Projective special linear group equals alternating group in only finitely many cases

Statement
The following are the cases where a fact about::projective special linear group is isomorphic to an fact about::alternating group:


 * 1) $$PSL(1,k)$$ is isomorphic to the alternating group on one letter, and on two letters, for any field $$k$$.
 * 2) $$PSL(2,3)$$ is isomorphic to the alternating group of degree four.
 * 3) $$PSL(2,4)$$ and $$PSL(2,5)$$ are both isomorphic to the alternating group of degree five.
 * 4) $$PSL(2,9)$$ is isomorphic to the alternating group of degree six.
 * 5) $$PSL(4,2)$$ is isomorphic to the alternating group of degree eight.

The case of degree more than two
We first consider the case where $$m \ge 3$$. In this case, we prove that the only solution is $$m = 4, q = 2, n = 8$$.

The case of characteristic two
We now consider the case of fields of characteristic two. In this case, the group is:

$$\! PSL(2,2^r)$$

Its order is given by:

$$\! (2^r + 1)2^r (2^r - 1)$$.

For this to be isomorphic to the alternating group $$A_n$$, we must have:

$$\! (2^r + 1)2^{r+1}(2^r - 1) = n!$$.

Note that both $$2^r + 1$$ and $$2^r - 1$$ are both odd, so the largest power of $$2$$ dividing $$n!$$ is $$r + 1$$. This yields:

$$\! r + 1 \le n - 1$$.

Thus, $$\! r \le n - 2$$. We thus have:

$$\! n! = 2^{r + 1} (2^{2r} - 1) \le 2^{n-1}(2^{2n - 4} - 1) \le 2^{3n - 5}$$.

This puts a small bound on $$n$$, namely, $$n \le 14$$.as well as on $$r$$, namely $$r \le 12$$. A hand calculation shows that the only solutions are $$r = 2, n = 5$$ and $$r = 3,n = 8$$. A further check shows that in this case, the groups are indeed isomorphic.

The case of odd characteristic
We now consider the case of an odd prime, so the group is:

$$\! PSL(2,p^r)$$.

The order of the group is:

$$\! p^r(p^r + 1)(p^r - 1)/2$$.

For this to be isomorphic to the alternating group $$A_n$$, we get:

$$\! p^r(p^r + 1)(p^r - 1) = n!$$.

Clearly, $$p^r + 1$$ and $$p^r - 1$$ are relatively prime to $$p$$. Thus, the largest power of $$p$$ dividing $$n!$$ is $$p^r$$. This yields:

$$r \le \frac{n}{p-1}$$.

Thus, we get:

$$n! \le p^{3n/(p-1)}$$.

Note that since $$\! p^{3/(p-1)} < 8$$ for all primes $$p$$, we get:

$$\! n! < 8^n$$.

This puts a bound $$n \le 19$$. A hand calculation now yields that we have the following solutions:


 * $$\! n = 4, p = 3, r = 1$$.
 * $$\! n = 5, p = 5, r = 1$$.
 * $$\! n = 6, p = 3, r = 2$$.