Page 101 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.

mož Moravec: Some Topics in the Theory of Finite Groups 89

Theorem 3.3.17 For n ≥ 2 and any field F , the group PSL(n, F ) is simple, except in the

two cases, n = 2, F = GF(2) or n = 2, F = GF(3).

We will only prove this theorem for n = 2, the proof for n > 2 is similar, but somewhat
technical. Our proof will rely on Iwasawa’s lemma applied to the action of G = SL(2, F ) on

1(F ). We will show in a series of steps that all the conditions of the lemma are satisfied.

Proposition 3.3.18 If n ≥ 2, then SL(2, F ) acts doubly transitively on 1(F ).

PROOF. Let span(v1) and span(v2) be two distinct 1-dimensional subspaces of F 2. For any
other pair span(w1) and span(w2) there exists a linear map that maps vi → wi , i = 1, 2.
One can modify this map to obtain one with determinant 1. We let the reader fill in the
details.

Let e1, e2 be a standard basis of F 2. Denote x = span(e1). The stabilizer of x is

stabG (x ) = {A ∈ SL(2, F ) | span(e1) = span(Ae1)}

= a b | a ∈ F×,b ∈ F .
0 1/a

There is an abelian normal subgroup of stabG (x ) given as follows:

U= 1 b |b ∈F .
0 1

Its elements are called transvections.

Proposition 3.3.19 The subgroup U and its conjugates generate SL2(F ).

PROOF. First we note that

0 1 −1 0 1 = 1 0 |b ∈F =U .
−1 0 −1 0 b 0
U

Now pick A = a b ∈ SL2(F ). Suppose first that b = 0. Then
c d

A= 1 0 1b 1 0 ∈ 〈U ,U 〉.
(d − 1)/b 1 01 (a − 1)/b 1

If c = 0, then

A= 1 (a − 1)/c 10 1 (d − 1)/c ∈ 〈U ,U 〉.
0 1 c1 0 1

Finally assume that b = c = 0. Then

A= 1 0 11 10 1 −1/a ∈ 〈U ,U 〉.
(1 − a )/a 1 01 (a − 1) 1 0 1

This proves the result.