Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently in the process of reading an article by D.Bundy The connectivity of commuting graphs. In section 3 (in the Preliminary Results) Bundy gives the following result:

$\mathbf{(3.1)}$ Let $G=\operatorname{Sym}(n)$ and $H$ be the stabilizer in $G$ of a system of imprimitivity with blocks of size $s$, for $1<s<n$. Then $H$ is a maximal subgroup of $G$.

Proof. Elementary. $\Box$

I'm afraid I fail to see how to prove this. Indeed, suppose that $1<s<n$ and that $st=n$ for some $1<t<n$. Then we have that $H\cong \operatorname{Sym}(s)\wr\operatorname{Sym}(t)$, so the result is equivalent to proving that if $1<t,s<n$ with $ts=n$, then the copy of $\operatorname{Sym}(s)\wr\operatorname{Sym}(t)$ contained in $\operatorname{Sym}(n)$ is maximal in $\operatorname{Sym}(n)$. Any help on seeing why this is true would be greatly appreciated.

share|cite|improve this question
up vote 6 down vote accepted

I'll write $S_n$ rather than ${\rm Sym}(n)$. Let $H$ be the natural copy of $S_s \wr S_t$ in $S_n$, and let $H < G \le S_n$. We want to prove that $G=S_n$.

The stabilizer $H_\alpha$ of a point $\alpha$ in $H$ has three orbits, or lengths 1, $s-1$ and $s(t-1)$. If the two nontrivial orbits are fused in $G$ to a single orbit, then $G$ is 2-transitive. But then, since $H$ contains transpositions (as elements of $S_s$), $G$ contains all transpositions and hence is equal to $S_n$.

Otherwise $G_\alpha$ has the same three orbits as $H_\alpha$. Since $H_\alpha$ acts as $S_{s-1}$ on the orbit of length $s-1$, the action of $G_\alpha$ on the orbit of length $t(s-1)$ must strictly contain that of $H_\alpha$. This is not possible when $t=2$ (since then $H_\alpha$ acts as $S_s$ on that orbit), so $t>2$, and we can assume by induction that $G_\alpha$ acts as $S_{t(s-1)}$ on that orbit. But now, if we consider $H_\beta$ and $G_\beta$ for a point $\beta$ in that orbit, the orbit of $H_\beta$ of length $s-1$ is contained within that orbit, and hence is strictly contained in an orbit of $G_\beta$. So $G_\beta$ has only two orbits, which is a contradiction, because $G_\alpha$ and $G_\beta$ are conjugate in $G$.

share|cite|improve this answer
Thanks for a nice, concise and helpful answer. – David Ward Jan 5 '13 at 9:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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