Page 99 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 99
mož Moravec: Some Topics in the Theory of Finite Groups 87
The remaining cases of Sn and An for 1 ≤ n ≤ 4 are somewhat exceptional, but easy
to deal with. We show here how to use GAP to examine these groups:
gap> for n in [ 1..4 ] do
> sn := SymmetricGroup( n );
> an := AlternatingGroup( n );
> Print("n = ", n, "\n");
> Print("A_n: ", StructureDescription( an ), " ", IsSimple( an ), "\n" );
> Print("S_n: ", StructureDescription( sn ), " ", NormalSubgroups( sn ), "\n" );
> od;
n=1
A_n: 1 false
S_n: 1 [ Group( () ) ]
n=2
A_n: 1 false
S_n: C2 [ SymmetricGroup( [ 1 .. 2 ] ), Group( () ) ]
n=3
A_n: C3 true
S_n: S3 [ SymmetricGroup( [ 1 .. 3 ] ), Group( [ (1,2,3) ] ), Group( () ) ]
n=4
A_n: A4 false
S_n: S4 [ SymmetricGroup( [ 1 .. 4 ] ),
Group( [ (2,4,3), (1,4)(2,3), (1,3)(2,4) ] ),
Group( [ (1,4)(2,3), (1,3)(2,4) ] ), Group( () ) ]
3.3.3 Simplicity of projective special linear groups
Unless stated otherwise, F will denote the Galois field GF(q ), where q is a prime power.
The projective space n−1(F ) is the set of all one-dimensional subspaces of F n . There are
q n − 1 non-zero vectors in F n , each of which spans a one-dimensional subspace. Each
such space is spanned by any of its q −1 non-zero vectors, hence | n−1(F )| = (q n −1)/(q −
1). The group GL(n , F ) acts on n−1(F ) from the left as follows: (A, span(v )) → span(Av ).
Proposition 3.3.15 The following conditions for A ∈ GL(n, F ) are equivalent:
1. A ∈ Z (GL(n, F ));
2. A is in the kernel of the action of GL(n , F ) on n−1(F );
3. A is a scalar matrix, i.e., A = λI for some λ ∈ F ×.
PROOF. Clearly (3) implies (1). To see that the converse holds, take A ∈ Z (GLn (F )). Then,
in particular, A has to commute with all matrices with 1 on the diagonal and the position
(i , j ), i = j , and zero elsewhere. Easy calculation then shows that A is a scalar matrix.
Let us prove that (2) and (3) are equivalent. Clearly every scalar matrix fixes all 1-
dimensional subspaces of F n . Conversely suppose that A fixes all 1-dimensional sub-
spaces. Let e1, . . . , en be a standard basis of F n . Then Aei = λi ei for some non-zero
λi ∈ F . Fix different i and j . There also exists λ ∈ F × such that A(ei + e j ) = λ(ei + e j ), and
this implies λ = λj = λi . Consequently, A is a scalar matrix.
The remaining cases of Sn and An for 1 ≤ n ≤ 4 are somewhat exceptional, but easy
to deal with. We show here how to use GAP to examine these groups:
gap> for n in [ 1..4 ] do
> sn := SymmetricGroup( n );
> an := AlternatingGroup( n );
> Print("n = ", n, "\n");
> Print("A_n: ", StructureDescription( an ), " ", IsSimple( an ), "\n" );
> Print("S_n: ", StructureDescription( sn ), " ", NormalSubgroups( sn ), "\n" );
> od;
n=1
A_n: 1 false
S_n: 1 [ Group( () ) ]
n=2
A_n: 1 false
S_n: C2 [ SymmetricGroup( [ 1 .. 2 ] ), Group( () ) ]
n=3
A_n: C3 true
S_n: S3 [ SymmetricGroup( [ 1 .. 3 ] ), Group( [ (1,2,3) ] ), Group( () ) ]
n=4
A_n: A4 false
S_n: S4 [ SymmetricGroup( [ 1 .. 4 ] ),
Group( [ (2,4,3), (1,4)(2,3), (1,3)(2,4) ] ),
Group( [ (1,4)(2,3), (1,3)(2,4) ] ), Group( () ) ]
3.3.3 Simplicity of projective special linear groups
Unless stated otherwise, F will denote the Galois field GF(q ), where q is a prime power.
The projective space n−1(F ) is the set of all one-dimensional subspaces of F n . There are
q n − 1 non-zero vectors in F n , each of which spans a one-dimensional subspace. Each
such space is spanned by any of its q −1 non-zero vectors, hence | n−1(F )| = (q n −1)/(q −
1). The group GL(n , F ) acts on n−1(F ) from the left as follows: (A, span(v )) → span(Av ).
Proposition 3.3.15 The following conditions for A ∈ GL(n, F ) are equivalent:
1. A ∈ Z (GL(n, F ));
2. A is in the kernel of the action of GL(n , F ) on n−1(F );
3. A is a scalar matrix, i.e., A = λI for some λ ∈ F ×.
PROOF. Clearly (3) implies (1). To see that the converse holds, take A ∈ Z (GLn (F )). Then,
in particular, A has to commute with all matrices with 1 on the diagonal and the position
(i , j ), i = j , and zero elsewhere. Easy calculation then shows that A is a scalar matrix.
Let us prove that (2) and (3) are equivalent. Clearly every scalar matrix fixes all 1-
dimensional subspaces of F n . Conversely suppose that A fixes all 1-dimensional sub-
spaces. Let e1, . . . , en be a standard basis of F n . Then Aei = λi ei for some non-zero
λi ∈ F . Fix different i and j . There also exists λ ∈ F × such that A(ei + e j ) = λ(ei + e j ), and
this implies λ = λj = λi . Consequently, A is a scalar matrix.
