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

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 $m\geq 3$. 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:

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

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

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

This puts a small bound on $n$, namely, $n\leq 14$.as well as on $r$, namely $r\leq 12$. 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:

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

Thus, we get:

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

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

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

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

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