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 $G_1$?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
11
(Sorry for missing the "$G$ stabilizes $\Gamma_1$" condition in the previous edit)
The answer is no. Consider $\Gamma = ([4], \{12, 13, 23, 14\})$. $G$ stabilizes $\Gamma_1$ = the edge $14$. $G_0 = G$, and $|G_1| = 2$, while for the semiderict product we must have $|N \rtimes H| = |N| \times |H|$.
answered Sep 16, 2017 at 13:34
-
$\begingroup$ in your example it seems G does mot stabilize $\Gamma_1$! $\endgroup$– maryamCommented Sep 16, 2017 at 13:50
-
$\begingroup$ In new example I think G stabilize an edge but I wrote vertex! $\endgroup$– maryamCommented Sep 16, 2017 at 14:21
-
$\begingroup$ I think "$G$ stabilizes a subgraph $\Gamma$" is typically understood as "sets of vertices and edges of $\Gamma$ are invariant w.r.t $G$" (this is the difference between stabilizer and point-wise stabilizer of the set). In any case, both endpoints of the edge are stabilized too. $\endgroup$ Commented Sep 16, 2017 at 14:25
-
$\begingroup$ I do not mean point wise stabilizer, who edit my question? $\endgroup$– maryamCommented Sep 16, 2017 at 14:30
-