Generation of permutation groups by fixed elements subgroups

Question

Suppose $(H,X)$ is a permutation group (with $H$ a group acting faithfully on the set $X$). Under what circumstances is $H$ generated by its subgroups $H_x$, where $H_x$ is the subgroup of $H$ fixing $x \in X$, and $x$ varies over $X$ ?

Of course, there are plenty of examples where this cannot work, such as sharply transitive permutation groups (and certainly many many more in the non-transitive case).

Side questions:

• If $H$ is transitive, but not sharply transitive, is the answer: "always" (even in the infinite case) ?

• How important are finiteness conditions ?

Answer 1

We looked at this question during our research retreat and obtained the following characterisation: If $H$ is transitive on $X$ then it will be generated by its point stabilisers if and only if it does not have a proper system of imprimitivity upon which it acts regularly. By proper I mean one with at least two parts and by regular I mean that any element of $H$ that fixes one block in the partition fixes all of the blocks in the partition.

The proof is as follows: Let $N$ be the normal subgroup of $H$ that is generated by all the point stabilisers. If $N\neq H$ then since $H_x\leqslant N< H$ for any $x\in X$ it follows that $N$ is intransitive and so its set of orbits forms a proper system of imprimitivity. If $B$ is one orbit of $N$ then $H_B$ acts transitively on $B$. Since $N$ also acts transitively on $B$ we have that $H_B=NH_x$ for some $x\in B$. Thus $H_B=N$ and so $H_B$ fixes each $N$-orbit, that is, $H$ acts regularly on the set of $N$-orbits. Conversely, suppose that $\mathcal{B}$ is a proper system of imprimitivity upon which $H$ acts regularly. Let $K\neq H$ be the kernel of this action. Let $x\in X$ and $B$ be the block of $\mathcal{B}$ containing $x$. Then $H_x\leqslant H_B=K$. Since $K$ is normal and all point stabilisers are conjugate to $H_x$ it follows that $K$ contains all point stabilisers and so the subgroup generated by all point stabilisers is contained in $K$ and so is not equal to $H$.

Answer 2

This does not answer your question exactly, but I would like to point out a Theorem of H. Wielandt of this general nature, which I find appealing.

Wielandt generalized the famous theorem of Frobenius as follows: let $G$ is a finite group and let $H$ be a proper subgroup of $G$. Suppose that $H$ has a normal subgroup $H_{0}$ such that $H \cap H^{g} \leq H_{0}$ for all $g \in G \backslash H$. Then there is a normal subgroup $G_{0}$ of $G$ such that $G = HG_{0}$ and $H \cap G_{0} = H_{0}.$ This generalizes Frobenius' theorem which is the case $H_{0} = 1$. The proof may be found in (Curtis and Reiner, Representation theory of finite groups and associative algebras, 1962).

A consequence of this is that if $G$ is a finite simple primitive permutation group acting on the set $\Omega$ with non-trivial point stabilizer $H = G_{\alpha}$, then $G_{\alpha}$ is generated by the two-points stabilizers $G_{\alpha \beta} : \beta \neq \alpha \in \Omega$ (here, $\alpha$ is fixed).

To see this, note that the group generated by the above two-point stabilizers is $H_{0} = \langle H \cap H^{g} : g \in G \backslash H \rangle$ ( as $G$ is simple and primitive, $H$ is maximal and $H = N_{G}(H)$). This is a normal subgroup of $H$. Then clearly $H \cap H^{g} \leq H_{0}$ for all $g \in G \backslash H$, so there is $G_{0} \lhd G$ with $G = HG_{0}$ and $H \cap G_{0} = H_{0}.$ If $H_{0}$ is proper in $H$, then $G_{0}$ is proper in $G$. Since $H$ is proper and $G = G_{0}H$, we see that $G_{0}$ is non-trivial. Hence the simplicity of $G$ is contradicted. Thus we must have $H_{0} = H.$

Answer 3

Theorem: Suppose that $p$ is a prime number and $|X|=p$. Then every subgroup $H$ of $\text{Sym}(X)$ is generated by $\bigcup_{x\in X}H_{x}$ except for the cyclic subgroup of $\text{Sym}(X)$ of order $p$.

Proof: Let $Z=\langle\bigcup_{x\in X}H_{x}\rangle$. Suppose that $|X|=p$ and $H\subseteq\text{Sym}(X)$. Then let $h\in H$ be an element where $h^{p}\neq e$. Suppose that $h=\alpha_{1}\dots\alpha_{r}$ where $\alpha_{1},\dots,\alpha_{r}$ are disjoint cycles and suppose that $o(\alpha_{i})=n_{i}$ for each $i$. If $h$ has a fixed point, then $h\in H_{x}$ for some $x$. If $h$ does not have a fixed point, then $n_{1}+\dots+n_{r}=p$. Now, if $n_{i}|m$, then $h^{m}$ has a fixed point, so $h^{m}\in H_{x}$ for some $x$. In particular, for each list of integers $a_{1},\dots,a_{r}$, we have $h^{a_{1}n_{1}+\dots+a_{r}n_{r}}\in Z$. Now since $r>1$ and $n_{1}+\dots+n_{r}=p$, we conclude that $n_{1},\dots,n_{r}$ have no common prime factor. Therefore, there are integers $b_{1},\dots,b_{r}$ with $b_{1}n_{1}+\dots+b_{r}n_{r}=1$, so $h=h^{b_{1}n_{1}+\dots+b_{r}n_{r}}\in Z$. Therefore, since $Z$ contains all $q$-Sylow subgroups for primes $q$ with $q\neq p$ and where $q$ divides the order of $H$, we conclude that either $H=Z$ or $[H:Z]=p$. If $H=Z$, then the proof is complete, so assume that $[H:Z]=p$.

Observe that $Z$ is a normal subgroup of $H$. Now let $h\in H\setminus Z$. Then $o(hZ)=p$. Therefore, $p$ is a factor of $o(h)$, so $o(h)=p$. We conclude that if $h\in H$, then $o(h)=p$ if and only if $h\not\in Z$.

Let $h\in H\setminus Z$. Without loss of generality, assume that $h=(1,...,p)$. Let $z\in Z\setminus\{e\}$. Then there are some $i,j$ where $z(i)=i+j\mod p$ and where $j\neq 0\mod p$. Therefore, $z(i)=h^{j}(i)$, hence $i=h^{-j}(z(i))$. However, since $h^{-j}z$ has a fixed point, we cannot have $o(h^{-j}z)=p$. We therefore conclude that $Z\setminus\{e\}=\emptyset$, so $H$ must be a cyclic group of order $p$. Q.E.D.