§ Simplicity of A5 using PSL(2, 5)

§ Presentation of A5

We take as faith A5 has the presentation:
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.

§ Step 1: PSL(2, 5) is isomorphic to A5

PSL(2, 5) consists of projective Mobius transformations with function composition as the group operation. Here, we freely use the linear algebraic relationship between transformations of the form (az + b)/(cz + d) and matrices [a b; c z].
a,b,c,d∈Z5,ad−bc=1f:Z5∪{∞}→Z5∪{∞}f(z)≡(az+b)/(cz+d) \begin{aligned} &a, b, c, d \in \mathbb Z5, ad - bc = 1 \\ &f: \mathbb Z5 \cup \{ \infty \} \rightarrow \mathbb Z5 \cup \{ \infty \} \\ &f(z) \equiv (az + b)/(cz + d) \\ \end{aligned}
  • We allow coefficients for the Mobius transform to be from Z5\mathbb Z5, and we allow the domain and codomain of the function to be projectivized: so we add a point at infinity to Z5\mathbb Z5.
  • We construct a map from PSL(2,5)PSL(2, 5) to A5A5 and then show that this map is an isomorphism. We exploit the presentation of A5A5 to find elements a,b∈PSL(2,5)a, b \in PSL(2, 5) such that p2=q3=(pq)5=Ip^2 = q^3 = (pq)^5 = I. We can link this to the presentation of A5 which requires precisely those relations.
  • For an element of order 3, we pick q(z) = 1/(1-z).
q(z)=1/(1−z)q(q(z))=11−11−z=1(1−z)−11−z=(1−z)−z=(z−1)z=1−1zq(q(q(z)))=1−1q(z)=1−(1−z)=z \begin{aligned} &q(z) = 1/(1-z) \\ &q(q(z)) = \frac{1}{1 - \frac{1}{1-z}} \\ & = \frac{1}{\frac{(1-z) - 1}{1-z}} \\ & = \frac{(1-z)}{-z} = \frac{(z-1)}{z} \\ & = 1 - \frac{1}{z} \\ &q(q(q(z))) = 1 - \frac{1}{q(z)} = 1 - (1 - z) = z \end{aligned}
  • I don't know of a principled way to arrive at this choice of q(z), except by noticing that az + b does not work, and neither does 1/z. The next simplest choice is things of the form 1/(1-z). If there is a nicer way, I'd love to know.
  • For a function of order 55, we have to use the structure of the finite field somehow. We can consider the function r(z) = 1 + z. On repeating this 5 times, we wil get 5 + z = z. However, it is hard to connect r(z) = 1 + zto the previous choice of q(z) = 1/(1-z).
  • We use the same idea for r(z), and pick r(z) = z - 1. This will allow us to accumulate -1s till we hit a -5 = 0.
  • To get 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.
To recap, we have achieved a set of functions: 
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.
  • 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.

§ References