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For a transitive finite $G$-set $X$ with at least two points, the augmentation subrepresentation of a permutation representation is irreducible if and only if the action is two-transitive. Indeed, writing the permutation character as $\pi =1+{\sum }_i{m}_i{\chi }_i$ givesThe augmentation subrepresentation is irreducible exactly when this norm is two. By the character norm of a permutation representation, that means that the diagonal and the off-diagonal are the only two orbits on $X\times X$, which is exactly two-transitivity.

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.

Assume $n\ge 2$. The place-permutation action on ${\mathbb{C}}^n$ preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of $S_n$ is two-transitive, so the irreducible augmentation criterion for a transitive group action makes $W$ irreducible. The two-row Young permutation module decomposition of $M^{(n-1,1)}\cong {\mathbb{C}}^n$ identifies its nontrivial summand as $V^{(n-1,1)}$. Hence $W\cong V^{(n-1,1)}$.

All standard Young tableaux of shape $(n-1,1)$ are $T_j$, $2\le j\le n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of $w_j$ with $w_l$, $j<l$, is zero because the coefficients of $w_j$ sum to zero. Directly summing the action of $(ar)$ gives $X_r{w}_j=c_{T_j}(r)w_j$.

For $s_i=(ii+1)$, $s_1{w}_2=-w_2$. For $2\le i\le n-1$, its only nontrivial two-dimensional block isEvery other $w_j$ is fixed, including all $w_j$ with $j>2$ when $i=1$. These formulas follow by swapping coordinates $i,i+1$ in the displayed vectors, and are the Young orthogonal form with axial distance $i$ for $T_i$.

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.