```
```

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]`

.
$\begin{aligned}
&a, b, c, d \in \mathbb Z5, ad - bc = 1 \\
&f: \mathbb Z5 \cup \{ \infty \} \rightarrow \mathbb Z5 \cup \{ \infty \} \\
&f(z) \equiv (az + b)/(cz + d) \\
\end{aligned}$

- 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 + z`

to 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`-1`

s 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.

```
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.