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

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Statement

The following are the cases where a Projective special linear group (?) is isomorphic to an 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.

Proof

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.

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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.