I have a finite group G which acts on a set X. I want to establish the number of distinct members of a subset Y $\subset$ X, however Y is not closed under G (I think that the correct terminology here is that Y is not an invariant subset of X?), so we can't use Burnside's Lemma.
The alternative, which is to count all the orbits of Y under G is ruled out because it is computationally out of reach: |G| = 3,359,232, while |Y| = 554,191,840,696.
My general question: is there any way we can reduce the complexity of the orbit-counting problem?
Supplementary question: I see this in the Wikipedia entry for "Burnside's Lemma":
The first step in the proof of the lemma is to re-express the sum over the group elements g $\in$ G as an equivalent sum over the set of elements x $\in$ X:
$\sum_{g \in G}|X^g| = |\{(g,x)\in G\times X : g.x = x\}| = \sum_{x \in X} |G_x|.$
Does this equivalence hold in general, ie would it apply to my subset Y, or does it only apply for invariant subsets, ie: when the Lemma itself is applicable?
My hope is that somehow I can still express the orbit-counting problem in terms of fixed points, which are computationally more accessible.