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Rather light stylistic improvements. One mathematical change: it seems wrong to me that the OP spoke of the 'kernel of the action on the vertex set of the graph whose vertex set is stabilized'. My new formulation seems standard, and what the OP really means.
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Peter Heinig
Peter Heinig
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Suppose $\Gamma$ is a simple graphgraph and $G=Aut(\Gamma)$$G=\mathrm{Aut}(\Gamma)$ is the automorphism group of $\Gamma$. If $G$ stabilizes a subgraph $\Gamma_1$,, and $G_0$ is the kernelpoint-wise stabiliser of the set $V(\Gamma_1)$ w.r.t. the action of $G$ on $\Gamma_1$$V(\Gamma)$, that is, $G_0=\cap_{x\in V(\Gamma_1)}Stab_G(x) $$G_0=\cap_{x\in V(\Gamma_1)}\ \mathrm{Stab}_G(x) $ and $G_1=Aut(\Gamma_1)$$G_1=\mathrm{Aut}(\Gamma_1)$, is it true that $G$ is the semidirect product of $G_0$ and $G_1$?

Suppose $\Gamma$ is a simple graph and $G=Aut(\Gamma)$ is the automorphism group of $\Gamma$. If $G$ stabilizes a subgraph $\Gamma_1$ and $G_0$ is the kernel of action of $G$ on $\Gamma_1$, that is, $G_0=\cap_{x\in V(\Gamma_1)}Stab_G(x) $ and $G_1=Aut(\Gamma_1)$, is it true that $G$ is the semidirect product of $G_0$ and $G_1$?

Suppose $\Gamma$ is a simple graph and $G=\mathrm{Aut}(\Gamma)$ is the automorphism group of $\Gamma$. If $G$ stabilizes a subgraph $\Gamma_1$,, and $G_0$ is the point-wise stabiliser of the set $V(\Gamma_1)$ w.r.t. the action of $G$ on $V(\Gamma)$, that is, $G_0=\cap_{x\in V(\Gamma_1)}\ \mathrm{Stab}_G(x) $ and $G_1=\mathrm{Aut}(\Gamma_1)$, is it true that $G$ is the semidirect product of $G_0$ and $G_1$?

fixed grammar
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Colin Reid
Colin Reid
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Group automorphism Automorphism group of a graph

Suppose $\Gamma$ is a simple graph and $G=Aut(\Gamma)$ is the group automorphism group of $\Gamma$. If $G$ stabilizestabilizes a subgraph $\Gamma_1$ and $G_0$ is the kernel of action of $G$ on $\Gamma_1$, I meanthat is, $G_0=\cap_{x\in V(\Gamma_1)}Stab_Gx $$G_0=\cap_{x\in V(\Gamma_1)}Stab_G(x) $ and $G_1=Aut(\Gamma_1)$, is it true $G=G_0:G_1$ wherethat :$G$ is the semidirect product of $G_0$ and $G_1$?

Group automorphism of a graph

Suppose $\Gamma$ is a simple graph and $G=Aut(\Gamma)$ is the group automorphism of $\Gamma$. If $G$ stabilize subgraph $\Gamma_1$ and $G_0$ is the kernel of action of $G$ on $\Gamma_1$, I mean $G_0=\cap_{x\in V(\Gamma_1)}Stab_Gx $ and $G_1=Aut(\Gamma_1)$, is it true $G=G_0:G_1$ where : is semidirect product?

Automorphism group of a graph

Suppose $\Gamma$ is a simple graph and $G=Aut(\Gamma)$ is the automorphism group of $\Gamma$. If $G$ stabilizes a subgraph $\Gamma_1$ and $G_0$ is the kernel of action of $G$ on $\Gamma_1$, that is, $G_0=\cap_{x\in V(\Gamma_1)}Stab_G(x) $ and $G_1=Aut(\Gamma_1)$, is it true that $G$ is the semidirect product of $G_0$ and $G_1$?

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maryam
maryam
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Group automorphism of a graph

Suppose $\Gamma$ is a simple graph and $G=Aut(\Gamma)$ is the group automorphism of $\Gamma$. If $G$ stabilize subgraph $\Gamma_1$ and $G_0$ is the kernel of action of $G$ on $\Gamma_1$, I mean $G_0=\cap_{x\in V(\Gamma_1)}Stab_Gx $ and $G_1=Aut(\Gamma_1)$, is it true $G=G_0:G_1$ where : is semidirect product?