Let $G$ be a permutation group acting transitively on $\Omega$ and suppose $N \unlhd G$ is a normal subgroup of $G$. Assume that for $g \in N_g(G_{\alpha})$ we have $$ G_{\alpha} \cap G_{\alpha}^g = 1 $$ for some point stabilizer $G_{\alpha}$. Also assume that every nontrivial element either has no fixed point or exactly $p$ fixed point for some prime $p$. Also suppose $N$ has order coprime to $p$.

The orbits of $N$ form a system of blocks, and we have a natural action of $G$ on these blocks. Under the stated assumptions, do we have that the kernel of this action on the orbits of $N$ is precisely $N$, i.e. $$ \forall \alpha \in \Omega: (\alpha^N)^g = \alpha^N \quad \Leftrightarrow \quad g \in N. $$ Does this hold and how to proof it? In general this is not true, as could be seen here.


The dihedral group of order $12$ in its natural action on $6$ points, $$G = \langle (1,2,3,4,5,6), (1,4)(2,3)(5,6) \rangle$$ with $p=2$ is a counterexample.

We can take $N = \langle (1, 5, 3)(2, 6, 4) \rangle$ of order $3$. The kernel of the action on the orbits of $N$ is the subgroup $\langle N,(2, 6)(3, 5) \rangle$ of order $6$.

More generally, for any distinct primes $p,q$ with $q>2$, the group $D_{2q} \times C_p$ has a transitive permutation representation of degree $pq$ and a normal subgroup $N$ of order $q$ such that the kernel of the action on the orbits of $N$ is $D_{2q}$.

| cite | improve this answer | |
  • $\begingroup$ Thanks you for your clear counterexample! Maybe for some context see my other recent question, why I thought this might hold (but as I know now it does not): math.stackexchange.com/questions/1567426/… $\endgroup$ – StefanH Dec 9 '15 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.