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The projections $P_T$ in the Gelfand–Tsetlin algebra satisfyConsequently the map sending a diagonal element to its scalar on each Gelfand–Tsetlin basis vector is a ring isomorphismEach factor is a simple module over this algebra, and the algebra is their direct sum as a module over itself. This proves that $G{Z}_n$ is a semisimple algebra. The argument uses its explicit diagonal structure; being a subalgebra of a semisimple algebra by itself would not suffice.

Work over the complex numbers. The chain $S_0\subset S_1\subset \cdots \subset S_n$ has multiplicity-free restriction: each irreducible representation restricts to a direct sum of pairwise inequivalent irreducible representations. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted Olshanskii centralizer lemma makes $C_{\mathbb{C}{S}_r}(\mathbb{C}{S}_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative matrix algebra factor. Thus no identification of the branching graph with Young diagrams is being assumed here.

Successively restricting an irreducible representation $V$ gives one-dimensional spaces $V_T$ indexed by paths of irreducible representations from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a Gelfand–Tsetlin basis. Define the Gelfand–Tsetlin algebra as the subalgebra of $\mathbb{C}{S}_n$ acting diagonally in all these bases. The Artin–Wedderburn theorem identifiesUnder this identification the Gelfand–Tsetlin algebra is the direct sum of the full diagonal matrix algebras in the indicated bases, so it is commutative.

To see that it is actually available inside the group algebra, let $e_{\alpha }^{(r)}$ be the central primitive idempotent selecting the irreducible representation $\alpha$ of $S_r$. For a path $T=({\alpha }_0,\dots ,{\alpha }_n)$ the productis the projection onto $V_T$ in its final irreducible representation and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal matrix units, and therefore span the proposed algebra.

If an element of the group algebra commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the Gelfand–Tsetlin algebra. Thuswhich proves that it is a maximal commutative subalgebra.

The local spectral interchanges preserve an irreducible representation, and standard Young tableaux of a fixed shape are connected by admissible adjacent interchanges. Different shapes have different multisets of Young-diagram cell contents. Their elementary symmetric functions in the Young–Jucys–Murphy elements are central: the identityproves this coefficient by coefficient. By the Schur lemma, one irreducible representation cannot therefore contain two different shape classes. The multiset determines a diagram because the counts on its positive and negative diagonals determine its arm and leg lengths at the diagonal cells.

Different irreducible representations cannot share a joint eigenvalue vector: the Gelfand–Tsetlin algebra is generated by the Young–Jucys–Murphy elements, so all its elements would act identically on that vector in both representations, whereas a central primitive idempotent distinguishes their two blocks. Thus distinct irreducibles give distinct shape classes.

The number of irreducible representations is the number $p(n)$ of conjugacy classes, each indexed by a partition of an integer. Every spectral vector gives a tableau by the preceding construction. Closure under admissible interchanges makes every shape class that occurs occur in full. The two counts $p(n)$ then force every shape to occur, with exactly one irreducible representation per shape. This supplies $\mathrm{Spec}(n)=\mathrm{Cont}(n)$.

On restricting to $S_{n-1}$, delete the last coordinate. For tableaux of shape $\lambda$, the entry $n$ occupies one Removable node of a Young diagram. Grouping by that node gives one copy of the corresponding module; the path decomposition has simple multiplicities. Hence the restriction branching rule for a symmetric group isThe different removable nodes produce distinct partitions. The multiplicity-free structural input defining the Gelfand–Tsetlin basis is distinct from identifying this graph with the graph of Young diagrams.

Use a Gelfand–Tsetlin basis in Young seminormal form.

For a standard Young tableau $T$, put $d=c_T(i+1)-c_T(i)$, using the Content of a Young-diagram cell. Choose the row-reading tableau $T_0$, and let $\ell (T)$ be the Coxeter length of the unique permutation sending $T_0$ to $T$. A Gelfand–Tsetlin basis can be chosen so that, when $R=s_{i}T$ is standard and $\ell (R)=\ell (T)+1$,If $R$ is not standard, the action is $+v_T$ for two consecutive entries in one row, and $-v_T$ for two in one column. This is one usual normalization of the Young seminormal form.

Here is a construction and proof of the normalization. Fix $v_{T_0}\ne 0$, let $P_T$ be the projection onto the tableau line, and defineThe permutation ${\pi }_T$ has a reduced expression consisting entirely of admissible swaps, by the reduced adjacent-swap path between linear extensions. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance $\ell (T)$ uses all $\ell (T)$ swaps; omitting a swap gives a shorter path. Thus $v_T\ne 0$. This also makes its definition independent of a chosen reduced expression, because ${\pi }_T$ itself is fixed.

The relation $s_i{X}_i{s}_i+s_i=X_{i+1}$ forces the coefficient of $v_T$ in $s_i{v}_T$ to be $1/d$, and forces every other component to lie on the swapped line. If length increases, project the identity ${\pi }_R=s_i{\pi }_T$ onto the line for $R$. The shorter terms of ${\pi }_T{v}_{T_0}$ cannot reach $R$ in one step, so the coefficient of $v_R$ is exactly one. Applying $s_i^2=1$ then gives the reverse coefficient $1-d^{-2}$ and diagonal coefficient $-1/d$. In the nonstandard cases the same relation gives $d=\pm 1$ and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.

The Gelfand–Tsetlin basis spans $V^{\lambda }$, and the scalar computed on its vectors depends only on the shape. Thusfor every $w\in V^{\lambda }$. The product is the sum of the $(n-1)!$ permutations in the conjugacy class of an $n$-cycle. Taking traces therefore gives $(n-1)!{\chi }^{\lambda }((n))$ equal to the displayed scalar times $\dim V^{\lambda }$.

For a hook partition, a standard Young tableau is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\dots ,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, and cancellation of the factorials yieldsThis uses the central character value of a conjugacy-class sum and tableau counting, without a character rule for removing rim hooks.

Let $T$ be a standard Young tableau, $s_i=(ii+1)$, and $d=c_T(i+1)-c_T(i)$. A suitable Gelfand–Tsetlin basis has, for an admissible pair $R=s_{i}T$ with tableau length increasing,For a nonadmissible swap the scalar is $+1$ in a row and $-1$ in a column. The diagonal coefficient follows from the Young–Jucys–Murphy element relation $s_i{X}_i{s}_i+s_i=X_{i+1}$, and $s_i^2=1$ forces the product of off-diagonal coefficients. One global normalization is $v_T=P_T{\pi }_T{v}_{T_0}$ from the row-reading tableau: a reduced admissible path makes this vector nonzero and gives coefficient one on every length-increasing edge.