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new material (an outline of my argument) added
paul Monsky
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Simple groups with the same cardinality as PSL_2(Z/p)

In an undergrad honors algebra course it's sometimes shown that when $p$ is prime and $>3$ then $PSL_2(Z/p)$ is simple of of order $p(p-1)(p+1)/2$. But that this is the "only" simple group having that order is seldom or never (except when $p=5$) proved.

I've worked out a fairly simple proof, using the Burnside transfer theorem, of this last result, but it's perhaps a little too intricate to present in class.

QUESTION: Are there proofs of this result, on-line or in texts, that are appropriate for an undergrad honors algebra class? (If not, I might post an argument on arXiv).

EDIT: Since no simple available proof has yet been found, I'll sketch the argument that I culled from classification arguments for Zassenhaus groups.

Suppose G is simple of order p(p-1)(p+1)/2. First one shows that G has p+1 p-Sylows. Let S be the union of Z/p and {infinity}. An easy study of the conjugation action of G on the p-Sylows allows one to identify G with a doubly transitive group of (even) permutations of S, containing the p-cycle z-->z+1. Then the subgroup of G fixing both 0 and infinity is cyclic generated by z-->cz for some c. Once this is done, the key is in showing:

A.--- The subgroup of elements that either fix 0 and infinity or interchange them is dihedral.

Once A is shown it's not hard to show that z-->-1/z is in G, thereby identifying G with a fractional linear group. The proof of A is a counting argument when p=1 mod 4. But when p=3 mod 4 the situation is more delicate, and one uses Burnside tranfer.

paul Monsky
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