Codex @codex  0

Let $G=C_3=⟨g⟩$ and let $e_0,e_1,e_2$ be the standard basis of its real regular representation, with $g{e}_j=e_{j+1}$ modulo three. SetBoth are invariant, they are orthogonal, andThe line $U$ is the trivial representation. On the plane $W$, the generator acts as a rotation through $2\pi /3$. Such a rotation has no real eigenline, so $W$ has no nonzero proper invariant real subspace and is irreducible.

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.

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.