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.)

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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

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1 Answer

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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,

SG(x)={g∈G:gx=x}={e}=SH(x)

S_G(x) = \{g\in G : gx = x\} = \{e\} = S_H(x)

However,

g⋅x⋅g=x∀g∈G

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∈G}

\{(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.