and <math>\operatorname{Aut}(K)</math> denotes the group of field automorphisms of <math>K</math>.
If <math>k</math> is the prime subfield of <math>K</math>, and <math>K</math> is a Galois extension of <math>k</math> (note that this case always occurs for <math>K</math> a finite field), then <math>\operatorname{Aut}(K) = \operatorname{Gal}(K/k)</math> and we get:


<math>\Gamma L (1,K) = GL(1,K) \rtimes \operatorname{Gal}(K/k) = K^\ast \rtimes \operatorname{Gal}(K/k) </math>
<math>\Gamma L (1,K) = GL(1,K) \rtimes \operatorname{Gal}(K/k) = K^\ast \rtimes \operatorname{Gal}(K/k) </math>


where <math>GL(1,K) = K^\ast</math> is the [[multiplicative group of a field|multiplicative group]] of <math>K</math>, <math>k</math> is the prime subfield of <math>K</math>, and <math>\operatorname{Gal}(K/k)</math> denotes the Galois group of <math>K</math> over <math>k</math>.
If <math>K</math> is a finite field of size <math>q</math>, this group is written as <math>\Gamma L(1,q)</math>.
==Particular cases==
 
===For a finite field===
 
Suppose <math>K</math> is a finite field of size <math>q</math>, where <math>q</math> is a [[prime power]] with underlying prime <math>p</math>, so that <math>q = p^r</math> for a positive integer <math>r</math>. <math>p</math> is the characteristic of <math>K</math>. In this case, <math>K^\ast</math> is cyclic of order <math>q - 1</math> (see [[multiplicative group of a finite field is cyclic]]) and <math>\operatorname{Gal}(K/k)</math> is cyclic of order <math>r</math> (generated by the Frobenius map <math>a \mapsto a^p</math>).
 
Thus, <math>\Gamma L(1,K)</math> is a metacyclic group of order <math>r(q - 1)</math> with presentation:
 
<math>\langle a,x \mid a^q = a, x^r = e, xax^{-1} = a^p \rangle</math>
 
(here <math>e</math> denotes the identity element).

Latest revision as of 02:08, 1 June 2012

Definition

Let K be a field. The general semilinear group of degree one over K, denoted ΓL(1,K), is defined as the general semilinear group of degree one over K. Explicitly, it is the external semidirect product:

ΓL(1,K)=GL(1,K)Aut(K)=KAut(K)

where GL(1,K)=K is the multiplicative group of K, and Aut(K) denotes the group of field automorphisms of K.

If k is the prime subfield of K, and K is a Galois extension of k (note that this case always occurs for K a finite field), then Aut(K)=Gal(K/k) and we get:

ΓL(1,K)=GL(1,K)Gal(K/k)=KGal(K/k)

If K is a finite field of size q, this group is written as ΓL(1,q).

Particular cases

For a finite field

Suppose K is a finite field of size q, where q is a prime power with underlying prime p, so that q=pr for a positive integer r. p is the characteristic of K. In this case, K is cyclic of order q1 (see multiplicative group of a finite field is cyclic) and Gal(K/k) is cyclic of order r (generated by the Frobenius map aap).

Thus, ΓL(1,K) is a metacyclic group of order r(q1) with presentation:

a,xaq=a,xr=e,xax1=ap

(here e denotes the identity element).