A Universe of Sorts

§ 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].

\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 $\mathbb Z5$ , and we allow the domain and codomain of the function to be projectivized: so we add a point at infinity to $\mathbb Z5$ .

We construct a map from $PSL(2, 5)$ to $A5$ and then show that this map is an isomorphism. We exploit the presentation of $A5$ to find elements $a, b \in PSL(2, 5)$ such that $p^2 = q^3 = (pq)^5 = I$ . 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).

\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 $5$ , 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.

§ Simplicity of A5 using PSL(2, 5)

§ Presentation of A5

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

§ Step 2: PSL(2, 5) is simple

§ References

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