P.J. Cameron, C.E. Praeger, J. Saxl and G.M. Seitz: On the Sims conjecture and distance transitive graphs, Bulletin of the London Mathematical Society, 15 (1983), pp. 499–506, http://citeseerx.ist.psu.edu/viewdoc/download?doi=

In this paper where the Sims Conjecture was first proved, when the author discussed the case of a regular socle. There are a few points I am not sure I entirely get.

Suppose $N$ ist the socle of a primitive group $G$. Here is what the paper says:

$G_{\alpha}$-action on $N-\{1\}$ by conjugation and on $\Omega-\{\alpha\}$ are isomorphic. Now $G$ is primitive, so there are no non-trivial $G_{\alpha}$-invariant subgroups of $N$. (why is this?) Hence any $G_{\alpha}$-orbits on $N-\{1\}$ generates $N$. It follows that $G_{\alpha}$ is faithful on each orbit (why is this?) and so $|G_{\alpha} \leq d!$


1 Answer 1


First, you've left off a pretty important assumption from the paper: $N$ is regular.

For your first question, note that a subgroup $K\le N$ that was $G_\alpha$-invariant would lead to the subgroup $G_\alpha K$, lying strictly between $G_\alpha$ and $G$. This contradicts primitivity of the action / maximality of $G_\alpha$.

So of course, if we take any set $S\subset N$ and look at the orbit of $G_\alpha$ on this set, it will generate a $G_\alpha$-invariant subgroup of $N$. By the preceding paragraph, this must be all of $N$.

Now let $x\in N$ be a non-trivial element. The orbit of this element under $G_\alpha$ is $G_\alpha x$, and by the above it generates the subgroup $N$. What if $G_\alpha$ was not faithful? That would mean there was a $g\in G_\alpha$ that fixed $G_\alpha x$ pointwise. Since this set generates $N$, $g$ would fix $N$ pointwise. But the action on $N$ and on $\Omega$ are the same; in other words, $g$ would fix every point of $\Omega$. This means the action of $G$ is not faithful, contradiction.

Finally, since $d$ is a subdegree of $G$, and $G_\alpha$ is faithful on every orbit, $G_\alpha$ is faithful on the orbit of size $d$. This means $G_\alpha$ injects into $S_d$, and so has cardinality bounded above by $d!$.


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