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For any partition of an integer $\lambda ⊢n$, the number $f^{\lambda }$ of standard Young tableaux of shape $\lambda$ isThe numerator is a factorial, and the denominator is the hook product of a partition, using every cell's hook length once. The empty shape has one empty standard Young tableau, and the formula still holds with $0!=1$ and an empty product equal to one. By the standard polytabloid basis, $f^{\lambda }$ is also the dimension of the Specht module $S^{\lambda }$ over the complex numbers, or equivalently the degree ${\chi }^{\lambda }(1)$ of its irreducible character. For a teaching derivation of this dimension formula, see Etingof et al., Introduction to Representation Theory, §5.17, Theorem 5.17.1.

In the Young diagram of $(3,2)$, the hook length array $[[4,3,1],[2,1]]$ givesThis agrees with the five explicit fillings in standard Young tableau. Its conjugate partition $(2,2,1)$ has the same hook lengths after transposition, hence also gives dimension five. At the extremes, for $n\ge 1$, a single row $(n)$ and a single column $(1^n)$ each have hook product $n!$ and exactly one standard Young tableau; their Specht modules over the complex numbers are the trivial representation and sign representation, respectively.

Why the formula is beautiful: global compatibility of all the row and column inequalities reduces to a product of local hook lengths. It is useful when enumeration is already expensive: computing one hook length per cell replaces testing up to $n!$ numbered Young tableaux. For $(3,2)$, the proportion of the $120$ bijective fillings that are standard Young tableaux is $5/120=1/24$, the reciprocal of the hook product. This product does not justify treating overlapping hook conditions as independent; Grinberg explains that tempting but invalid argument in §1.3 of his teaching slides.

Historically, the modern hook-product statement was published by J. S. Frame, G. de B. Robinson and R. M. Thrall in 1954: ^[ref]. Their manuscript was received on 1 September 1953, and their proof reorganizes an earlier irreducible character degree formula into the product of the hook lengths. The theorem unifies the counting and representation viewpoints developed earlier by Young. For proof routes within this wiki, see trace computation of Specht module dimension, determinant formula for Specht module dimension, and Greene–Nijenhuis–Wilf hook walk.

With $s=(12)$, the idempotents $e_{\pm }=(1\pm s)/2$ split the right regular module into two three-dimensional projective modules. Their tops are respectively the trivial representation and sign representation. The nilpotent ideal $J(k{S}_3)$ ensures that any nonzero direct summand has nonzero top, so the one-dimensional tops make these projective modules indecomposable. Their successive radical series of a module factors are trivial, sign, trivial and sign, trivial, sign. This decomposition concerns right modules; the block of S3 in characteristic three remains a single algebra.

A one-dimensional complex group representation acts on a one-dimensional complex vector space, and is equivalently a group homomorphism into the multiplicative group ${\mathbb{C}}^{\times }$. It is automatically an irreducible representation. Its character of a representation is its scalar value, a one-dimensional character. A change of basis cannot change those scalar values, so two such group representations are equivalent exactly when their homomorphisms agree. For a symmetric group $S_n$ with $n\ge 2$, all transpositions are conjugate, square to the identity and generate the group. Their common scalar image is therefore $1$ or $-1$, giving exactly the trivial representation and the sign representation.

Since $S^{\mu }\subset M^{\mu }$, Question 2 shows that every composition factor of $S^{\mu }$ is $D^{\lambda }$ for a regular partition $\lambda ⊵\mu$. If $\mu$ is regular, its radical quotient $D^{\mu }$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^{\mu }$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.

Define $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. Order the row labels, all partitions of an integer of $n$, and the column labels, the regular partitions of an integer, by the same total order refining decreasing dominance. ThenThe matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.

This is the decomposition matrix, not just an abstract composition factor table. The standard polytabloid basis makes $S_{\mathbb{Z}}^{\mu }$ a free invariant integral lattice whose extension to characteristic zero is the ordinary irreducible $S^{\mu }$. Reducing this same lattice modulo $p$ gives exactly the modular $S^{\mu }$. On p-regular elements, reduction of its ordinary character is the sum of the Brauer characters of its composition factors, with coefficients $d_{\mu \lambda }$. These are the defining decomposition numbers; the resulting class in the modular representation ring is independent of the chosen invariant lattice.

For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In characteristic zero, or $p>3$, all three Specht modules are irreducible by Maschke's theorem, andFor $p=2$ the regular labels are $(3),(2,1)$ andIndeed the sign representation equals the trivial representation. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point permutation module. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root eigenvalues there, and a transposition exchanges their lines. Neither line is invariant under $S_3$, proving absolute irreducibility.

For $p=3$ the regular labels are $(3),(2,1)$ andIn the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the determinant of the sum-zero representation is sign, and the determinant of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple head $D^{(2,1)}$. The bottom Specht module is itself sign. This proves every entry of the three possible matrices.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.

On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the polytabloid obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the standard Young tableaux of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every field. In characteristic zero, Maschke's theorem splits the filtration and gives the direct sum. Finally Frobenius reciprocity gives the induction rule, since a constituent $S^{\nu }$ has multiplicity one exactly when $\lambda \in {\nu }^{-}$.

The hook-length formula statesThe hook of a Young diagram consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary irreducible dimensions; the standard polytabloid basis also makes the dimension of a modular Specht module the same integer.

All transpositions are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary representation their common scalar is therefore either $1$ or $-1$. These choices give precisely the trivial representation $S^{(n)}$ and the sign representation $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has dimension $n-1$: its standard Young tableau is specified by choosing the bottom entry from $2,\dots ,n$. Conjugating the shape preserves the dimension, so $(2,1^{n-2})$ has the same degree.

We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed hook-length formula, are summarized below. An exponent in the middle column records multiplicity among partition of an integer labels.

For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.

Induct on $n\ge 9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^{\lambda }\ge 2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge 2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times 2$ rectangle. Therefore two-step branching gives $f^{\lambda }\ge 2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.

Consequently the least non-linear ordinary degree of a symmetric group isFor degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.

For $n\ge 2$, a cycle type $\alpha ⊢n$ lies in $A_n$ exactly when $n-\ell (\alpha )$ is even. Its $S_n$ conjugacy class is either one $A_n$ class or two equal-sized classes. Splitting occurs exactly when its centralizer in $S_n$ contains no odd permutation. An even-length cycle is itself odd; two equal odd-length cycles can be interchanged by an odd permutation. Conversely, for distinct odd cycle lengths the centralizer is a product of cyclic groups of odd order, all contained in $A_n$. Thus the classes that split are exactly the partitions of an integer into distinct odd parts. Repeated fixed points count as repeated parts of length one.

For $n=1$ the group is trivial, with its single trivial representation. Assume $n\ge 2$ for the index-two argument that follows. For ordinary representations work over $\mathbb{C}$. Tensoring $S^{\lambda }$ by the sign representation gives $S^{{\lambda }^{\prime }}$. The index-two restriction identity, obtained by Frobenius reciprocity, isIf $\lambda \ne {\lambda }^{\prime }$, restriction is irreducible, and the conjugate pair gives the same irreducible. If $\lambda ={\lambda }^{\prime }$, the norm is two, so restriction is the sum of two distinct irreducibles. An odd permutation interchanges them, so both have degree $f^{\lambda }/2$. Every irreducible of $A_n$ occurs in some such restriction: induce it to $S_n$ and choose an irreducible constituent, then apply reciprocity. The displayed inner product distinguishes all the listed constituents. HenceThe first family has degree $f^{\lambda }$, the second degree $f^{\lambda }/2$. A choice of labels for the two split classes and constituents fixes the otherwise interchangeable signs.

The one-dimensional representations are characters of the abelianization. For $n\ge 5$, simplicity and noncommutativity of $A_n$ make its abelianization trivial. For $n=4$, its commutator subgroup is the Klein four-group and its abelianization is $C_3$, giving the three characters obtained by sending a quotient generator to $1,\omega ,{\omega }^2$. For $n=3$, $A_3=C_3$ has the same three characters; for $n=1,2$ the group is trivial and has just one. The two additional linear characters at $n=4$ are the split constituents of shape $(2,2)$; at $n=3$ they come from $(2,1)$.

Here is a dimension argument that also proves the asserted uniqueness without assuming a classification of small character degrees. Write $f^{\lambda }$ for the number of standard tableaux. Removing the cell containing the largest label givesHook lengths, or direct Young tableau counting, give the following small cases: for $S_5,S_6$ every non-linear degree is at least $4,5$, respectively. Among non-self-conjugate shapes excluding the row, column and standard pair, the minimum degree at $n=5,6,7$ is respectively $5,5,14$. The self-conjugate shapes at these sizes have half-degrees $3,8,10$, respectively, from $(3,1^2),(3,2,1),(4,1^3)$.

Inductively, for every $n\ge 7$, a shape other than the row, column or standard pair has all its one-cell predecessors non-linear. If it has at least two removable corners, each predecessor has degree at least $n-2$, so its degree is at least $2(n-2)$. If it has only one corner, it is a nontrivial rectangle $(a^b)$. Remove that corner and then either of the two corners of $(a^{b-1},a-1)$. For $n\ge 8$ these two size-$n-2$ shapes are non-linear, so its degree is at least $2(n-3)$. At $n=7$ there is no nontrivial rectangle. Together with the size-5 and size-6 checks this proves by induction that the standard pair is the only pair of shapes of degree $n-1$ for $n\ge 7$, and that every other non-linear degree is larger.

We must additionally exclude a self-conjugate shape of degree $2(n-1)$, since its restriction splits. For $n\ge 8$, if such a shape has at least three corners, the preceding lower bound gives $f^{\lambda }\ge 3(n-2)>2(n-1)$. With two corners, its predecessors are a conjugate pair and are not standard shapes; the bounds just proved give $f^{\lambda }\ge 4(n-4)>2(n-1)$. With one corner it is a square $(a^a)$. For $a\ge 4$, the two size-$n-2$ predecessors obtained after two removals are nonstandard, giving $f^{\lambda }\ge 4(n-5)>2(n-1)$. The remaining square $(3^3)$ has degree $42>16$ by the hook-length formula. The size-7 self-conjugate case has degree $20>12$. This excludes both smaller split degrees and equality with the standard degree.

It follows that the lowest non-linear ordinary degree is $3$ for $A_4$ and $n-1$ for every $n\ge 6$. For $A_1,A_2,A_3$ no non-linear irreducible exists. The excluded case $A_5$ has lowest degree $3$, although its standard degree-$4$ representation is unique. At $A_6$ there are two degree-$5$ irreducibles, from the standard conjugate pair and the pair $(3,3),(2,2,2)$.

Finally, for every $n\ge 7$ the preceding proof gives a unique degree-$n-1$ irreducible, namely the restriction of $S^{(n-1,1)}$. The small cases $n=4,5$ have unique degree $3,4$, respectively, and $A_2$ has its unique trivial degree-$1$ representation. These prove exactly the requested uniqueness range; the omitted cases $n=3,6$ genuinely fail.

The conjugacy classes are specified by cycle type, with sizes $1,6,3,8,6$ for $1,(12),(12)(34),(123),(1234)$. The trivial and sign representations give the first two rows. The permutation representation on four letters contains a trivial line; its orthogonal complement has character equal to the number of fixed letters minus one. Tensoring that representation by the sign gives another row. Finally the quotient $S_4/V_4\cong S_3$ supplies the two-dimensional standard representation. The resulting table isWith the class-size-weighted inner product, these characters have norm one and are pairwise orthogonal, so all are irreducible and inequivalent. Their squared dimensions sum to $1+1+9+9+4=24=|S_4|$, proving completeness.

Since $X_1=0$, every joint weight has $a_1=0$. In $S_2$, $X_2=(12)$ acts by $1$ on the trivial representation and by $-1$ on the sign representation, which exhaust the irreducible representations. ThusThe content vector of a standard Young tableau conditions similarly require the second coordinate to be a neighbour of $0$, soThese vectors correspond respectively to the single-row tableau of shape $(2)$ and the single-column tableau of shape $(1,1)$.

For $\lambda =(1^n)$, there is exactly one standard Young tableau, the column with entries $1,2,\dots ,n$ from top to bottom. Its content vector of a standard Young tableau isEvery adjacent distance is $-1$, so Young orthogonal form gives $s_r{w}_T=-w_T$ for every $r$. Thereforethe sign representation.

The single-column Young diagram has exactly one standard Young tableau, so $V^{(1^n)}$ has dimension one. Consecutive labels are in one column, and the Young seminormal form gives $s_{i}w=-w$ for every adjacent transposition $s_i$. These generators determine the representation, and the sign representation sends every one of them to $-1$. ConsequentlyAt $n=1$ there are no adjacent generators and both modules are the trivial representation.

The conjugate partition is ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$, obtained by transposing the Young diagram. Transposition sends a standard Young tableau $T$ to a standard tableau $T^{\prime }$ and negates every Content of a Young-diagram cell.

Use the Young orthogonal form. For an admissible pair $R=s_{i}T$ and axial distance $d=c_T(i+1)-c_T(i)$, the action on $(w_T,w_R)$ isIn the transposed pair it is $A_{-d}$, whereas in the tensor product of group representations with the sign representation it is $-A_d$. Their diagonal entries agree, and their off-diagonal entries differ by a sign. Fix a reference tableau and let ${ϵ}_T$ be the sign of the unique label permutation from it to $T$. For every admissible swap ${ϵ}_R=-{ϵ}_T$, sointertwines these two actions. In a nonadmissible pair, transposition exchanges row and column and changes the scalar from $+1$ to $-1$ or conversely, also agreeing with the sign twist. Since adjacent transpositions generate $S_n$, this is an isomorphism:The parity choice is globally well defined because the label permutation is unique; no arbitrary edge-by-edge phase choices are required.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

The Jacobson radical is $J(A)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals of $A$; equivalently, it is the intersection of the annihilators of all simple modules. It is a two-sided ideal. A projective module has the lifting property against every surjective R-module homomorphism. A finitely generated module is a projective module precisely when it is a direct summand of a finitely generated free module. An indecomposable module is nonzero and admits no direct sum decomposition into two nonzero submodules.

To calculate the top of an indecomposable projective module, use the right Artinian ring hypothesis, namely the descending chain condition on right ideals. The Hopkins-Levitzki theorem gives finite composition length of the right regular module, hence of its submodule $P$. Also $J=J(A)$ is a nilpotent ideal, and $A/J$ is a semisimple ring. Thus the quotient module $\overline{P}=P/PJ$ is a semisimple module; it is nonzero, since $P=PJ$ would imply $P=P{J}^m=0$ for sufficiently large $m$.

Suppose $\overline{P}$ were not a simple module. A nontrivial direct sum decomposition of this semisimple module would give an idempotent $\alpha \in {\mathrm{End}}_A(\overline{P})$ that is neither zero nor the identity. Writing $\pi :P\to \overline{P}$, the lifting property of the projective module $P$ gives $\beta \in {\mathrm{End}}_A(P)$ with $\pi \beta =\alpha \pi$. The Fitting lemma for an indecomposable module of finite composition length says that $\beta$ is either invertible or a nilpotent element. Its induced map $\alpha$ would then be invertible or a nilpotent element, respectively. Neither is possible for a nontrivial idempotent. Therefore $P/PJ(A)$ is simple. This argument does not assume that an embedded projective module automatically splits off from the ambient module.

A block of an Artinian algebra is a nonzero two-sided direct summand $Ae$ determined by a primitive central idempotent $e$: $e$ cannot be written as a sum of two nonzero orthogonal central idempotents. Its identity is $e$. For a finite-dimensional associative algebra, the blocks of an Artinian algebra give its unique decomposition as a finite product of indecomposable algebras, or equivalently as a direct sum of two-sided ideals.

For the block of S3 in characteristic three, put $A=k{S}_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The group algebra has basis $r^i,r^{i}s$ for $0\le i<3$. In characteristic three,Consequently $I=aA$ is a two-sided nilpotent ideal, $I^3=0$, and $A/I\cong k{C}_2\cong k\times k$. A nilpotent ideal lies in the Jacobson radical, and a quotient that is a semisimple ring forces the reverse inclusion. HenceDefine orthogonal idempotents $e_{+}=(1+s)/2$ and $e_{-}=(1-s)/2$. They sum to one, so the right regular module decomposes asThe two summands are the indecomposable projectives of S3 in characteristic three. Each direct summand is a projective module, with basis $e_{\pm },e_{\pm }r,e_{\pm }{r}^2$. Its quotient module modulo multiplication by $J(A)$ is one-dimensional: the trivial representation for $e_{+}A$, and the sign representation for $e_{-}A$. Each is an indecomposable module, since two nonzero direct summands would each have nonzero quotient module modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the radical series of a module are the trivial representation, sign representation, trivial representation for $e_{+}A$, and the sign representation, trivial representation, sign representation for $e_{-}A$. Indeed, modulo $J^2$ the relation $sa=-as+a^{2}s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.

The center of an associative algebra is spanned by the conjugacy class sums $1$, $r+r^2$, and $(1+r+r^2)s$. Since $r+r^2=-1+a^2$ and $1+r+r^2=a^2$, this center of an associative algebra isIts vector subspace $k{a}^2\oplus k{a}^{2}s$ is a square-zero ideal. If $d1+n$ is a central idempotent, then $d^2=d$ and $(2d-1)n=0$. Thus $d=0$ or $1$ and $n=0$. There is no nontrivial central idempotent, so the whole group algebra is its single block. The two three-dimensional projective modules above are a decomposition of the regular module, not two blocks of an Artinian algebra.

Conjugating a tableau exchanges row symmetrization with column antisymmetrization. The resulting module is the original Specht module twisted by the sign representation, giving the conjugate Specht characterTaking traces yields

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

A group homomorphism $D:G_1\to G_2$ satisfies $D(gh)=D(g)D(h)$ for all $g,h\in G_1$. Its group kernel is

For the matrices, put $X=R(x)$, $Y=R(y)$ and $z=e^{2\pi i/3}$. Since $z^3=1$, $z\ne 1$ and $1+z+z^2=0$, direct matrix multiplication givesThese verify the listed matrices and the defining relations, with $R(e)=I_2$. To check every product explicitly through a single calculation, write elements in the unique form $x^{ϵ}{y}^j$, with $ϵ=0,1$ and $j=0,1,2$. The relations giveThe displayed matrices obey $Y^{j}X=X{Y}^{-j}$, so their product has exactly the same form:All exponents are reduced modulo two or three respectively. Thus this verifies all 36 products, and $R$ is a two-dimensional complex group representation. It is given to be an irreducible representation.

For a one-dimensional complex group representation $U$, let $u=U(x)$ and $v=U(y)$, which are nonzero complex numbers. The relations requireSince scalar multiplication commutes and $u\ne 0$, the last relation gives $v=v^2$, hence $v=1$. Then $u=\pm 1$. The choice $u=1$ is the trivial representation; the unique nontrivial choice is the sign representation:Its group kernel isA faithful representation is injective and has trivial kernel. Here the kernel has three elements, so $U$ is not faithful.

Both one-dimensional representations are irreducible representations, and they are inequivalent to each other and to the given two-dimensional one. The sum of squares of irreducible representation dimensions givesThese exhaust all complex irreducible representations. Their number equals the number of conjugacy classes, so there are three conjugacy classes. Explicitly they areThe last two sets consist of the 3-cycles and transpositions respectively.

The two one-dimensional representations are the trivial representation, ${\rho }_{+}(g)=1$, and the sign representation, ${\rho }_{-}(g)=\mathrm{sgn}(g)$. The first respects multiplication immediately. For the second, the sign of a permutation is the determinant of its permutation matrix, and multiplicativity of the determinant proves ${\rho }_{-}(gh)={\rho }_{-}(g){\rho }_{-}(h)$. They are inequivalent since a transposition has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.

Their group homomorphism kernels areBoth are subgroups, indeed normal subgroups, because they are kernels. In the second case $A_4$ is the alternating group of even permutations, with twelve elements.

The characters are their scalar values, since the trace of a one-dimensional matrix is its only entry. In the class order from part (b),The character inner product is the group average of ${\chi }_{+}\overline{{\chi }_{-}}$, soEach has squared norm $(1+6+3+8+6)/24=1$, in agreement with character orthogonality for distinct irreducible representations.