To recap, we have achieved a set of functions:
That is, we have found a way elements in
p(z) = -1/z [order 2]
q(z) = 1/(1-z) [order 3]
r(z) = (z - 1) [order 5]
r = -1/[1/(1-z)] = p . q
PSL(2, 5) such that
p^2 = q^3 = (pq)^5 = 1.
This gives us the [surjective ] map from
PSL(2, 5) into
- By a cardinality argument, we know that the size of
PSL(2, 5) is 60. Hence, since
PSL(2, 5) and
A5 have the same number of elements, this map must be a bijection.
§ Step 2: PSL(2, 5) is simple
TODO! I'm still reading Keith Conrad's notes.