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For the symmetric group on five letters, $S^{(4,1)}$ is the four-dimensional augmentation representation. In characteristic three it is simple, since its only possible proper nonzero submodule would be the constant line, which is excluded by $5\ne 0$ in the field. Thus it is $D^{(4,1)}$. If this module occurs as a submodule of $S^{(3,2)}$, the quotient has dimension one, since the standard polytabloid basis gives $\dim S^{(3,2)}=5$.

A transposition has trace one on $S^{(3,2)}$, by the Murnaghan–Nakayama rule, and trace two on $S^{(4,1)}$, by subtracting the trivial line from the five-point permutation representation. These integral traces reduce to the same elements in characteristic three. The quotient trace is therefore minus one, so its one-dimensional action is the sign representation. In particular an arbitrary permutation acts on it by its sign. This identifies the quotient action without asserting that the original short exact sequence splits.

Use the standard polytabloids for tableaux with rows $12\mid 34$ and $13\mid 24$, as an ordered basis $e_1,e_2$. Adjacent transpositions act byIn characteristic three their common fixed line is $F(e_1+e_2)$. On its quotient every transposition acts by minus one, giving the sign representation. Thus the displayed sequence has the trivial representation and sign as its composition factors, each once. It is nonsplit: the minus-one eigenspace of $\rho (12)$ is $F{e}_2$, which is not preserved by $\rho (23)$.

The natural tabloid form restricts to Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$. Its radical in characteristic three is exactly that fixed line. Therefore $D^{(2,2)}$ is the sign quotient, while the trivial factor is $D^{(4)}$. This gives an explicit small failure of semisimplicity in modular symmetric-group representation theory.

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.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

The irreducible representations can all be specified on the generators:The first is the trivial representation and the second is the sign representation. To see directly that these exhaust the one-dimensional representations, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.

Taking traces gives the character table, in the conjugacy class order of solution:Here $\alpha +{\alpha }^2=-1$ gives the middle entry in the last row. The weighted character inner products of distinct rows vanish, and the norm of each row is one. Together with solution, this proves completeness of the table.

Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

The specified linear maps form a permutation representation, sinceThe trace of a permutation matrix counts fixed coordinate basis vectors. On the identity, three-cycles and transpositions, respectively, this gives ${\chi }_R=(3,0,1)$. Comparing with solution,For an explicit direct sum,Both summands are invariant subspaces under coordinate permutations. The first is fixed pointwise and hence gives the trivial representation; the second has character $(2,-1,0)$ and hence gives $D$. The sign representation has multiplicity zero.

The symmetric group $S_3$ has three conjugacy classes, determined by cycle type: the identity, three transpositions, and two three-cycles. Hence it has three complex irreducible representations. Two are the trivial representation $1$ and the sign representation $\epsilon$. Their dimensions are one, so the sum of squares of irreducible representation dimensions gives $1+1+d^2=6$, whence $d=2$ for the remaining representation $D$.

Write ${\chi }_D=(2,a,b)$ on these classes. Orthogonality to $1$ and $\epsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the character table isThe character of a tensor product representation is the product of the factor characters. Thus ${\epsilon }^2=1$, $\epsilon \otimes D=D$, and ${\chi }_{D\otimes D}=(4,0,1)={\chi }_1+{\chi }_{\epsilon }+{\chi }_D$. These determine all the Clebsch-Gordan decomposition of group representations:

For permutations $\sigma ,\tau$, the prescribed linear maps satisfySince this holds on a basis, $R(\sigma )R(\tau )=R(\sigma \tau )$. Also $R(e)=I$ and $R(\sigma {)}^{-1}=R({\sigma }^{-1})$. Hence $R$ is a representation.

The trace of a permutation matrix counts its fixed coordinate basis vectors. There are respectively $3,1,0$ fixed vectors on the three conjugacy classes, so ${\chi }_R=(3,1,0)={\chi }_1+{\chi }_D$. ConsequentlyIn particular the sign representation is absent.

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation $1$ and sign representation $\epsilon$ are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leavesCall the remaining irreducible representation $2$. The permutation representation on ${\mathbb{C}}^3$ has character equal to the number of fixed objects, namely $(3,1,0)$ on the ordered conjugacy classes $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector trivial representation gives $(2,0,-1)$, whose character inner product with itself is one. This constructs $2$ and gives the character tableTo obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular, ${\chi }_{\epsilon }^2={\chi }_1$, ${\chi }_{\epsilon }{\chi }_2={\chi }_2$ andThus all ordered pairs of irreducible representations are covered by

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

Order the conjugacy classes of the symmetric group $S_4$ by representatives $1,(12),(12)(34),(123),(1234)$; their sizes are $1,6,3,8,6$. For a three-dimensional rotation through an angle $\theta$, the eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so the trace of a three-dimensional rotation is $1+2\cos \theta$.

Under the identification with the rotational symmetry group of a cube, elements of order two give half-turns, elements of order three give turns through $\pm 2\pi /3$, and elements of order four give turns through $\pm \pi /2$. A half-turn can have an axis through opposite edges or opposite faces, but both kinds have trace $-1$. ThusIts character inner product with itself isBy character orthogonality, a nonzero complex representation has character inner product one with itself exactly when it is an irreducible representation. Therefore $C$ is an irreducible representation of dimension three.

For comparison useful in part (e), the standard representation of the symmetric group has character equal to the number of fixed letters minus one, namely $(3,1,-1,0,-1)$ in this order. Multiplication by the sign representation changes this character to $(3,-1,-1,0,1)$. Thus $C$ is the tensor product of the standard representation of the symmetric group with the sign representation.

In the order of conjugacy classes used in part (c), the elements of $H$ have the following types:By part (d), only $1$ and $r^2$ have nonzero character values. Formula (b), or equivalently the element-sum version proved there, therefore giveswith zero on the other three conjugacy classes. Here the centralizer order is $24/3=8$. Thus ${\chi }_V=(6,0,-2,0,0)$.

Let $\epsilon$ denote the sign representation and take the explicit two-dimensional $G$-moduleIts character is ${\chi }_U=(2,0,2,2,0)$. The character of a tensor product of group representations is the pointwise product of their characters, because $\mathrm{tr}(M\otimes N)=\mathrm{tr}(M)\mathrm{tr}(N)$. ConsequentlyBy Maschke's theorem both complex representations are semisimple representations, and characters distinguish finite-dimensional complex semisimple representations. ThereforeThis is the induction of the planar square representation to the symmetric group on four letters.

The dimension of the fixed-point subspace of a group action is the multiplicity of the trivial representation. Its character inner product here isHence $V^G=0$. Equivalently, Frobenius reciprocity identifies this multiplicity with $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.

On a one-dimensional representation, $s_i$ acts by a nonzero scalar ${\epsilon }_i$. Its square is one. The braid relation becomes ${\epsilon }_i^2{\epsilon }_{i+1}={\epsilon }_{i+1}^2{\epsilon }_i$, and hence ${\epsilon }_{i+1}={\epsilon }_i$. Thus all adjacent transpositions act by one common scalar $\epsilon$ with ${\epsilon }^2=1$.

In characteristic different from two, $\epsilon =1$ or $-1$. Since adjacent transpositions generate the group, these choices give exactly the trivial representation and sign representation. For $n\le 1$ they coincide, leaving only the trivial representation.

The standard polytabloid basis is defined over the integers and remains a basis after reduction modulo three. Thus $\dim S^{(3,2)}=f^{(3,2)}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)-\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$. The given submodule has dimension four by part (d), so the quotient is one-dimensional and part (c) leaves only the trivial or sign action.

A transposition has ordinary trace one on $S^{(3,2)}$: the Murnaghan–Nakayama rule has only the horizontal bottom-row two-strip, leaving shape $(3)$ with dimension one. Reduction of its integral action matrix preserves this trace in $F$. Its trace on $D^{(4,1)}=S^{(4,1)}$ is $3-1=2$, since the natural five-point permutation representation has three fixed coordinates and contains the trivial line. Therefore its trace on the one-dimensional quotient is $1-2=-1$, distinct from one in characteristic three. This gives the characteristic-three sign quotient of the Specht module of shape three two. The quotient is the sign representation:

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.

For the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.