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For complex class functions on a finite group $G$, the character inner product isIf $\psi$ is irreducible, this inner product is its multiplicity in the representation with character $\chi$.

If $\chi$ is an irreducible character of a finite group $G$ and $H\le G$, thenEquality holds exactly when $\chi$ vanishes on $G\setminus H$.

By character orthogonality, the squared multiplicities are the character inner product norm of the restriction of a character:Since $\chi$ is an irreducible character, its norm on $G$ is one, and henceThe omitted summands are nonnegative, so the character restriction norm bound is an equality precisely when

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.

Burnside lemma says that the number of orbits of a finite group $G$ on a finite set $X$ isIndeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives ${\sum }_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the orbit-stabilizer theorem, proving the formula.

Let $\pi (g)=|X^g|$ be the character of the permutation representation. Thenis, by Burnside's lemma, the number of orbits on $X\times X$. A two-transitive group action has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity also gives $⟨\pi ,1_G⟩=1$. If $\pi =1_G+\sum m_i{\chi }_i$ is its irreducible character decomposition, then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.