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For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

Two permutations in the same symmetric group are conjugate if $\tau =\pi \sigma {\pi }^{-1}$ for some permutation $\pi$. Conjugation relabels every permutation cycle, so it preserves cycle type. Conversely, equal cycle types permit a relabelling between corresponding cycles, establishing conjugacy.

In a finite group, a cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function. Each element generates exactly one cyclic subgroup, so dividing the number of elements of order $m$ by $\phi (m)$ counts these subgroups. Include the identity subgroup using $m=1$ and $\phi (1)=1$. In a symmetric group, element counts come from cycle type and order of a group element is the least common multiple of cycle lengths.

The full symmetry group of a cube acts transitively and faithfully on its twelve edges. For vertices $(\pm 1,\pm 1,\pm 1)$, its signed permutation matrices act transitively on the edges. The stabilizer subgroup of $\{(1,1,z):-1\le z\le 1\}$ consists of independently swapping $x,y$ and reversing $z$, and is a Klein four-group. The orbit-stabilizer theorem gives order $12\cdot 4=48$. Fixing every edge fixes every vertex, as each vertex is the unique intersection of its three incident edges, so the group action is faithful. The resulting subgroup of $S_{12}$ has index $12!/48$ and is not normal: central inversion has cycle type $2^6$, whose conjugacy class in $S_{12}$ has $12!/(2^{6}6!)=10395>48$ elements.

An group automorphism preserves element orders and sizes of conjugacy classes. A nonidentity involution in $S_n$ has cycle type $2^k{1}^{n-2k}$ and class sizeFor its class to have the same size as the transposition class, it must satisfyFor fixed $k\ge 2$ the left side grows strictly with $n$. When $k=2$, it is $(n-2)(n-3)$, which is two at $n=4$ and at least six at $n=5$, never the required four. For $k=3$ equality first occurs at $n=6$, since $4!=2^2\cdot 3!$, and thereafter the left side is larger. For $k\ge 4$ it is already larger at $n=2k$: the ratio $(2k-2)!/[2^{k-1}k!]$ is $15/4$ at $k=4$, and increases with $k$, its successive ratio being $k(2k-1)/(k+1)>1$. This proves the symmetric-group involution class-size collision result. Therefore, for $n\ne 6$, every group automorphism maps transpositions to transpositions.

Now recover the underlying points from transpositions. Two distinct transpositions fail to commute exactly when their two-element supports meet. A maximal family of pairwise intersecting edges of a complete graph is either a vertex star or a triangle. Indeed, after selecting edges $\{1,2\}$ and $\{1,3\}$, every edge either contains 1 or is $\{2,3\}$; if the latter occurs, all edges lie in that triangle. For $n\ge 5$, the stars have size $n-1\ge 4$ and triangles size three, so stars are recognized purely by the multiplication structure of $S_n$.

An group automorphism must thus permute the stars, giving a permutation $\tau$ of the points. The unique transposition shared by stars $i,j$ is $(ij)$, so its image is $(\tau (i)\tau (j))$. Since transpositions generate $S_n$, the whole group automorphism is conjugation by $\tau$.

At $n=4$, stars and triangles both have three elements, but their generated subgroups distinguish them: the transpositions in a star generate all of $S_4$ (since $(ij)=(ai)(aj)(ai)$ for its center $a$), while a triangle generates the $S_3$ fixing the fourth point. The same reconstruction follows. At $n=3$, conjugation by $S_3$ induces every permutation of its three transpositions (a transposition swaps the other two, and a three-cycle rotates all three), so a transposition-preserving group automorphism is again inner. At $n=2$ the only group automorphism of $C_2$ is the identity, and $n=0,1$ are trivial (if the empty-set degree is included). ThusThe small-degree adjustments are necessary: a cardinality-only star argument would not cover $n=4$.

Use the six-point matching geometry. A duad is an edge of the complete graph on six points, a syntheme is a perfect matching of its three pairs, and a pentad is a one-factorization: five pairwise edge-disjoint synthemes covering all fifteen duads.

First count the pentads without assuming their number. There are fifteen synthemes: the partner of the first point has five choices, the least remaining point then has three possible partners, and the last pair is forced. For a fixed syntheme $M$, eight others share no edge with it: inclusion-exclusion gives $15-3\cdot 3+3\cdot 1-1=8$. Two edge-disjoint synthemes $M,M^{\prime }$ have union a six-cycle. The complement of that cycle is a triangular prism, consisting of two triangles with three corresponding cross edges. It has precisely four perfect matchings: the all-cross matching and three matchings using one cross edge plus one edge from each triangle. The all-cross matching leaves two odd triangles and therefore cannot be completed by two matchings. The other three matchings are pairwise edge-disjoint and partition the prism. Hence $M,M^{\prime }$ extend to exactly one pentad.

Each pentad through $M$ uses four of its eight disjoint partners, and every partner gives exactly one completion. Thus $M$ lies in two pentads. Counting incidences gives $15\cdot 2/5=6$ pentads in total. Distinct pentads cannot share two synthemes because two disjoint synthemes determine their unique completion. Every syntheme therefore identifies a distinct pair of pentads; there are fifteen such pairs, so this incidence map is a bijection.

Permuting the six original points permutes the six pentads, defining a group homomorphism $\Phi :S_6\to \mathrm{Sym}(\text{pentads})$. It is faithful. If an element fixes every pentad, it fixes every syntheme by the pair-incidence bijection. A duad is the unique common edge of two distinct synthemes containing it, so every duad is fixed as well. A permutation fixing all two-element subsets fixes every point, by intersecting $\{i,j\}$ and $\{i,k\}$ for distinct $i,j,k$. The kernel of a group homomorphism is therefore trivial. Both groups have order $6!$, so labeling the pentads produces an group automorphism of $S_6$.

It remains to prove this group automorphism is outer. A transposition, say $(12)$, fixes no pentad. If it fixed one, the unique syntheme in that pentad containing the duad $\{1,2\}$ would be fixed. Every other syntheme $M^{\prime }$ in the pentad contains 1 and 2 in separate edges. Swapping them changes those two edges but leaves the third edge unchanged. Hence $(12)M^{\prime }$ is a distinct syntheme sharing an edge with $M^{\prime }$, impossible for two members of the same pentad. Thus $(12)$ acts without fixed points on the six pentads and becomes a product of three disjoint transpositions.

Inner automorphisms preserve cycle type, whereas $\Phi$ sends a single transposition to a triple transposition. ConsequentlyThis realizes the exceptional class-size coincidence from part (a). It is the full symmetric-group version of the pentad construction of the exceptional alternating-group automorphism, with faithfulness proved directly from incidence rather than borrowed from a simplicity theorem.

Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

The rotation $g$ has order six. The possible cycle types in the alternating group $A_4$ are the identity, a three-cycle, and a product of two disjoint transpositions; their orders are one, three and two. Since a group isomorphism preserves the order of a group element, $D_{12}$ is not isomorphic to $A_4$.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.

Undo conjugation by $\pi$. The remaining group automorphism $\beta$ fixes each cyclic $3$-subgroup, so it fixes or inverts each generator. For adjacent supports, inverting exactly one generator changes the order of their product between two and three; inverting both preserves that order. Hence the two inversion choices must agree. The triple-support graph is connected, since one can replace differing points one at a time, so the choice is uniform. Inverting every $3$-cycle is impossible: take $c=(123)$, $d=(142)$, whose product is the $3$-cycle $(143)$. Then $\beta (cd)=(cd{)}^{-1}=d^{-1}{c}^{-1}$, whereas the group homomorphism property would give $c^{-1}{d}^{-1}$; these are unequal because $c,d$ do not commute. Therefore $\beta$ fixes every $3$-cycle and is the identity. Conversely, conjugation by $S_n$ acts faithfully on $A_n$, since centralizing all $3$-cycles fixes every triple and every point. The automorphisms of alternating groups from triple supports are exactly

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.

Use the induction product of symmetric-group characters: for $a\ge 0$, $[a]$ is the trivial character of $S_a$, $[0]$ is the unit and $[a]=0$ for $a<0$. Products mean induction of external tensor products. The determinantal form iswhere $r\ge \ell (\lambda )$. Each determinant term is a virtual character of $S_n$, because its indices sum to $n$.

Here is a proof. The Frobenius characteristic map sends $[a]$ to the complete homogeneous symmetric function $h_a$, and induction products to multiplication. By Young's rule its value on $[\lambda ]$ is the Schur function $s_{\lambda }$, the generating function of semistandard tableaux of shape $\lambda$. To identify that function with $\det (h_{{\lambda }_i-i+j})$, work in $m$ variables and take starting points $A_j=(-j,1)$ and ending points $B_i=({\lambda }_i-i,m)$. A path from $A_j$ to $B_i$ has east and north steps; each east step at height $k$ has weight $x_k$ and each north step has weight one. Its east-step count is ${\lambda }_i-i+j$, so its weight sum is $h_{{\lambda }_i-i+j}$. Expanding the determinant sums signed systems of paths with permuted endpoints. For a system with an intersection, exchange the two tails after the first intersection. This preserves its monomial weight and reverses the endpoint permutation's sign, so all intersecting systems cancel in pairs. Nonintersecting systems necessarily have the identity endpoint permutation and correspond precisely to weak rows with strictly increasing columns, namely semistandard tableaux of shape $\lambda$. Their signs are positive. This proves $s_{\lambda }=\det (h_{{\lambda }_i-i+j})$, the Jacobi–Trudi identity, and injectivity of the Frobenius characteristic map gives the character formula above. The path sum may be taken in finitely many variables first and then stabilized, so no infinite formal sum is required in the argument.

A removable rim hook, or border strip, of length $k$ is a connected skew diagram $\lambda /\nu$ of $k$ cells with no $2\times 2$ square, where removal leaves a partition of an integer. Its leg length is the number of occupied rows minus one. The Murnaghan–Nakayama rule statesA proof sketch follows directly from the alternant form of Schur functions. Multiply an alternant by $p_k={\sum }_j{x}_j^k$. In each term one alternant exponent increases by $k$. Equal exponents cancel; otherwise reordering the exponents contributes the sign of the number crossed. In the beta-set description, increasing one exponent is adding a $k$-rim hook, and the number crossed is its leg length. Therefore $p_k{s}_{\nu }={\sum }_{\lambda }(-1{)}^{\mathrm{leg}(\lambda /\nu )}{s}_{\lambda }$. Taking its adjoint in the Hall inner product of symmetric functions, where Schur functions are orthonormal and power-sum symmetric polynomials have squared norm $z_{\rho }$, reads off the value at a $k$-cycle and the remaining cycle type. This gives the stated rule and explains its sign.

The printed shape is $(4^2,3)=(4,4,3)$, of size 11; the converted TeX's $(4,2,3)$ is incorrect and would have the wrong size. There are two removable length-5 rim hooks. They leave $(3,2,1)$ with leg length 2 and $(4,2)$ with leg length 1. ThusThe staircase $(3,2,1)$ has no removable length-4 rim hook, so its term is zero. The unique removable length-4 rim hook of $(4,2)$ leaves $(1,1)$ and has leg length 1. The remaining length-2 rim hook in $(1,1)$ also has leg length 1, giving ${\chi }^{(4,2)}(4,2)=(-1)(-1)=1$. HenceAs a separate determinant check, the only term whose row sizes can be unions of cycles of lengths $5,4,2$ is the transposition term with sizes $(4,5,2)$. Its permutation character has value one and its determinant sign is negative.

Taking $k=1$ in the Murnaghan–Nakayama rule removes a single corner and every leg length is zero. Evaluating on a permutation fixing the distinguished point therefore givesThis is the restriction branching rule for a symmetric group; ordinary semisimple representation theory converts the character equality into a direct-sum decomposition. Frobenius reciprocity gives the corresponding add-one-cell induction rule.

For a positive prime $p$, the weight of a partition $\lambda$ at prime modulus $p$, denoted $w$, is the number of length-$p$ rim hooks removed in reaching its partition core $\kappa$, so $n=|\kappa |+pw$. On a prime-modulus partition abacus write its partition quotient as $({\lambda }^{(0)},\dots ,{\lambda }^{(p-1)})$ and put $w_i=|{\lambda }^{(i)}|$, with ${\sum }_i{w}_i=w$. Removing a length-$p$ rim hook removes one corner from one quotient partition of an integer. Complete removal sequences therefore numberwhere $H_{\alpha }$ is the hook product of a partition and $H_{\varnothing }=1$. Each component's standard tableaux encode its possible removal orders, and the multinomial coefficient interleaves the components.

All these sequences have the same total sign ${\epsilon }_p(\lambda )$. To see this, label the beads on each runner in their preserved order. Each slide crosses exactly the beads counted by its rim-hook leg. Its sign is the change in the total order of bead labels, and the product telescopes to the permutation from the initial ordering to the final packed-core ordering, independently of the slides chosen. Repeating the Murnaghan–Nakayama rule on the $w$ distinguished $p$-cycles givesHere $\pi$ acts on the complementary $n-pw$ letters; no $p$-regularity hypothesis on $\pi$ is necessary. If its cycle type contains a part divisible by $p$, the partition core character is zero, since its partition abacus admits no hook removal of a length divisible by $p$. For $w=0$ the coefficient and sign are one.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.

Thus equal residue multisets yield equal characters of the entire modular center. Different residue multisets are separated already by the coefficients of the central polynomial ${\prod }_k(t+J_k)$: its scalar on the shape $\lambda$ reduces to ${\prod }_r(t+r{)}^{c_r(\lambda )}$, whose unique factorization records the multiplicities exactly, even in characteristic $p$. In a finite-dimensional split algebra, two simple modules are in the same block exactly when their central characters agree: the center of each block is local, while different block idempotents separate different blocks. It follows that equal cores give one block and different cores give distinct blocks. This proves both directions rather than just observing that core is constant on known blocks.

Finally, each ordinary Specht lattice has a single reduced central character, so all constituents of its reduction lie in that one block. Question 1 supplies every simple module and its label; the regular labels therefore identify all the simples in each core block. The constructions use integer coefficients and absolutely irreducible prime-field modules, so the block description descends to the arbitrary characteristic-$p$ field in the question. This completes the main steps of the block proof without presupposing semisimplicity in positive characteristic.

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.

For any permutation $\tau$, conjugating a permutation cycle relabels its entries:Indeed, the conjugate sends $\tau (a_j)$ to $\tau (a_{j+1})$ and fixes the other relabeled entries exactly as the original permutation cycle does. Therefore conjugation preserves the multiset of permutation cycle lengths. Conversely, if two permutations have the same cycle type, pair their permutation cycles of each length and let $\tau$ carry the $j$th entry of each first permutation cycle to the $j$th entry of its partner. Include one-cycles for fixed points. This defines a bijection of all labels, and the displayed relabeling identity makes the permutations conjugate. ThusFor $n\ge 2$, the alternating conjugacy class splitting criterion says that an even permutation has the same class in $A_n$ and $S_n$ precisely when its type is not a collection of pairwise distinct odd lengths. Equivalently, there is an even-length permutation cycle or a repeated permutation cycle length, counting each fixed point as a length-one permutation cycle. Distinct odd lengths instead give two alternating-group classes. For $n=1$, $A_1=S_1$ and the sole class is the same; the index-two splitting criterion is not applicable.

Squaring an odd-length permutation cycle keeps one permutation cycle of that length, since advancing by two visits all of its entries. Squaring an even-length permutation cycle splits it into two permutation cycles, each of half that length. Thus any even-length permutation cycle strictly increases the total permutation cycle count on squaring, while odd permutation cycles leave that count unchanged. The cycle type can agree with its square exactly when every length is odd. This proves permutation conjugate to its square:The final equivalence follows because the order of a group element that is a permutation is the least common multiple of its permutation cycle lengths.

The possible even types on five labels are the identity, a three-cycle, a product of two disjoint transpositions and a five-cycle. The identity and the double transpositions are their own inverses. For a three-cycle $(abc)$ with unused labels $d,e$, the even permutation $(bc)(de)$ conjugates it to its inverse. For a five-cycle $(abcde)$, the even permutation $(be)(cd)$ reverses its order. Therefore every element of $A_5$ is conjugate to its inverse within $A_5$. Inverting a five-cycle uses two transpositions, in agreement with inversion of an odd cycle in an alternating group.

A small counterexample elsewhere isThe alternating group $A_3=\{e,(123),(132)\}$ is cyclic and therefore abelian. Every conjugacy class in an abelian group is a singleton, while $\sigma \ne {\sigma }^{-1}$. Thus these two elements are not conjugate in $A_3$.

For a permutation $\sigma$ of a finite set, start with a point $x$. Repeatedly applying $\sigma$ eventually returns to $x$: a repeated point in the finite sequence can be cancelled backwards using injectivity. The distinct successive points before that return form a cycle. If any point has not yet been used, repeat the construction there. The resulting orbits are disjoint and cover the set, so their cycles, including one-cycles for fixed points, give a cycle decomposition of a permutation. Disjoint cycles commute because they act on different points.

Conjugation relabels a cycle:Thus conjugate permutations have the same cycle type. Conversely, if two decompositions have equal numbers of cycles of each length, pair their equally long cycles and define $\gamma$ to take corresponding entries, in order, to one another. Since the cycles partition the set, this defines a bijection, and the displayed identity proves the conjugacy. Hence cycle type characterizes conjugacy in the symmetric group.

For $\tau =(ab)$, use the convention that the rightmost permutation acts first. If $a,b$ are in different cycles, write those cycles asLeft multiplication by $\tau$ changes the arrow into $a$ to an arrow into $b$, and the arrow into $b$ to an arrow into $a$. Following these arrows gives the single cycleNo other cycles change, so $\ell (\tau \sigma )=\ell (\sigma )-1$. If $a,b$ are in the same cycle, write it in that last form; the same two arrow changes split it into the two cycles in the preceding display. Thus $\ell (\tau \sigma )=\ell (\sigma )+1$. Empty lists are allowed and produce one-cycles. This proves transposition changes the cycle count by one in all cases:

A simple group has no normal subgroup except the identity subgroup and the whole group. To prove the Simplicity of the alternating group A5, we compute its conjugacy classes. Since the sign homomorphism is onto, $|A_5|=5!/2=60$. Its possible nonidentity cycle types are a three-cycle, a product of two disjoint transpositions, and a five-cycle.

There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles. The centralizer in $S_5$ of $(123)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its centralizer in $A_5$ therefore has order three, giving one conjugacy class of size $60/3=20$ by the orbit-stabilizer theorem.

There are $5\cdot 3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(12)(34)$, a commuting permutation fixes $5$ and may interchange the two pairs and swap within them, giving a centralizer of order eight in $S_5$. It contains an odd element, $(12)$, so exactly half its elements are even. The centralizer in $A_5$ has order four, and these fifteen elements form one conjugacy class of size $60/4=15$.

Finally, there are $4!=24$ five-cycles. A permutation commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the centralizer in $A_5$ has order five. Each such conjugacy class has size $60/5=12$, giving two classes of size twelve.

A normal subgroup must be a union of conjugacy classes and must include the identity. Its order therefore has the formThe resulting distinct possibilities areOf these, only $1$ and $60$ divide $60$. Lagrange's theorem rules out every other possible order. Hence

Every finite cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function: if $g$ has order $m$, then $g^k$ generates precisely when $\gcd (k,m)=1$. Each element generates exactly one cyclic subgroup. Thus counting cyclic subgroups by their generators amounts to counting elements by order of a group element and dividing by $\phi (m)$.

Use the cycle decomposition of a permutation in $S_5$. A permutation with $a_j$ cycles of length $j$ occurs $5!/{\prod }_j(j^{a_j}{a}_j!)$ times: order the five letters, then divide by the cyclic choices of starting point and the orderings of equal-length cycles. Its order is the least common multiple of the cycle lengths. The resulting count isThe two subgroups $⟨(12)⟩$ and $⟨(12)(34)⟩$ are both isomorphic to $C_2$ by a group isomorphism. They are not conjugate subgroups: conjugating their unique nonidentity elements would have to preserve cycle type, whereas a transposition and a product of two disjoint transpositions have different cycle types.

Conjugating a cycle merely relabels its entries:Thus conjugate permutations have the same cycle type. Conversely, if $\sigma$ and $\tau$ have the same cycle type, match the entries of each cycle of $\sigma$ bijectively and in cyclic order with those of a cycle of $\tau$ of the same length. Extending these matches to a permutation $g$ gives $g\sigma g^{-1}=\tau$.

Let $C_{S_n}(\sigma )$ be the centralizer of $\sigma$. Its $S_n$-conjugacy class remains one $A_n$-conjugacy class exactly when $C_{S_n}(\sigma )$ contains an odd permutation. Indeed, in that case $C_{A_n}(\sigma )$ has index two in $C_{S_n}(\sigma )$, so the orbit-stabilizer theorem givesIf the centralizer contains only even permutations, the $A_n$ class has half the size and the $S_n$ class splits into two $A_n$ classes.

The even cycle types in $S_5$ areThe classes of the identity, a three-cycle, and a double transposition do not split: their centralizers contain an odd permutation. The centralizer of a five-cycle is its cyclic subgroup of order five, which lies in $A_5$, so that class splits in two. Therefore

An element of order three in $S_7$ has cycle type $31^4$ or $3^{2}1$. There aresingle three-cycles. For two disjoint three-cycles, choose the two unordered three-element supports and an orientation on each:Hence $S_7$ contains $\boxed{350}$ elements of order three.