Solutions To Matrix Groups Problems:

 1. For the group PSL(2,z) show that the identity element (e)

= s ⁴ =   t ³ .

Solution:

The identity element e =

[1 0]
[01]


While s =

(0 1)
(-1 0)

And t =

(0 -1)
(1 -1)

One therefore simply needs to multiply s x s x s x s in order to obtain e (using the rules of matrix multiplication) and t x t x t, to find e in the same.

2. (a) For the group sl(2) show that:(a) [h.f] = h*f - f*h = -2f

The "Casimir element", C, of sl(2) is defined according to:

h ²  / 2 + h + 2f*e

find the element

Solutions:

Again, using the elements h and f as defined, we multiply:

h*f - f*h =

(0 0)
(-2 0)

And since f =

(0 0)
(1 0)

It means that the relation: h*f - f*h = -2f


(b) By matrix multiplication, we determine:  h ²  / 2  =

(½ 0)
(0 ½)

And adding the matrix h yields:

(1½   0)
(0    -½)


Now, adding 2fe =

(0 0)
(0 2) 

Therefore:  C =

(1½    0)
(0    1½)

3. Show that the Klein Viergruppe, V4, is Abelian.

Solution:

This is easily shown by matrix multiplication for which we will find:

a*b = b*a =

(-1 0)
(0 -1)

And: a*c = c*a =

(-1 0)
(0 1)

And finally, b*c = c*b =

(1 0)
(0 -1)