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The projective special linear groups $PSL(2,4)$, $PSL(2,5)$ and $PSL(2,9)$ have the property that their orders equal the order of an alternating group. They are also isomorphic to the respective alternating groups.

In this case, we have that $|PSL(2,4)| = |PSL(2,5)| = |A_5|\ $ and $|PSL(2,9)| = |A_6|$. Let $F$ be a finite field. The order of $|PSL(2,F)|$ is given by $(2^n - 1)2^n(2^n + 1)$ when $F$ is of characteristic $2$. Otherwise it is equal to $\frac{1}{2}(p^n - 1)p^n(p^n + 1)$, where $p$ is the characteristic of $F$.

I've been wondering about the following question: when is $|PSL(2,F)| = |A_k|$? In other words, for which $n$ and $k$ the equations

\begin{align*} & 2^{n+1}(2^n - 1)(2^n + 1) = k!\\ &(p^n - 1)p^n(p^n + 1) = k! \text{, where p is an odd prime} \end{align*}

have solutions? Are there only finitely many solutions? And to generalize, what about $PSL(m, F)$?

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One other silly example is PSL(2,3) = A4, which is not simple so not as interesting, but still comes up a fair amount: SL(2,3) is horribly weird and avoiding it can make a group much easier to understand. – Jack Schmidt Nov 28 '11 at 21:09
up vote 5 down vote accepted

This is exactly the topic of Artin (1955a). You'll find proofs of Ted's assertions there. Artin (1955b) handles the other simple groups known at the time. Garge (2005) is handy as it also handles semi-simple groups (so in particular, one can mostly identify a chief factor solely by its order, other than the known problem of |Bn|=|Cn| and |A8| = |PSL(3,4)|).

  • Artin, Emil. "The orders of the linear groups." Comm. Pure Appl. Math. 8, (1955). 355–365. MR70642 DOI:10.1002/cpa.3160080302
  • Artin, Emil. "The orders of the classical simple groups." Comm. Pure Appl. Math. 8 (1955), 455–472. MR73601 DOI:10.1002/cpa.3160080403
  • Garge, Shripad M. "On the orders of finite semisimple groups." Proc. Indian Acad. Sci. Math. Sci. 115 (2005), no. 4, 411–427. MR2184201 DOI:10.1007/BF02829803
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What about PSL(3,4) and A8? – user641 Nov 28 '11 at 18:47
@Steve: bah, I thought Ted had given the "right" one like a mind-reader, rather than giving the isomorphic ones like in the original question. – Jack Schmidt Nov 28 '11 at 21:06
Thank you both for answering. The first paper by Artin is exactly what I was looking for.. too bad it is behind a paywall. – Mikko Korhonen Nov 30 '11 at 18:42

In addition to the ones you found, there is $|PSL(4,2)| = |A_8|$.

Edit (based on comment by Jack Schmidt):

There is also $|PSL(3,4)| = |A_8|$, and $|PSL(2,3)| = |A_4|$.

$PSL(4,2) \cong A_8$, and $PSL(2,3) \cong A_4$, but $PSL(3,4) \not \cong A_8$.

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@Jack Schmidt : Updated. – Ted Nov 29 '11 at 3:32

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