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 $|G_{\alpha} \leq d!$


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 $d!$.


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.