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By Maschke's theorem, the complex regular representation of a finite group is a direct sum of irreducibles. Its kernel is the intersection of their kernels. If the group is simple, each kernel is either trivial or the whole group; faithfulness of the regular representation therefore forces at least one irreducible constituent to be faithful.

Every nonfaithful finite-dimensional complex representation of the infinite dihedral group is a direct sum of subrepresentations of dimension at most two. Its kernel contains a nonzero power of the rotation, so the representation factors through a finite dihedral group. Maschke's theorem makes that quotient representation semisimple, and all irreducible complex representations of a finite dihedral group have dimension at most two.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

Maschke's theorem makes the group algebra $\mathbb{C}G$ semisimple, since $G$ is a finite group and the ground field has characteristic zero. By the Artin–Wedderburn theorem, write $\mathbb{C}G\cong {\prod }_{i=1}^r{M}_{d_i}(\mathbb{C})$. Its simple modules give exactly the $r$ nonisomorphic irreducible complex group representations.

The center of each matrix algebra consists of scalar matrices, so ${\dim }_{\mathbb{C}}Z(\mathbb{C}G)=r$. On the other hand, an element ${\sum }_g{a}_{g}g$ is central precisely when its coefficients are constant on conjugacy classes. The sums of the elements in the separate conjugacy classes are therefore a basis of the center. Hence the number of irreducible complex representations equals the number of conjugacy classes.

If the two-dimensional group representation were reducible, Maschke's theorem would split it into a direct sum of two one-dimensional group representations. In a basis respecting that decomposition, every representing matrix would be diagonal. All would therefore commute, contradicting the specified pair. Hence the group representation is irreducible. Complete reducibility over $\mathbb{C}$ is essential to this argument.

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.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.

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.

Maschke's theorem says that if $H$ is a finite group and the characteristic of a field $F$ does not divide $|H|$, then every finite-dimensional $F$-representation of $H$ is semisimple.

To prove it, let $W$ be an invariant subspace of a representation $V$, and choose any linear projection $P:V\to W$. Average its conjugates:The scalar $|H{|}^{-1}$ exists by the characteristic assumption. Reindexing the sum shows that $\overline{P}$ commutes with every $\rho (g)$. Since each summand restricts to the identity on $W$, so does $\overline{P}$; hence $\overline{P}$ is an equivariant projection onto $W$. Its kernel is an invariant complement, and induction on $\dim V$ decomposes $V$ into irreducible representations.

The isometry group of $\mathbb{Z}$ is the infinite dihedral groupLet $\rho$ be a nonfaithful finite-dimensional complex representation. Its nontrivial normal kernel contains $t^m$ for some $m>0$, by the stated hint, so $\rho$ factors through the finite quotientMaschke's theorem decomposes this quotient representation into irreducibles. To bound their dimensions, choose a $t$-eigenvector $v$ with eigenvalue $\lambda$ in an irreducible constituent. The relation givesso $\mathrm{span}\{v,sv\}$ is invariant under both generators. Irreducibility makes it the whole constituent, whose dimension is therefore at most two. This proves the result on nonfaithful representations of the infinite dihedral group.

For the requested representation of $(\mathbb{Z},+)$, let its generator act on the polynomials of degree at most one by translation:In the basis $(1,x)$,This is a nontrivial Jordan block, so it is not diagonalizable. A direct sum of one-dimensional subrepresentations would make $T$ diagonalizable, proving that this nonsemisimple translation representation of the infinite cyclic group has the required property.

Finally, on $V=\mathbb{C}[x{]}_{\le 2}$ defineThen $S^2=I$ and $STS=T^{-1}$, so these operators define a quadratic-polynomial representation of the infinite dihedral group. In the basis $(1,x,x^2)$ they areHere $(T-I{)}^2\ne 0$ but $(T-I{)}^3=0$, so $T$ consists of one size-three Jordan block. If $V$ were a direct sum of subrepresentations of dimensions at most two, then $T$ would be block diagonal with Jordan blocks of sizes at most two, a contradiction. Thus this representation has the required failure of decomposition.

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.