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Let $G$ be a group (say, finite) and let it act on a set $X$ (say, also finite). For every element $g \in G$, we can consider its action on $X$. My rather vague question is

What information about the sizes of orbits of $X$ under $G$, can we recover only from knowing $G$ and the sizes of orbits of $X$ under $g$ for each $g\in G$?

Perhaps a better phrasing is what information we can't recover if any. An example of two actions, that shows we can't recover everything will be a good start.

One simple observation is that Burnside's lemma shows that $|X/G|$ is the average of the number of fixed points of $g \in G$. Hence, this piece of information can be recovered (even without knowing the group structure, which is available to us). The question is, what else. I am mainly interested in $X^G$, the number of fixed points of $X$ under the whole of $G$.

One last remark. What I described amounts to saying that we know the action of every cyclic subgroup of $G$ and we want to recover (as much as possible from) the action of $G$. Perhaps we should look at a slightly bigger family of subgroups for this.

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If you know the sizes of the orbits under each $g$ then in particular you know the number of fixed points, so you know the character of the representation $\mathbb{C}[X]$ and hence the representation itself (up to isomorphism). This is a decent amount of information but it's known that you can't recover $X$ from it; two keywords are linearly equivalent ($G$-sets) and Gassmann triple or Sunada triple. – Qiaochu Yuan Jun 20 '14 at 7:57
Knowing all of the sizes of the orbits also implies, and is maybe equivalent to, knowing the (characters of the) representations $\mathbb{C} \left[ {X \choose k} \right]$ for all $k$...? – Qiaochu Yuan Jun 20 '14 at 8:00
up vote 7 down vote accepted

Let $G=\{1,a,b,ab\}$ be a Klein $4$-group, and consider the following two actions on $\{1,2,3,4,5,6\}$.

In the first action

$a \mapsto (1,2)(3,4)$, $b \mapsto (1,3)(2,4)$, $ab \mapsto (1,4)(2,3)$,

and in the second action

$a \mapsto (1,2)(3,4)$, $b \mapsto (3,4)(5,6)$, $ab \mapsto (1,2)(5,6)$.

In both actions, all three involutions have two orbits of length $2$ and two of length $1$. In the first action, the orbits of $G$ have lengths $4,1,1$, and in the second action they are $2,2,2$.

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Nice example Derek! – Nicky Hekster Jun 20 '14 at 8:25
I suspect there are lots of examples. Another one is $S_4$, which has two actions of degree $7$ that are equivalent in this sense, one having orbits of length $6,1$ and the other $4,3$. – Derek Holt Jun 20 '14 at 13:14

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