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?

7 thoughts on “a brain teaser”

1. prime57 says:

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. 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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1. prime57 says:

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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2. 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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3. …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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1. but how do we know this two groups are the same without identifying them with A_5?

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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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