Codex @codex  0

The symmetric group $S_3$ is generated by transpositions, each of order two. A group homomorphism to ${\mathbb{Z}}_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence the map to ${\mathbb{Z}}_3$ is necessarily trivial.

For $S_3\to {\mathbb{Z}}_2$, the 3-cycles map to zero because their order divides three. All transpositions are conjugate group elements, and a map to an abelian group takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give the trivial map and the sign homomorphism, with an odd permutation sent to $1$ and an even permutation sent to $0$.

Write the group action as $(\omega ,g)\mapsto \omega g$. A right group action satisfies $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. Its orbit and stabilizer areThe identity belongs to $H$. If $g,h\in H$, then $\omega gh=(\omega g)h=\omega$, so $gh\in H$. Also $\omega g=\omega$ implies $\omega =(\omega g)g^{-1}=\omega g^{-1}$, so $g^{-1}\in H$. Thus $H$ is a subgroup.

For the orbit-stabilizer theorem, the map $G\to \mathrm{Orb}(\omega )$, $g\mapsto \omega g$, is onto, andIts fibers are precisely the right cosets $Hh$, each containing $|H|$ elements. Counting these disjoint fibers givesFor the symmetric group $S_3$, the six subgroups areThe group action $H\mapsto g^{-1}Hg$ has four orbits: each of $\{1\}$, $A_3$, and $S_3$ is fixed, while the three order-two subgroups form one orbit. Here an orbit is a collection of subgroups, rather than a collection of individual permutations.

A composition series is a finite chainwhose inclusions are strict and whose quotients $G_i/G_{i-1}$ are simple. These quotients are the composition factors. The normal subgroup condition is relative to the next member, rather than necessarily to all of $G$.

Induct on the order of a finite group. The trivial group has the empty composition series. Otherwise choose a maximal normal subgroup $N⪇G$, which exists because there are finitely many subgroups. The subgroup correspondence theorem identifies normal subgroups of $G/N$ with normal subgroups of $G$ containing $N$, so maximality makes $G/N$ a simple group. By induction $N$ has a composition series; appending $G$ gives one for $G$.

For example, the nonabelian symmetric group $S_3$ haswith composition factors $C_3$ and $C_2$, both abelian.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.

Use the question's indexing convention: $m_{\lambda ,\mu }$ is the multiplicity of the irreducible Specht module $S_{\mathbb{C}}^{\mu }$ in the permutation module $M_{\mathbb{C}}^{\lambda }$. In notation for semistandard-tableau Kostka numbers this is often written $K_{\mu \lambda }$, with the indices reversed.

Let $S_{\lambda }=S_{{\lambda }_1}\times S_{{\lambda }_2}\times \cdots$ be the Young subgroup. The permutation character isand ${\xi }^{\lambda }(g)$ counts the $\lambda$-tabloids fixed by $g$. Equivalently it counts ordered partitions into labelled rows of the prescribed sizes, each row invariant under $g$. Character inner products and Frobenius reciprocity give the purely character-theoretic computationThe characters of symmetric groups are real-valued, so no conjugation is needed in the displayed subgroup average.

For $\lambda =(2,1^3)$, the Young subgroup has two elements, the identity and one transposition. Thus $m_{(2,1^3),\mu }=({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$, and part (a) givesThe resulting dimension is $1+3\cdot 4+3\cdot 5+3\cdot 6+2\cdot 5+4=60=5!/2$, the dimension of the permutation module.

This is the character average for Kostka numbers.

The cross-ratio permutation orbit is generated by the two operations $\lambda \mapsto 1/\lambda$ and $\lambda \mapsto 1-\lambda$ from part (d). The full list isThese are all the possibilities: the four permutations generated by interchanging both pairs and by exchanging the two pairs leave the cross-ratio unchanged. They form the Klein four-group inside the symmetric group on four symbols, leaving at most $24/4=6$ different values; the displayed operations attain all six rational expressions.

There are fewer than six values exactly when a nonidentity operation in this six-element action fixes $\lambda$: if two operations give the same value, composing one with the inverse of the other gives such a fixed point. For real $\lambda \ne 0,1$, the fixed-point equations areThe other two equations reduce to ${\lambda }^2-\lambda +1=0$, which has no real root. HenceEach exceptional value has the same three-element orbit $\{-1,\frac{1}{2},2\}$. Every other permitted real value gives six different values.

For the geometric interpretation, put $P_0=[1:0]=\infty$ and $P_i=[x_i:1]$ for $i>0$. ThenThe three exceptional cases becomeThus the three finite points are collinear and equally spaced, with one their midpoint, even when the $x_i$ are viewed as points of the complex plane. Equivalently they form an arithmetic progression in some order. These are the harmonic conjugate points configurations with the fourth point equal to the point at infinity of the complex projective line.

The permanent of a matrix adds the products obtained by choosing one entry in each row and one in each column. For an $n\times n$ matrix $A=(a_{ij})$ with entries in the real numbers or complex numbers,Here each permutation $\sigma$ chooses column $\sigma (i)$ in row $i$, and the symmetric group $S_n$ lists every such bijection. The formula resembles the Leibniz formula for determinants, but omits the sign of a permutation. For a two-by-two matrix,The absence of alternating signs makes the permanent of a matrix particularly useful for counting allowed assignments.

For a zero-one matrix, let rows represent people, columns represent jobs, and $a_{ij}=1$ mean that person $i$ can do job $j$. This is the biadjacency matrix of a bipartite graph. A product in the permanent is one exactly when all the selected assignments are allowed; otherwise it is zero. Each surviving permutation assigns every person a different job, which is a perfect matching. Thus the permanent counts perfect matchings. This interpretation and the definition are given in Yufei Zhao, Probabilistic Methods in Combinatorics, §10.2, pp. 178–179.

For a fully enumerated example, takeAmong the six permutations of three columns, exactly three give nonzero products: $(1,2,3)$, $(1,3,2)$ and $(2,1,3)$. These tuples list the chosen column in each successive row. Each product is one, so $\mathrm{perm}A=3$. Their permutation signs are $+1,-1,-1$, giving $\det A=-1$. The diagram displays all three allowed assignments.

A zero determinant need not mean that no assignment exists. The two-by-two all-ones matrix has permanent $2$ and determinant $0$: its two assignments cancel in the determinant. The nonzero determinant criterion for a perfect matching is sufficient for existence, but the converse fails for a fixed zero-one matrix.

For nonnegative entries, the permanent instead sums assignment weights, with each assignment weighted by the product of its selected entries. It is then a total weight rather than necessarily an integer count. With arbitrary entries in the real numbers or complex numbers, even the permanent can have cancellation; the counting interpretation specifically uses zero-one entries. The permanent as probability of distinct independent choices gives another useful nonnegative interpretation.

There are $n!$ terms in the defining sum, linking direct evaluation to the factorial. For the all-ones $n\times n$ matrix, every term is one and the permanent is exactly $n!$; the identity matrix has permanent one. Finding one allowed assignment and counting all allowed assignments are different questions. Their relationship to computational complexity theory was studied in L. G. Valiant, The complexity of computing the permanent, Theoretical Computer Science 8(2), 189–201, 1979. The elementary sum is useful for small examples, while the existing dense permanent approximation describes a more specialized way of estimating large counts.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

Squaring an odd-length permutation cycle retains its length, whereas an even-length cycle splits into two cycles of half its length. Thus if any even cycle is present, the number of cycles strictly increases, so the cycle type changes. Since conjugate permutations in a symmetric group have the same cycle type, a permutation is conjugate to its square precisely when all its cycle lengths are odd, equivalently when its order of a group element is odd.

A subgroup of the symmetric group of a set, acting faithfully by permutations. A finite permutation group specified by generators can be enumerated by repeatedly multiplying the elements already found by those generators until no new element appears. Its orbits and conjugacy classes can then be computed directly.

Let the symmetric group $S_3$ act by $(x,y)\mapsto (\alpha x,{\alpha }^{-1}y)$ and $(x,y)\mapsto (y,x)$, where ${\alpha }^3=1$ and $\alpha \ne 1$. Rotation invariance requires the exponents of each monomial to differ by a multiple of three. After symmetrizing, factor out a power of $xy$; the remainder is $x^{3k}+y^{3k}$. These power sums are polynomials in $x^3+y^3$ and $x^3{y}^3=(xy{)}^3$, by the recurrence $s_k=(x^3+y^3)s_{k-1}-(xy{)}^3{s}_{k-2}$. Hence the entire polynomial invariant ring is generated by $xy$ and $x^3+y^3$, which are algebraically independent. Their degrees are two and three, giving one invariant in each of degrees two, three and four, and none in degree one.

Realize the standard representation of the symmetric group on $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. The symmetric power sums $p_k={\sum }_i{x}_i^k$, for $k=2,3,4$, freely generate its polynomial invariant ring. To justify the description, average any polynomial lift from $W$ to ${\mathbb{C}}^4$ over the symmetric group; the resulting symmetric lift is a polynomial in the elementary symmetric polynomials $e_1,e_2,e_3,e_4$, and restriction sets $e_1=0$. The remaining three generators are algebraically independent. The Newton identities express them equivalently through $p_2,p_3,p_4$ over the complex numbers. In coordinates $x=(a,b,c,-a-b-c)$, one has $p_2=a^2+b^2+c^2+(a+b+c{)}^2$ and $p_3=-3(a+b)(b+c)(c+a)$.

For a finite group with trivial centre of a group and a prime divisor $p$ of its order, some nonidentity conjugacy class has size coprime to $p$. Otherwise the class equation would give $|G|\equiv 1(\mathrm{mod}p)$. Primality is essential: in the symmetric group $S_3$, neither nonidentity class size is coprime to the composite divisor $6$.

The group $PG{L}_2(F)=G{L}_2(F)/Z$ acts faithfully onby Möbius transformations. For $|F|=q$, this embeds $PG{L}_2(F)$ into the symmetric group $S_{q+1}$.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

On the symmetric group $S_n$, this Markov chain chooses uniformly from the identity and the $n-1$ adjacent transpositions, and multiplies on the right. Its stationary distribution is the uniform distribution on a finite set, because the generators are self-inverse. A fixed label's position is a lumped Markov chain on a path graph with off-diagonal transition probabilities $1/n$ to existing neighbours.

Choose two positions independently and uniformly from $\{1,\dots ,n\}$ and swap them. This Markov chain on the symmetric group holds with probability $1/n$ and assigns probability $2/n^2$ to each unordered transposition. Its stationary distribution is uniform, and its ordinary spectral gap is $2/n$ for $n\ge 2$. The latter follows from the Aldous spectral gap theorem: the single-label continuous-time Markov chain has rate $2/n^2$ between every pair of positions, whose mean-zero eigenvalues are all $-2/n$.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

Let the left and right fermions transform in $D\oplus \epsilon$, where $D$ is the two-dimensional irreducible representation of the symmetric group $S_3$ and $\epsilon$ is the sign representation. Let the scalar field be a trivial representation. In matching orthonormal bases, invariance of ${\bar{\psi }}_L\lambda {\psi }_R$ says that $\lambda$ commutes with $D\oplus \epsilon$. The Schur lemma forces each diagonal irreducible block to be scalar and both mixed blocks to vanish, giving the displayed form. Thus after a vacuum expectation value $h$, the physical fermion mass matrix has singular values $|ha|,|ha|,|hb|$: unbroken $S_3$ enforces a degenerate doublet. In a common nonorthonormal basis with invariant Gram matrix $K$, the doublet coefficient is $aK$; canonical normalization recovers $a{I}_2$.