# The kernel of an action on blocks, specifically the action on the orbits of normal subgroup

Let $G$ be a permutation group acting transitively on some set $\Omega$ and suppose we have a normal subgroup $N \unlhd G$. Then the orbits of $N$ form a system of blocks, and if $\Delta$ is such an $N$-orbit, then for its setwise stabilizer $$G_{\{\Delta\}} = \{ g \in G : \Delta^g = \Delta \}$$ we have $G_{\{\Delta\}} = G_{\alpha}N$ for some $\alpha \in \Delta$. As $\Delta = \alpha^N$ the inclusion $G_{\alpha}N \le G_{\{\Delta\}}$ is clear, for the other inclusion note that if $\Delta^g = \Delta$, then $\alpha^g = \alpha^n$, hence $g \in G_{\alpha}\cdot n$. Now $G$ acts on the $N$-orbits, as they form a system of blocks, and the kernel of this action is $$\bigcap_{\Delta \mbox{ some N-orbit}} G_{\{\Delta\}} = \bigcap_{\alpha \in \Omega} G_{\alpha} N.$$

Do we have $$\bigcap_{\alpha\in \Omega} G_{\alpha} N = N?$$

Surely as the action is faithful we have $\bigcap_{\alpha \in \Omega} G_{\alpha} = 1$, and we have $g \in \bigcap_{\alpha\in \Omega} G_{\alpha} N$ iff $$\forall \alpha \in \Omega \exists n \in N : \alpha^g = \alpha^n$$ i.e. for each $\alpha \in \Omega$ we have that $\alpha^g$ lies in the same $N$-orbits as $\alpha$.

## 1 Answer

We do not necessarily have that the kernel of the action on blocks is equal to $N$.

A small counterexample is $D_8$ (or $D_4$ if you prefer!) = $\langle (1,2,3,4),(2,4) \rangle \le S_4$. Let $N=Z(G) = \langle (1,3)(2,4) \rangle$, which has two orbits of length $2$. The kernel of the action on blocks has order $4$ and is $\langle (1,3), (2,4) \rangle$.

Note that in this example the kernel is equal to the block stabilizer $NG_1$.

Of course you get more obvious counterexamples in which $N$ is transitive, and then the kernel of the action is the whole of $G$.

• Thanks you for your example. I would also like to add that the point stabilizers $G_1 = G_3 = \langle (2 ~ 4) \rangle, G_2 = G_4 = \langle (1 ~ 3) \rangle$ have trivial intersection, so this would not be a sufficient condition to imply that precisely $N$ is the kernel. – StefanH Dec 8 '15 at 16:24