Page 100 - 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. 100
3.3 Finite simple groups
We define the projective general and projective special linear groups by
PGL(n, F ) = GL(n, F )/Z (GL(n, F ))
and
PSL(n, F ) = SL(n, F )Z (GL(n, F ))/Z (GL(n, F )).
Therefore the projective groups are the images of ther linear group counterparts in the
action on the projective space, so we can think of them as subgroups of Sym n−1(F ). We
see that | PGL(n,q )| = | GL(n,q )|/(q − 1) = | SL(n ,q )|.
Proposition 3.3.16 | PSL(n,q )| = | SL(n,q )|/ gcd(n,q − 1).
PROOF. The kernel of the action of SL(n ,q ) on the corresponding projective space con-
sists of scalar matrices with determinant one, i.e., matrices of the form λI with λn = 1.
The multiplicative group of GF(q ) is cyclic of order q − 1, so the number of solution of
λn = 1 is gcd(n ,q − 1).
If we restrict to the case n = 2, we see that 1(F ) has q + 1 points, so PGL(2,q ) and
PSL(2,q ) are subgroups of Sq+1. Let us consider some small cases:
q = 2: PSL(2, 2) = PGL(2, 2) is a subgroup of S3 of order 6, hence PSL(2, 2) =∼ S3.
q = 3: PGL(2, 3) is a subgroup of S4 of order 24, hence PGL(2, 3) = S4. The group PSL(2, 3)
is a subgroup of index 2 in PGL(2, 3), hence PSL(2, 3) ∼= A4.
q = 4: PGL(2, 4) = PSL(2, 4) is a subgroup of S5 of order 60, so it is isomorphic to A5; one
can double-check this with GAP:
gap> StructureDescription(PSL(2,4));
"A5"
q = 5: PSL(2, 5) ∼= A5:
gap> StructureDescription(PSL(2,5));
"A5"
We also remark here that there is another way of interpreting the actions of PGL(2, F ) and
PSL(2, F ) on the projective line. The one-dimensional subspaces of F 2 can be spanned
by either a unique vector of the form (1, x ), where x ∈ F , or the vector (0, 1). We iden-
tify points of the first type with F , and the point of the second type with ∞. Then the
elements of PGL(2, F ) can be identified with linear fractional maps
az +b
z→ ,
cz +d
where a ,b, c , d ∈ F , a d − b c = 0. The group PSL(2, F ) then consists of those linear frac-
tional maps with a d − b c = 1.
We will prove the following result:
We define the projective general and projective special linear groups by
PGL(n, F ) = GL(n, F )/Z (GL(n, F ))
and
PSL(n, F ) = SL(n, F )Z (GL(n, F ))/Z (GL(n, F )).
Therefore the projective groups are the images of ther linear group counterparts in the
action on the projective space, so we can think of them as subgroups of Sym n−1(F ). We
see that | PGL(n,q )| = | GL(n,q )|/(q − 1) = | SL(n ,q )|.
Proposition 3.3.16 | PSL(n,q )| = | SL(n,q )|/ gcd(n,q − 1).
PROOF. The kernel of the action of SL(n ,q ) on the corresponding projective space con-
sists of scalar matrices with determinant one, i.e., matrices of the form λI with λn = 1.
The multiplicative group of GF(q ) is cyclic of order q − 1, so the number of solution of
λn = 1 is gcd(n ,q − 1).
If we restrict to the case n = 2, we see that 1(F ) has q + 1 points, so PGL(2,q ) and
PSL(2,q ) are subgroups of Sq+1. Let us consider some small cases:
q = 2: PSL(2, 2) = PGL(2, 2) is a subgroup of S3 of order 6, hence PSL(2, 2) =∼ S3.
q = 3: PGL(2, 3) is a subgroup of S4 of order 24, hence PGL(2, 3) = S4. The group PSL(2, 3)
is a subgroup of index 2 in PGL(2, 3), hence PSL(2, 3) ∼= A4.
q = 4: PGL(2, 4) = PSL(2, 4) is a subgroup of S5 of order 60, so it is isomorphic to A5; one
can double-check this with GAP:
gap> StructureDescription(PSL(2,4));
"A5"
q = 5: PSL(2, 5) ∼= A5:
gap> StructureDescription(PSL(2,5));
"A5"
We also remark here that there is another way of interpreting the actions of PGL(2, F ) and
PSL(2, F ) on the projective line. The one-dimensional subspaces of F 2 can be spanned
by either a unique vector of the form (1, x ), where x ∈ F , or the vector (0, 1). We iden-
tify points of the first type with F , and the point of the second type with ∞. Then the
elements of PGL(2, F ) can be identified with linear fractional maps
az +b
z→ ,
cz +d
where a ,b, c , d ∈ F , a d − b c = 0. The group PSL(2, F ) then consists of those linear frac-
tional maps with a d − b c = 1.
We will prove the following result:
