Stabilizer of two compatible actions

If we have HH and GG that acts on a set XX on right and left, this is h.x∈Xh.x\in X and x.g∈Xx.g\in X (with h∈Hh\in H and g∈Gg\in G) that are compatible, this is h.(x.g)=(h.x).gh.(x.g)=(h.x).g with g∈Gg\in G, h∈Hh\in H and x∈Xx\in X.

So if we pick an x∈Xx\in X what can we say about: {(g,h)∈G×H|h.x.g=x}\{(g,h)\in G\times H | h.x.g=x\}

It contains the product of stabilizers SG(x)×SH(x)S_G(x)\times S_H(x) but we have the equality. If not how can we write the set above?

(We may assume that HH and GG and XX are all finite.)




In your compatibility relation, do you mean h⋅(x⋅g)=(h⋅x)⋅gh\cdot (x\cdot g) = (h\cdot x)\cdot g?
– Prahlad Vaidyanathan
2 days ago



Yes, sorry my mistake, thank you.
– João Dias
2 days ago


1 Answer


One example to show that equality may not hold in general: Let G=HG = H be an abelian group acting on itself by
h⋅x:=hx and x⋅g:=g−1x
h\cdot x:= hx \text{ and } x\cdot g := g^{-1}x

Then for any x∈Gx\in G,
S_G(x) = \{g\in G : gx = x\} = \{e\} = S_H(x)

g\cdot x\cdot g = x \quad\forall g\in G

and so the stabilizer you have written down contains the diagonal set
\{(g,g) : g\in G\}

As for how to write this set in general, I am not sure. If you have a specific example in mind, then do include that in the question, and perhaps that might have a meaningful description.