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Let $G$ be a finite group acting on a finite set $X$, and $\chi$ be the corresponding permutation character over the field $\mathbb{C}$.

1)If $G$ acts transitively, then $\chi=\mathbb{1}+\theta$, ($\mathbb{1}$ is trivial character of $G$); moreover the trivial character appears only once in $\chi$.

2)If $G$ acts doubly transitively, then the character $\theta$ is irreducible.

(Ref. Linear Representations of finite groups -Serre)

Question: What can be said about $\chi$ when the action is primitive, $3$-transitive, or in general $n$-trnasitive?

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If the action is primitive, this means only that a point stabilizer is a maximal subgroup. I do not think that this puts much restriction on $\theta,$ but others may be able to say more than I can. In general, the character inner product $1 + \langle \theta, \theta \rangle$ is the rank of $G$ as a permutation group (when $G$ is transitive). This is the number of orbits that a point stabilizer $G_{\alpha}$ has on the points other than $\alpha.$ Higher transitivity puts further restrictions on how $\theta$ restricts to $G_{\alpha}$, but I at least do not know of many easily described properties in terms of $k$-transitivity for $k >2$ which can be attributed to $\theta$.

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While, as Geoff has said in his answer, the only nice properties of $\chi$ that come to mind if the action is $3$-transitive, say, are about what happens if $\chi$ is restricted to a point stabiliser, $k$-transitive sets are nice for a different reasons. Namely, you can produce more irreducible characters than just $\chi$ out of them. E.g. you can consider the set of unordered pairs of elements of $X$ (or more generally, the set of $l$-element subsets for $l\lt k$). Of course, the action is not quite transitive, because e.g. the pair $(x,x)$ can only be sent to another pair with equal entries, but you can analyse this defect and obtain explicit descriptions of other irreducible characters.

For a guided solution with nice results in the case of symmetric groups, see exercise 7 on one of my representation theory exercise sheets.

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  • $\begingroup$ Thanks! To compute the inner products, does one just compute the characters by counting fixed points, and then sum up? I computed the characters on cycles, but duty called before I could organize the rest, and I'm not sure I am skilled enough at binomial coefficients to simplify the inner product. $\endgroup$ – Jack Schmidt Aug 29 '11 at 16:30
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    $\begingroup$ @Jack Thanks, I am glad you like it! Note that $\langle \pi_k,\pi_l\rangle = \langle \pi,1\rangle$, where $\pi$ is the permutation character corresponding to the set $X_k\times X_l$. Now, this inner product is just the number of $G$-orbits on that latter set. It remains to count the orbits in a clever way. I will not spoil the rest of the fun for you. If you do it this way, no manipulation of binomial coefficients is necessary, all you need is a small but clever insight to count the orbits. Feel free to email me if you would like me to tell you the rest of the solution. $\endgroup$ – Alex B. Aug 30 '11 at 4:09

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