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The Young permutation module $M^{(n-k,k)}$ is the symmetric-group subset permutation representation on the $k$-element subsets of $\{1,\dots ,n\}$: the second row of a tabloid specifies the subset. For $k\ge 1$, restriction of a character to $S_{n-1}$ fixing $n$ splits these subsets into two orbits of a group action, according as they contain $n$.

The corresponding permutation representations are on the $k$-subsets and the $(k-1)$-subsets of $\{1,\dots ,n-1\}$. Their Young permutation modules have two row sizes $n-1-k,k$ and $n-k,k-1$, respectively. Sort each pair into decreasing order and omit any zero part; swapping the two row labels gives an isomorphic permutation representation. This sorting matters at $n=2k$.

We use Young's rule: over the complex numbers, the multiplicity of $S^{\rho }$ in $M^{\eta }$ is the Kostka number $K_{\rho \eta }$, and it can be nonzero only if $\rho$ dominates $\eta$ in the dominance order on partitions. Since each $\eta$ above has at most two parts and size $n-1$, dominance givesThus $\rho$ has at most two nonzero parts. The character inner product measures this constituent multiplicity, soIf $k=0$, the original and restricted modules are the trivial representations, giving the same conclusion. For $n=1$, there is no partition of $n-1=0$ with three parts, so the assertion is vacuous.

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.

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.

Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional permutation representation splits as the rigid-rotation line spanned by $(1,1,1)$ and the two-dimensional sum-zero standard representation of the symmetric group.

The two displayed vibrational vectors merely choose a basis of that sum-zero plane. A permutation generally mixes them, but their common frequency leaves the whole plane invariant. This is precisely the degenerate normal modes from permutation symmetry required by the physical symmetry.

The Burnside lemma states that for a finite group $G$ acting on a finite set $X$,To prove it, countCounting by $g$ gives ${\sum }_g|X^g|$. Counting by $x$ givesusing the orbit-stabilizer theorem on each orbit $\mathcal{O}$. Dividing by $|G|$ proves the formula.

Let $\pi$ be the character of the permutation representation $\mathbb{C}[X]$. Then $\pi (g)=|X^g|$, and a second application of Burnside's lemma givesFor a two-transitive action there are exactly two orbits on $X\times X$: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity says the trivial representation occurs once. Since the squared multiplicities of the irreducible constituents sum to two, there is exactly one further constituent, with multiplicity one, and it is irreducible and nontrivial. Thusas in the permutation representation of a two-transitive action.