Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 15F / b

If $G$ is $n$-transitive, it is transitive by extending one-element tuples to $n$-tuples. Apply $n$-transitivity to tuples beginning with $\omega$ to see that the stabilizer subgroup $G_{\omega }$ is $(n-1)$-transitive on the remaining elements.

Conversely, given two ordered $n$-tuples of distinct elements, transitivity first supplies $g$ carrying their first entries to each other. Conjugating the given stabilizer hypothesis shows that the stabilizer of the target first entry is $(n-1)$-transitive. An element of that stabilizer then maps the remaining entries of the image tuple to their targets. Composing the two elements proves $n$-transitivity. For $n=1$, the condition on zero-tuples is vacuous.

For finite nonempty $\Omega$, let $e_{\omega }$ be the basis vectors of the permutation representation. PutThe line $\mathbb{C}u$ is fixed by $G$, and $V$ is the augmentation subrepresentation of a permutation representation. Since $|\Omega |\ne 0$ in $\mathbb{C}$, they have zero intersection and their dimensions add to $|\Omega |$. Hence $\boxed{\mathbb{C}\Omega \cong {\mathbb{C}}_G\oplus V}$.

The printed formulation does not explicitly require $\Omega$ to be finite. For an infinite nonempty $G$-set, interpret $\mathbb{C}\Omega$ as finite-support vectors. Choose one finite orbit $O$ (all orbits are finite because $G$ is finite), replace $u$ by ${\sum }_{\omega \in O}{e}_{\omega }$, and use the same augmentation kernel. This again splits off a trivial summand. The nonempty-set assumption is needed; the empty permutation module contains no trivial summand.

As printed, the equivalence needs a transitivity hypothesis. A group acting trivially on two points has $\mathbb{C}\Omega ={\mathbb{C}}_G\oplus {\mathbb{C}}_G$, so $V$ is one-dimensional and irreducible, although the action is not even transitive.

Here is the intended result for a transitive action. The set is then finite, and, for a point stabilizer $H$, its permutation character is $\chi ={\mathrm{Ind}}_H^{G}1$. Frobenius reciprocity givesThe last equality follows because an $H$-fixed vector has constant coefficients on each orbit. There is exactly one trivial constituent, since the action is transitive. Therefore, writing $\chi =1+\psi$ for the character of $V$,By character orthogonality, $V$ is irreducible exactly when this equals $1$. One stabilizer orbit is $\{\omega \}$, so this happens exactly when $H$ is transitive on $\Omega \setminus \{\omega \}$, equivalently when $G$ is two-transitive. Moreover $V$ is not trivial, since a transitive action has only one independent fixed vector in the whole permutation module.

The complete correction without assuming transitivity is also precise: $V$ is irreducible exactly for a two-transitive action or the exceptional action on two fixed points. Indeed for a finite set with $r$ orbits, the fixed subspace of $V$ has dimension $r-1$. If $r>1$ and $V$ is irreducible, $V$ must itself be a one-dimensional trivial module, forcing $|\Omega |=2$ and $r=2$. If $\Omega$ is infinite, $V$ is infinite-dimensional and cannot be irreducible: any nonzero vector generates a submodule of dimension at most $|G|$.