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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.

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1 Answer 1

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 $G$.

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

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