Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 19I / b

On the orbit $G/H$, an element $g\in G$ fixes the coset $xH$ exactly whenThe character of a representation of the corresponding permutation representation therefore isThis is the permutation character of a coset action.

The vectoris fixed by every element of $G$, so $\mathbb{C}{1}_X\cong {\mathbb{C}}_G$ is a trivial representation. The coefficient-sum subspaceis also $G$-invariant, and every vector has a unique decomposition into a constant vector and an element of $V$. HenceHere $V$ is the augmentation subrepresentation of a permutation representation.

Let $D$ represent $H\backslash G/H$. Combining Frobenius reciprocity with the Mackey's theorem givesThus

Work with the selected transitive orbit $X\cong G/H$. By Maschke's theorem, decompose its permutation character aswhere the ${\chi }_j$ are distinct nontrivial irreducible characters. Character orthogonality givesConsequently the augmentation summand $V$ is irreducible exactly when this inner product equals $2$.

The double cosets $H\backslash G/H$ are the $H$-orbits on $G/H$. Because $G$ acts transitively on $X$, these in turn correspond to the $G$-orbits on $X\times X$: move the first coordinate to the base point $H$, after which its stabilizer is $H$. One orbit is the diagonal. There are exactly two orbits precisely when $H$ is transitive on $X\setminus \{H\}$, equivalently when the action is two-transitive. Part (ii) therefore provesThis is the irreducible augmentation criterion for a transitive group action.

Finally, transitivity implies that the fixed-vector space $\mathbb{C}[X{]}^G$ consists only of constant vectors and has dimension one. If the nonzero representation $V$ were a trivial representation, it would contribute another fixed vector, a contradiction. Thus $V$ is not the trivial representation.