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Consider the group $S_3=\{1,x,x^2,y,xy,x^2y\}$ where $x=(123),\ y=(12)$. This group acts on the set $\mathcal P_3(S_3)$ of subsets of $S_3$ of cardinality 3 by left multiplication. Describe the orbits and the stabilizers of elements of $\mathcal P_3(S_3)$ under this action.

So there are 20 elements in $\mathcal P_3(S_3)$. Consider one of them, say $\{a,b,c\}$.

Assume its stabilizer is non-trivial. So it contains a non-trivial element $g$. Then $g$ cannot fix any of the elements $a,b,c$. So $g$ is a 3-cycle of order 3. Thus the stabilizer contains the subgroup $A_3=\{e,g,g^3\}$. It it true that the stabilizer is precisely $A_3$? How to show this? How to describe the orbits in this case?

If the stabilizer of $\{a,b,c\}$ is trivial, then its orbit has cardinality 6. How to describe the orbits explicitly in this case?

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To see that the stabilizer is precisely $A_3$: If the stabilizer were any larger, it would be $S_3$, and then this $\{a,b,c\}$ would be a fixed point of the action. But we can just pick some element $d \neq a,b,c$, and set $g = da^{-1}$, to get that $g\{a,b,c\}$ contains $d$, so $g\{a,b,c\} \neq \{a,b,c\}$. That is to say, this action does not have any fixed points. So the stabilizer can't be $S_3$.

If there is a $3$-cycle $g$ so that $g\{a,b,c\} = \{a,b,c\}$ then (up to relabeling elements) it must be $b = ga$ and $c = g^2a$. Then $\{a,b,c\} = \{a,ga,g^2a\} = \{e,g,g^2\}a = \langle g \rangle a$, the right coset of the subgroup $\langle g \rangle = A_3$. That should narrow down the possible orbit("s") in this case.

You should be able to figure out how many $\{a,b,c\}$ have stabilizer $A_3$ and hence orbit of size $2$. All other orbits have size $6$. The sizes of all orbits have to add up to $20$, so you can figure out how many $6$-element orbits there will be.

For any more information than that, I suspect you probably have to actually write out the elements.

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