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If a group of order 15 acts on a set of order 22 and there are no fixed points. How many orbits are there?

I know the group action corresponds to a homomorphism from G into $S_{22}$

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WHatare the possible sizes of the stabilizers? –  Mariano Suárez-Alvarez Jan 8 '12 at 20:23
    
Well stabilizer is a subgroup of G right so 1,3,5 or 15, but if it was 15 then we would have a fixed point which we don't. So 1,3, or 5. –  user9352 Jan 8 '12 at 20:35
    
Ok, so what are the possible sizes of the orbits, then? –  Mariano Suárez-Alvarez Jan 8 '12 at 20:37
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15,5 or 3 so must be 4 orbits of order 3 and 2 of order 5? –  user9352 Jan 8 '12 at 20:45
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Indeed. You have just answered your own question. You should write in more detail what you did in the answer box. –  Mariano Suárez-Alvarez Jan 8 '12 at 21:00

1 Answer 1

Mariano's comment is the key point, but maybe you need more of a push. Let's call the group $G$ and the set $S$.

$S_{22}$ is a pretty large group, so my mind wouldn't jump there. Rather, the orbits of the action partition $S$, and if $s \in S$ then the size of the orbit $Gs$ containing $s$ is the index $(G : G_s)$ of the stabilizer subgroup of $s$. Using Lagrange's theorem and the fact that $G$ has no fixed points (What restriction does this place on the order of $G_s$?) we get three possibilities for the index.

Now this becomes a familiar problem: how many ways can you make $22$ using addition and these three numbers? There should be only one.

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In the last sentence I mean "up to reordering". –  Dylan Moreland Jan 8 '12 at 20:42

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