Consider the group , by which I mean quotient of by its center. So this group has order . If an interested reader looks at the classification of groups of order :”apparently”, says this reader, “this group is isomorphic to !”

Now naturally acts on a set of 5 elements. On the other hand, naturally acts on a set of 6 elements, namely the points of .

The brain teaser is: how to find a natural action of on a set of 6 elements, and how to find a natural action of on a set of 5 elements? I couldn’t find these in a nice way. For instance, we can let act on (the projectivization of) degree 4 homogeneous polynomials (with coefficients in ). The orbit of consists of 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?

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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Sorry that I didn’t make myself clear. I was trying to figure out a “natural” action of:

1. A_5 on 6 elements;

2. PSL_2(F_5) on 5 elements.

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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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A5 contains the dihedral group D10 as a subgroup of index 6, and you can let A5 act on the quotient space…

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…and for your other question, you can use the exceptional isomorphism between SL2(F_4) and PSL2(F_5, and then just let SL2(F_4) act on P^1(F_4).

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but how do we know this two groups are the same without identifying them with A_5?

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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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