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The identity $X_i={\Omega }_i-{\Omega }_{i-1}$ from part (a) shows that each Young–Jucys–Murphy element belongs to the Gelfand–Tsetlin algebra. Conversely, $Z_r$ lies in the centralizer of $\mathbb{C}{S}_{r-1}$ in $\mathbb{C}{S}_r$. Applying the preceding Olshanskii centralizer lemma at level $r$ givesInduction, beginning with $Z_1=\mathbb{C}1$, proves $Z_r\subseteq \mathrm{alg}(X_1,\dots ,X_r)$. ThusHere every generated algebra is unital; $X_1=0$ causes no problem.

To deduce simple branching, use Maschke's theorem: complex representations of finite groups are semisimple modules. The Artin–Wedderburn theorem identifieswhere $L$ runs over the irreducible $S_r$-modules. Write their restrictions asBy Schur lemma, the centralizer in the displayed algebra iswith zero multiplicities omitted. Part (i) makes this algebra commutative, whereas a full matrix algebra of size at least two is noncommutative. Consequently every restriction multiplicity is $0$ or $1$. This proves multiplicity-free restriction at each step, with no use of the branching rule in the centralizer proof.

Use the following normalization for Fourier analysis on a finite group. Choose one unitary irreducible representation $\rho :G\to U(d_{\rho })$ from each equivalence class, including the trivial representation. For a scalar function $f:G\to \mathbb{C}$, putThis convention uses $\rho (x)$, rather than $\rho (x{)}^{\ast }$, in the Fourier transform on a finite group; it makes the normalized convolution on a finite group preserve multiplication order.

The needed representation theory consists of Maschke's theorem and unitarization of a finite-group representation, together with the Schur orthogonality relations:The regular representation contains $d_{\rho }$ copies of each $\rho$, so ${\sum }_{\rho }{d}_{\rho }^2=|G|$. Thus the scaled matrix coefficients $\sqrt{d_{\rho }}\rho (x{)}_{ij}$, and also their complex conjugates, form an orthonormal basis of all scalar functions on $G$. These facts imply Fourier inversion on a finite group and the Parseval identity on a finite group in the formsand henceIn particular, the transform is an isomorphism onto the direct sum of the matrix algebras $M_{d_{\rho }}(\mathbb{C})$, with the displayed weighted Hilbert-Schmidt inner product.

Define the normalized convolution on a finite group bySubstituting $x=yz$ and using the group representation identity yields the convolution theorem on a finite groupUnlike normalized convolution on a finite group on an abelian group, this product need not commute. If $\tilde{f}(x)=\overline{f(x^{-1})}$, then $\hat{\tilde{f}}(\rho )=\hat{f}(\rho {)}^{\ast }$. For left translation of a group function and right translation of a group function $L_{a}f(x)=f(a^{-1}x)$ and $R_{a}f(x)=f(xa)$,For an abelian group, every irreducible representation is one-dimensional; this reduces to Fourier analysis on a finite abelian group with characters relabelled by their inverses. These formulas establish the basic scalar theory, with all normalizations and multiplication orders fixed.

Now suppose every nontrivial irreducible representation has $d_{\rho }\ge m$. If $f$ is a mean-zero function, its component at the trivial representation is zero. The Parseval identity on a finite group gives, for each other $\rho$,Using the convolution theorem on a finite group, the Hilbert-Schmidt norm inequality $\Vert AB{\Vert }_{\mathrm{HS}}\le \Vert A{\Vert }_{\mathrm{op}}\Vert B{\Vert }_{\mathrm{HS}}$, and the Parseval identity on a finite group once more gives the product mixing in a quasirandom group estimateWrite $a,b,c$ for the subset density values of $A,B,C$, respectively, and let $f=1_A-a$, $g=1_B-b$ be balanced subset indicators. Since both are mean-zero functions, $1_A\ast 1_B=ab+f\ast g$. Their squared norms are $a(1-a)$ and $b(1-b)$. Also $\mathbb{E}(f\ast g)=0$, so the Cauchy-Schwarz inequality yieldsIf $abc>1/m$, the final bound is strictly smaller than $abc$. Thus the normalized number of pairs $(x,y)\in A\times B$ with $xy\in C$ is positive. Equivalently,This is the desired conclusion for a quasirandom group; the strict inequality ensures positivity rather than merely a nonnegative lower bound.

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.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.

A nonzero central idempotent is primitive central if it is not a sum of two nonzero orthogonal central idempotents. For an Artinian ring, it determines a block of an Artinian algebra. This differs from a primitive idempotent in the whole ring: for a matrix algebra of size greater than one over a field, $1$ is primitive central but is a sum of diagonal idempotents.

For a field of characteristic other than two and nonzero $a,b$, the quaternion algebra has basis $1,i,j,ij$ with the displayed relations. It is a four-dimensional central simple algebra. After adjoining a square root of $a$, diagonalizing $i$ and taking $j=\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$ identifies it with $M_2$. Quaternion conjugation fixes scalars and negates $i,j,ij$, giving norm $x_0^2-a{x}_1^2-b{x}_2^2+ab{x}_3^2$. An element is invertible exactly when its norm is nonzero. Thus the quaternion algebra is either a division algebra or a split matrix algebra. The familiar quaternion algebra is $(-1,-1{)}_{\mathbb{R}}$.

If $a=N_{L/K}c$, let $L$ act on itself by multiplication and let the extra generator act by $c\sigma$. Its $n$th power is multiplication by $a$, and these operators give the cyclic algebra as ${\mathrm{End}}_K(L)$. Conversely, if the algebra is a matrix algebra on an $n$-dimensional vector space, its embedded degree-$n$ field $L$ makes that space one-dimensional over $L$. The extra generator is then $c\sigma$ in an $L$-basis, so its $n$th power forces $a=Nc$.

A field $k$ is a splitting field for a finite group $G$ if every simple module over its group algebra $kG$ is absolutely simple: it remains simple over every extension field. Equivalently, the quotient by the Jacobson radical is a product of full matrix algebras over $k$. In characteristic $p$, containing all $m$th roots of unity, where $|G|=p^{a}m$ and $p\nmid m$, is sufficient. This is the modular splitting-field theorem; it is substantially stronger than merely having eigenvalues for one chosen group element.