a brain teaser

Consider the group G = PSL_2(\mathbb{F}_5), by which I mean quotient of SL_2(\mathbb{F}_5) by its center. So this group has order 60. If an interested reader looks at the classification of groups of order 60:”apparently”, says this reader, “this group is isomorphic to A_5!”

Now A_5 naturally acts on a set of 5 elements. On the other hand, G naturally acts on a set of 6 elements, namely the \mathbb{F}_5 points of \mathbb{P}^1.

The brain teaser is: how to find a natural action of A_5 on a set of 6 elements, and how to find a natural action of G on a set of 5 elements? I couldn’t find these in a nice way. For instance, we can let G act on (the projectivization of) degree 4 homogeneous polynomials (with coefficients in \overline{\mathbb{F}_5}). The orbit of x^4 - 2\sqrt{2} x^2 y^2 + y^4 consists of 5 elements (okay, let me not prove this here). So this must be it. But this is not satisfying. So the brain teaser is really: how can you realize these actions in a nice way?

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7 thoughts on “a brain teaser”

  1. If you allow an embedding A_4 –> A_5, then we can regard A_4 as a subgroup of G. The coset space G/A_4 is acted by G in a natural way, and it has 5 elements.

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      1. The one I wrote down is the most “natural” one you can get for (2), that is, you pick a maximal subgroup of index=5 of G=PSL_2(F_5), then you let G act on the corresponding quotient space. (This can be made canonical by viewing the space as the set of all subgroups of index=5.)

        You cannot construct other actions on 5 elements, because what you really get is “the” unique Steinberg representation.

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      1. Eh, I dunno man. Up to isomorphism, there’s only one group of order 60 with more than one Sylow 5-subgroup, and you can check that A_5, SL2(F_4), PSL2(F_5) all have this property…

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