If I find a nice proof of this isomorphism, or some other way to derive
the fact that PSL(2, 5)
is isomorphic to A5
, I will fill this up.
PSL(2, 5)
is isomorphic to A5
(az + b)/(cz + d)
and matrices [a b; c z]
.
q(z) = 1/(1-z)
. q(z)
, exceptby noticing that az + b
does not work, and neither does 1/z
. The nextsimplest choice is things of the form 1/(1-z)
. If there is a nicer way,I'd love to know.r(z) = 1 + z
. On repeating this 5times, we wil get 5 + z = z
. However, it is hard to connect r(z) = 1 + z
to the previous choice of q(z) = 1/(1-z)
. r(z)
, and pick r(z) = z - 1
. This will allowus to accumulate -1
s till we hit a -5 = 0
.r(z) = (z - 1)
, we need to compose q(z) = 1/(1-z)
with p(z) = -1/z
.This p(z)
is of order 2.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
That is, we have found a way elements in PSL(2, 5)
such that p^2 = q^3 = (pq)^5 = 1
.
This gives us the [surjective] map from PSL(2, 5)
into A5
.
PSL(2, 5)
is 60. Hence,since PSL(2, 5)
and A5
have the same number of elements, this mapmust be a bijection.