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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.

Throughout, modules are finite-dimensional over $k$, or finitely generated $\mathcal{O}$-free lattices when the coefficient ring is $\mathcal{O}$. The p-modular system has a complete discrete valuation ring $\mathcal{O}$, maximal ideal $(\pi )$, fraction field $K$ of characteristic zero and residue field $k$ of characteristic $p$.

A block of a group algebra is a nonzero, indecomposable two-sided ideal direct summand of $A=RG$. Equivalently it is $Ae$ for a primitive central idempotent $e$. To see the equivalence, suppose $A=B\oplus C$ as two-sided ideals and write $1=e+f$ with $e\in B$, $f\in C$. Products between $B$ and $C$ vanish because they lie in their zero intersection. Thus $e$ is the identity on $B$, zero on $C$, and is a central idempotent, with $B=Ae$. Conversely $A=Ae\oplus A(1-e)$. Splitting $Ae$ further is equivalent to splitting $e$ into orthogonal nonzero central idempotents; indecomposability is exactly primitivity. If $1=\sum e_i$ is the block decomposition, $M=⨁e_{i}M$. A nonzero indecomposable module therefore belongs to exactly one block of a group algebra, namely the one with $e_{i}M=M$.

For modular reduction of block idempotents, the conjugacy class sums are an $R$-basis of $Z(RG)$. Hence $Z(\mathcal{O}G)/\pi Z(\mathcal{O}G)\cong Z(kG)$. In the complete commutative algebra $Z(\mathcal{O}G)$, idempotent lifting gives a unique lift of each idempotent of $Z(kG)$; uniqueness follows because commuting idempotents congruent modulo the Jacobson radical are equal. Orthogonal decompositions lift too, so primitivity is preserved. Reduction gives a bijection between the integral and modular blocks.

Here is a definition of defect groups of a block which also proves the required relative-projectivity assertion. Give $RG$ its conjugation action and, for $H\le G$, define the relative trace and its transfer ideal of conjugation-fixed elements byA defect group of a block $RGe$ is a $p$-subgroup $D$ minimal with $e\in I_D$. Such a subgroup exists: for a Sylow subgroup $P$, $[G:P]$ is a unit in $R$ and $e={\mathrm{Tr}}_P^G(e/[G:P])$. The defect of a block is $d$ where $|D|=p^d$.

For completeness, this definition is independent of a choice up to conjugacy. Multiplying two relative traces and grouping the pairs of cosets by double cosets givesThis identity follows by writing each pair of cosets as a simultaneous $G$-orbit; its stabilizer is $D\cap {}^{g}E$, and its contribution is a trace from that stabilizer. If $D,E$ are both minimal for $e$, then $e=e^2$ belongs to the displayed sum. The commutative block center $eZ(RG)$ is a local ring: in the field case it is a commutative Artinian ring with no nontrivial idempotents; in the integral case its reduction has this property and completeness detects units. A sum of proper ideals cannot contain its identity. Consequently $e\in I_{D\cap {}^{g}E}$ for some $g$, forcing $D={}^{g}E$ by minimality. The same argument with an arbitrary $E$ shows that $e\in I_E$ precisely when $E$ contains a conjugate of $D$. This is the trace criterion for defect groups.

To relate this to the usual modular definition, the Brauer morphism deletes the coefficients outside the centralizer:It is an algebra homomorphism: in the coefficient of an element centralizing $P$ in a product, $P$ acts on the contributing pairs; nonfixed orbits have size divisible by $p$, while fixed pairs have both entries in $C_G(P)$. Its kernel is the sum of proper-subgroup transfers, because the nonfixed conjugation orbit sums form a basis of that kernel. If $D$ is minimal for $e={\mathrm{Tr}}_D^G(a)$ and ${\mathrm{Br}}_D(e)=0$, replace $a$ by $ea$. The kernel description gives $e\in {\sum }_{Q<D}{I}_Q$, contradicting the local-center argument. Also ${\mathrm{Br}}_P(I_D)=0$ unless $P$ is conjugate into $D$: group the cosets of $G/D$ into $P$-orbits, which have no fixed point otherwise. Thus the minimal transfer subgroups are exactly the maximal $p$-subgroups with nonzero Brauer morphism of $e$.

Now take $M=eM$ and write $e={\mathrm{Tr}}_D^G(a)$. Multiplication by $a$ is a $D$-linear endomorphism $\alpha$ of $M$, andHere is the splitting behind the D. Higman criterion. The natural map $\epsilon :RG{\otimes }_{RD}M\to M$, $g\otimes m\mapsto gm$, has a $G$-linear sectionChanging a representative by an element of $D$ leaves this tensor unchanged, $G$ permutes the terms, and $\epsilon s={\mathrm{id}}_M$. HenceThis is relative projectivity for $D$, proving the Green relative projectivity theorem; indecomposability was needed only to speak of a single block of a group algebra or vertex of an indecomposable module.

Finally let $\hat{e}$ lift $e$. A trace expression for $\hat{e}$ reduces to one for $e$, so a modular defect group of a block is contained, up to conjugacy, in an integral one. Conversely choose a modular defect group $D$ and $a\in (kG{)}^D$ with $e={\mathrm{Tr}}_D^G(a)$. Conjugation orbit sums lift $a$ to $\hat{a}\in (\mathcal{O}G{)}^D$. The elementreduces to its block identity, hence is a unit in that block center. Its inverse there is central, soThe integral defect is therefore contained in the modular one. The corresponding blocks have the same conjugacy class of defect groups and the same defect.

The principal block $B_0(G)$ is the unique block of a group algebra containing the trivial $kG$-module. Its central character of a block is augmentation $\epsilon :Z(kG)\to k$, and its block idempotent $e_0$ is the unique primitive central idempotent with $\epsilon (e_0)=1$. Its defect groups of a block are Sylow subgroups: if $e_0={\mathrm{Tr}}_D^G(a)$, augmentation gives $1=[G:D]\epsilon (a)$, so $p\nmid [G:D]$.

The Brauer third main theorem states that, if $b$ is a block of $kH$ with defect group $D$ and $D{C}_G(D)\le H\le G$, thenWe give a proof, including the normal-subgroup ingredient needed for the difficult implication.

First, block idempotents centralize a normal p-subgroup. If $P◃L$ is a p-subgroup, the ideal $J(kP)kL$ is nilpotent: conjugation preserves $J(kP)$, so its powers are $J(kP{)}^{m}kL$, eventually zero. For a central idempotent $e$ of $kL$, the difference $e-{\mathrm{Br}}_P(e)$ is a sum of nonfixed $P$-conjugation orbit sums. Such an orbit maps to a single element of $L/P$ multiplied by its size, which is zero in $k$. Thus the difference belongs to $J(kP)kL$. Since $P$ is normal, its centralizer is normal too, and the Brauer image is $L$-invariant; consequently it is itself a central idempotent of $kL$. Commuting central idempotents congruent modulo a nilpotent ideal coincide: their off-diagonal products are nilpotent idempotents, hence zero. ThereforeIt follows that central block idempotents of $kL$ are exactly the sums over $L$-orbits of primitive central idempotents of $k{C}_L(P)$. The principal block idempotent of $k{C}_L(P)$ is invariant under every automorphism preserving augmentation, so its singleton orbit is the principal block idempotent of $kL$.

In particular, if $P{C}_G(P)\le K\le L\le N_G(P)$, the groups $K,L$ have the same centralizer $C_G(P)$. Their principal block idempotents are both the principal block idempotent of $k{C}_G(P)$. By the definition using the central character of a block of block induction, a block of $K$ inducing to the principal block of $L$ must be principal: its idempotent, a sum of centralizer block idempotents, has nonzero overlap with this single principal idempotent, and is therefore precisely that idempotent. The forward implication also holds, by the same observation.

For an arbitrary subgroup $H$ in the theorem, use the Brauer first main theorem inside $H$ to replace $b$ by its correspondent $\beta$ in $K=N_H(D)$. This preserves principalness. Indeed a Brauer image preserves augmentation on central elements, since nonfixed $D$-orbits have sizes divisible by $p$; hence the principal block of $H$, when it has defect $D$, corresponds to the principal block of $K$. Conversely, if $\beta$ is principal, its induced central character of a block isso ${\beta }^H$ is principal. Transitivity gives ${\beta }^G=b^G$. Thus we may work with $D{C}_G(D)\le K\le N_G(D)$ and a block $\beta$ of defect $D$.

The forward implication is already proved by this augmentation calculation, now with $H$ replaced by $G$. For the converse, fix $G$ and suppose there is a counterexample with $\beta$ nonprincipal but ${\beta }^G$ principal. Choose one whose defect group $D$ has maximal order. Put $N=N_G(D)$ and $c={\beta }^N$. To verify the transitivity needed here, write $B={\beta }^G$. Its defining character gives $e_{\beta }{\mathrm{Br}}_D(e_B)=e_{\beta }$. The image ${\mathrm{Br}}_D(e_B)$ is $N$-invariant and central in $kN$, so the definition of $c$ gives $e_c{\mathrm{Br}}_D(e_B)=e_c$. If $E\ge D$ is a defect group of $c$, applying ${\mathrm{Br}}_E$ gives ${\mathrm{Br}}_E(e_c){\mathrm{Br}}_E(e_B)={\mathrm{Br}}_E(e_c)$, since $C_G(E)\subseteq C_G(D)$. The character defining $c^G$ therefore selects $e_B$. Thus $c^G={\beta }^G$. The normal-subgroup observation shows that $c$ is nonprincipal, since otherwise $\beta$ would be principal. Every defect group of $c$ contains the normal $p$-subgroup $D$ up to conjugacy: ${\mathrm{Br}}_D(e_c)=e_c\ne 0$. Choose a defect group $E\ge D$ of $c$. Then $C_G(E)\le C_G(D)\le N$, so $(G,N,c)$ is another admissible counterexample, now with defect group $E$. Maximality forces $E=D$.

The first main theorem applied to $N=N_G(D)$ therefore implies that the principal block $c^G$ has defect $D$. Thus $D$ is a Sylow subgroup of $G$, and also of $N$. The principal block of $N$ then has defect $D$ and, by the forward implication, induces to the principal block of $G$. Injectivity in the first main theorem gives $c=B_0(N)$, contradicting the established nonprincipalness of $c$. This proves the third main theorem.

A useful equivalent consequence, with its justification, isLet $a$ be a block idempotent of $k{C}_G(Q)$ occurring in the left side, and let $E$ be a defect group of $a$ containing the central $p$-subgroup $Q$. Then $C_G(E)\le C_G(Q)$, so $a^G$ is defined. Applying ${\mathrm{Br}}_E$ to $a{\mathrm{Br}}_Q(e_0)=a$ shows that the central character of a block defining $a^G$ selects $e_0$: deleting coefficients outside $C_G(E)$ can be done after deleting those outside $C_G(Q)$. Hence $a^G=B_0(G)$, and the theorem makes $a$ principal. The image is nonzero because its augmentation is $1$, so it is exactly the principal centralizer idempotent.

Now put $C=C_G(x)$ and $P=O_p(C)$, the p-core of a finite group. The hypothesis $O_{p^{\prime }}(C)=1$, using the p-prime core of a finite group, and the definition of a p-constrained group say $C_C(P)\le P$. The normal-subgroup observation puts every block idempotent of $kC$ inside $k{C}_C(P)\subseteq kP$. A group algebra of a p-group in characteristic p is local, so its only idempotents are $0,1$. Thus $kC$ has a single block, its principal block.

Write $Q=⟨x⟩$. By the trace and Brauer characterization of defect in Question 1,For the forward direction one can conjugate $Q$ into $D$; the nonzero image at $D$ then ensures the image at $Q$ is nonzero. The reverse direction follows because every subgroup with nonzero image is conjugate into a defect group. Since $kC$ has only one block, a nonzero image must equal $1=e_0(C)$. The preceding local-block argument then gives $B=B_0(G)$ by the third main theorem. Conversely, if $B$ is principal, its defect group is a Sylow subgroup and every $p$-element is conjugate into it. ThereforeThis includes $x=1$: in that case the hypotheses force $kG$ itself to have only its principal block.

For each involution $t\in G$, its conjugacy class has size fifteen, so its centralizer $C_G(t)$ has order four. A group of order four is a cyclic group or a Klein four-group; the cyclic possibility would contain an element of order four, excluded by the geometric enumeration. Hence each $C_G(t)$ is a Klein four-group containing three involutions.

If $u$ is any of the three involutions in $C_G(t)$, the latter group is an abelian group and hence is contained in $C_G(u)$. Both centralizers have order four, so they are equal. It follows that the fifteen involutions partition into five triples, with five distinct centralizers. Conjugation by $G$ permutes these five subgroups and gives a group homomorphism $\theta :G\to S_5$. This action is nontrivial: otherwise each of these order-four subgroups would be a normal subgroup, contrary to simplicity. The kernel of $\theta$ is normal; since $G$ is simple and the action nontrivial, its kernel is trivial. Thus $\theta$ is injective.

The sign of $\theta (g)$ defines a homomorphism $G\to \{\pm 1\}$. If it were nontrivial, simplicity would make it injective, impossible for a group of order sixty mapping into one of order two. Hence every image permutation is even, and $\theta (G)\subseteq A_5$. Both groups have order sixty, so they are equal. This constructs the requested group isomorphism:

Write $G=\mathrm{Rot}(D)$ for the rotational symmetry group of a regular dodecahedron. Centre the solid at the origin. A rotation in three dimensions is determined by the image of a chosen face and one of its vertices, so there are at most $12\cdot 5=60$ rotations. All these choices occur. The outward unit normal to a face and the direction from its centre to the chosen vertex determine an oriented orthonormal frame. The proper rotation mapping one such frame to another maps the whole regular pentagonal face. At each edge, the common dihedral angle and convexity then determine the adjacent face uniquely; propagating across the connected face-adjacency graph shows that the rotation maps the entire solid to itself. Hence $|G|=60$, and the action is transitive on faces, vertices and edges.

Its rotations can be counted geometrically. There are six axes joining opposite face centres, each allowing four nonidentity rotations of order five; ten axes joining opposite vertices, each allowing two of order three; and fifteen axes joining opposite edge midpoints, each allowing a half-turn. The orbit-stabilizer theorem also verifies these possibilities: vertex stabilizers have order $60/20=3$ and edge stabilizers have order $60/30=2$, and their nonidentity elements rotate about the corresponding vertex or edge-midpoint axis. Together with the identity this accounts for all $1+24+20+15=60$ elements. In particular there are no elements of order four.

We now determine the conjugacy classes. The group is transitive on edges, so all fifteen edge half-turns are conjugate. For a rotation whose angle is neither zero nor $\pi$, a commuting rotation must preserve its oriented axis: conjugation carries the axis and its rotation angle to their images, and reversing the axis would reverse a non-half-turn angle. Thus its centralizer consists of rotations about the same axis. For a face rotation this centralizer has order five, and for a vertex rotation it has order three. The orbit-stabilizer theorem for conjugation therefore gives class sizes $60/5=12$ and $60/3=20$. The twenty vertex rotations form one class, and the twenty-four face rotations form two classes. The class sizes are $1,15,20,12,12$.

A normal subgroup is a union of these classes and contains the identity. Its possible orders areBy Lagrange's theorem its order must divide $60$. Only $1$ and $60$ in this list do so. Therefore $\boxed{G\text{ is simple of order }60}$.

Projection onto the subring in part (b), along the two-sided ideal, is the Brauer homomorphism:It is a unital algebra homomorphism because it is the quotient by that ideal followed by the displayed subring identification.

Put $N=N_G(D)$. The projection is $N$-equivariant, since $N$ preserves the centralizer. Its restricted kernel isEquivalently it is spanned by the sums of the $N$-conjugation orbits on $G$ disjoint from $C_G(D)$. Normality of $D$ in $N$ ensures that an $N$-orbit either lies wholly inside that centralizer or wholly outside it.

Put $b={\mathrm{Br}}_D(e)$. By equivariance of the Brauer homomorphism, $b$ is a central idempotent of $kK$, since $C_G(D)\le K\le N_G(D)$. The Nagao module theorem asserts that, if $eM=M$, thenand every indecomposable summand of $(1-b)M$ is relatively projective for a $p$-subgroup of $K$ not containing $D$. Because $D$ is normal in $K$, its module vertices cannot contain $D$ even after $K$-conjugation.

Here is a direct proof. As an element of $(kG{)}^K$, $e-b$ is a linear combination of $K$-conjugation orbit sums supported outside $C_G(D)$. An orbit represented by $x$ has sumIts stabilizer $C_K(x)$ does not contain $D$. On $T=(1-b)M$, the element $e-b$ acts as the identity. Map the orbit-sum expression into ${\mathrm{End}}_k(M)$ by left multiplication, and compose on both sides with the $K$-linear projection onto $T$. It follows thatReplace each centralizer by a Sylow $p$-subgroup of it using trace transitivity and its invertible prime-to-$p$ index. Project this identity further onto an indecomposable summand $U$ of $T$. Its endomorphism ring is local by Krull–Schmidt theorem. A sum of nonunits cannot be the identity in a local ring, so one of the trace terms is a unit $u$ in ${\mathrm{End}}_{kK}(U)$. Multiplying the endomorphism inside that trace by $u^{-1}$ gives ${\mathrm{id}}_U$ as a single relative trace. The D. Higman criterion makes $U$ relatively projective for that Sylow subgroup $Q\le C_K(x)$, and $D\nleqq Q$. This proves the assertion rather than just invoking the theorem's name.

A composition series is a subnormal chain whose nontrivial successive quotients are simple. The Jordan–Hölder theorem asserts that any two composition series of a finite group have the same length and the same multiset of simple factors, up to isomorphism. Existence follows by successively choosing maximal proper normal subgroups until the identity is reached.

For uniqueness, induct on the group order. Let $A,B$ be the penultimate terms of two series. If $A=B$, apply induction to that subgroup and add the common top factor. If $A\ne B$, their normality and maximality give $AB=G$. Put $C=A\cap B$. The isomorphism theorems giveBoth are simple. Take any composition series of $C$. Appending $A$ to it gives a series for $A$, so induction identifies the factors of the original series inside $A$ with the factors of $C$ together with $G/B$. Similarly the factors inside $B$ are those of $C$ together with $G/A$. Adding the respective top factors produces the same multiset in both cases. This proves uniqueness, including multiplicities and length.

A normal subgroup $K$ of order two is central because its automorphism group is trivial. By Jordan–Hölder, a composition series through $K$ leaves the simple quotient $G/K\cong A_5$. The image of $Z(G)$ in this quotient is central and hence trivial, since $A_5$ is nonabelian simple. ThusThe symmetric group $S_5$ has trivial group center: conjugation of each transposition by a central permutation must preserve its two-element support, forcing the permutation to fix every point. Since its Jordan–Hölder factors are $A_5$ and $C_2$, a normal subgroup of order two would equal that trivial group center, a contradiction.

Identify $C_2=\{1,-1\}$ and put $D=1\times A_5$ in $C_2\times S_5$. Its three maximal normal subgroups areThey are the three kernels of nonzero maps from the abelianization $C_2^2$ to $C_2$. There is no quotient $A_5$: a central $C_2$ must map trivially, while $S_5$ has no quotient $A_5$. The latter follows from its normal subgroups $1,A_5,S_5$, which can themselves be obtained by intersecting with the simple subgroup $A_5$ and using its trivial centralizer. The complete list of composition series isWithin $M_1$ the two direct factors are the only maximal normal subgroups; within each of $M_2,M_3\cong S_5$ the only such subgroup is its copy of $A_5$. Hence there are exactly four series.

For $S_5\times S_5$, put $D=A_5\times A_5$, $X=A_5\times 1$, $Y=1\times A_5$ andAgain these are exactly the maximal normal subgroups: they are the three cyclic quotient kernels, and there is no simple quotient $A_5$ of the whole product. All series aretogether withTo verify exhaustiveness, $M_L$ has maximal normal subgroups $D$ and $1\times S_5$, and $M_R$ has $D$ and $S_5\times 1$. A quotient $A_5$ of a direct product can receive a nontrivial image from only one simple factor, since the images commute and $A_5$ has trivial group center. The remaining maximal-normal possibility is $M_D$. Its derived subgroup is $D$, because both alternating factors are perfect. A hypothetical quotient $A_5$ would project one alternating factor isomorphically and kill the other, but conjugation by an odd diagonal element induces an outer automorphism on that surviving factor, impossible inside $A_5$. Odd conjugation is outer because otherwise an odd permutation times an even one would centralize $A_5$; a permutation commuting with every three-cycle fixes every three-element support and is the identity. Thus $D$ is the only maximal normal subgroup of $M_D$. Finally the only maximal normal subgroups of $D$ are $X,Y$: normal subgroups of a product of two centerless simple groups are products of the factors, as follows by commutating with each factor. This proves exactly eight series and the classification of composition chains in products with S5.

For $S{L}_2(5)$, commuting with both elementary upper and lower transvections forces a central matrix to be scalar. A scalar $\lambda I$ has determinant ${\lambda }^2$, soIf $g^2=I$, characteristic five makes $g$ diagonalizable with eigenvalues in $\{1,-1\}$. Determinant one rules out one of each, so the unique nonidentity involution is $-I$. An index-two subgroup would have order $60$ and, by the Cauchy theorem for groups, contain that involution and hence the group center. Its image in $PS{L}_2(5)$ would have index two. The simplicity and identification $PS{L}_2(5)\cong A_5$ are established by the independent matrix argument in Solution 4; a simple group has no proper index-two subgroup. Therefore $S{L}_2(5)$ has no subgroup of index two.

The special linear group over a finite field $S{L}_n(q)$ consists of the determinant-one invertible matrices over ${\mathbb{F}}_q$. Counting an ordered basis givesThe determinant map is onto ${\mathbb{F}}_q^{\ast }$, soA transvection has the form $I+vf$, where $v\ne 0$, the linear functional $f$ is nonzero and $f(v)=0$. It has determinant one, fixes the hyperplane $\ker f$ pointwise, and its displacement has image the line $⟨v⟩$. In particular $E_{ij}(t)=I+t{e}_{ij}$ for $i\ne j$ is an elementary transvection when $t\ne 0$.

Row addition is multiplication by such a matrix. Gaussian elimination reduces a determinant-one matrix to diagonal form using row additions and determinant-one pivot exchanges. The latter, and all remaining diagonal factors, are themselves products of transvections: within two coordinates,A diagonal matrix with product of entries one is a product of these two-coordinate diagonal matrices. Thus transvections generate $S{L}_n(q)$. This generation statement actually holds for every $n\ge 2$ and every finite field; the exclusions in the question are needed for the subsequent simplicity assertion.

Commuting with all elementary matrices forces the group center to consist of scalar matrices $\lambda I$ with ${\lambda }^n=1$, so its order is $d=\gcd (n,q-1)$. The projective quotient acts faithfully on the one-dimensional subspaces. Its action is two-transitive: send an ordered pair of independent representative vectors to any other pair and adjust the determinant by rescaling a representative without changing its line. Therefore it is primitive.

The Iwasawa simplicity lemma says that in a faithful primitive action, if a point stabilizer has an abelian normal subgroup whose conjugates generate the group, every nontrivial normal subgroup contains the derived subgroup. In particular a nontrivial perfect group satisfying these conditions is simple. No proof of the lemma is required here.

For the point $\ell =⟨v⟩$, the subgroupis abelian, is normal in the line stabilizer, and its conjugates contain all transvections. Its image in the projective quotient has the same properties. Perfectness follows explicitly from elementary commutators. If $n\ge 3$, choose a third index $r$ to obtainIf $n=2,q>3$, choose $t$ with $t^2\ne 1$ and set $h(t)=\mathrm{diag}(t,t^{-1})$. ThenEvery upper transvection is a commutator, and conjugation gives the lower ones. The group and its projective quotient are perfect. Iwasawa's simplicity lemma now proves

The order of $PS{L}_2(4)$ is $4(4^2-1)=60$. Its faithful projective-line action has degree five, so its image is an index-two subgroup of $S_5$, necessarily $A_5$. Thus $PS{L}_2(4)\cong A_5$.

For $PS{L}_2(5)$ the order is $5(5^2-1)/2=60$. We construct a faithful degree-five action. In $S{L}_2(5)$ takeThey satisfy $i^2=j^2=-I$ and $ij=-ji$. Their quaternion subgroup projects to a Sylow 2-subgroup $V\cong C_2^2$. Its projective centralizer is $V$: a lift of a centralizing element conjugates each of $i,j$ to itself or its negative. In the matrix basis $I,i,j,ij$, those four sign choices each leave a one-dimensional space, and determinant one leaves just its two quaternion representatives. The matrixhas order three and cyclically permutes the three nonidentity elements of $V$. The normalizer quotient embeds in $S_3$ and has order dividing both $6$ and $60/4=15$, so it has order exactly three. Hence $|N(V)|=12$ and there are five Sylow 2-subgroups. Their conjugation action is nontrivial and transitive, and simplicity makes it faithful. Its image again has order $60$ in $S_5$, proving

Finally,In characteristic two an involutory matrix has form $I+N$ with $N^2=0$. Its Jordan blocks have size at most two. A projective involution in $PS{L}_3(4)$ has a unique involutory lift to $S{L}_3(4)$: its square is central, and squaring is an automorphism of the scalar center of order three, allowing a unique scalar adjustment. The nonidentity Jordan form in dimension three is only $(2,1)$. This GL-class remains a single SL-class, since the centralizer contains $\mathrm{diag}(I_2,c)$ of every determinant $c\in {\mathbb{F}}_4^{\ast }$. Thus $PS{L}_3(4)$ has one involution class.

In $PS{L}_4(2)=S{L}_4(2)=G{L}_4(2)$ there are two forms, $(2,1,1)$ and $(2,2)$, so there are two involution classes. More concretely, the first group's single class has $21\cdot 5\cdot 3=315$ elements, counting image line, kernel plane through it and nonzero map between the quotient and image. The rank-one class in dimension four over ${\mathbb{F}}_2$ has $15\cdot 7=105$ elements; the rank-two class has $35\cdot |G{L}_2(2)|=210$. Isomorphisms preserve conjugacy classes, so

A transitive permutation group is primitive when it preserves no nontrivial partition into blocks. A point stabilizer is maximal precisely when the action is primitive. The orbits of any normal subgroup form a block system, so every nontrivial normal subgroup of a faithful primitive group is transitive. This turns the classification of primitive groups into a study of their minimal normal subgroups and their group socle.

The socle of a finite group is generated by its minimal nontrivial normal subgroups. A minimal normal subgroup is characteristically simple, hence a direct power of a simple group. Distinct minimal normal subgroups commute. If a transitive normal subgroup is abelian it is regular: its point stabilizer, being the same at all points, lies in the group homomorphism kernel. In that case it is elementary abelian, and the ambient group is affine. If two minimal normal subgroups occur, each centralizes the other; the centralizer of a transitive permutation group is semiregular, so both are regular and the two regular groups are mutual left/right partners. These observations are central to the O'Nan–Scott theorem.

The theorem organizes finite primitive groups into the following socle/action types. In the affine type HA, the group socle is $C_p^d$, degree $p^d$, and $G=C_p^d⋊H$ with $H\le G{L}_d(p)$ irreducible. In almost-simple type AS the group socle is one nonabelian simple group $T$ and $T\le G\le \mathrm{Aut}(T)$; a maximal subgroup not containing $T$ specifies the primitive coset action.

In simple-diagonal type SD the group socle is $T^r$, $r\ge 2$, and its intersection with a point stabilizer is a full diagonal copy of $T$. The degree is $|T{|}^{r-1}$, and the action includes the appropriate permutation of the simple factors. Compound-diagonal type CD replaces this one diagonal by a product of diagonals in disjoint collections of factors; it is a product version of diagonal type. In product-action type PA one embeds in a wreath product with a primitive nonregular almost-simple base, acting on a Cartesian power ${\Delta }^r$, so the degree is $|\Delta {|}^r$ and the group socle is a corresponding power of $T$.

Types HS and HC have two regular minimal normal subgroups, each isomorphic respectively to $T$ or to $T^r$ with $r>1$. Their socle consists of the left and right regular groups; the degrees are $|T|$ and $|T{|}^r$. Finally TW, twisted-wreath type, has a unique regular nonabelian minimal normal subgroup $T^r$ and a point stabilizer acting through a twisted permutation of its factors. This has degree $|T{|}^r$; primitivity imposes restrictive conditions, in particular $r\ge 6$. These eight modern labels refine the older affine/almost-simple/diagonal/product description. The theorem reduces the problem to simple groups, their maximal subgroups, irreducible linear actions and the specified product constructions; it does not make every primitive group maximal in a symmetric group.

For maximal subgroups of $S_n$ there are first two elementary families. An intransitive maximal subgroup is the stabilizer $S_a\times S_{n-a}$ of an unordered distinguished subset, with $1\le a<n/2$; replacing the subset by its complement gives the same class. When $a=n/2$, this subgroup sits properly inside the stabilizer of the two equal blocks. A transitive imprimitive maximal subgroup is a full block-system stabilizer $S_a\wr S_b$, with $ab=n$, $a,b>1$, in its imprimitive action.

The alternating group $A_n$ is itself maximal. A remaining primitive maximum must contain odd permutations. The primitive part of O'Nan–Scott places it among full affine, diagonal, product-action, or almost-simple overgroups. Full affine groups have degree $p^d$; full diagonal groups have degree $|T{|}^{r-1}$ and socle $T^r$; full product-action wreath groups have degree $d^r$. Almost-simple cases require the actual simple-group permutation degree and its full permutation normalizer. Regular nonabelian types are contained in the corresponding holomorph/diagonal or product overgroups, so do not supply extra maximal families. Parity and exceptional containments must be checked before calling any of these candidates maximal.

In degree sixteen the complete list, up to $S_{16}$-conjugacy, isThe three wreath products here act on blocks, not Cartesian coordinates. The other proper primitive groups of this degree are affine and lie in $AG{L}_4(2)$. A nonzero translation of its sixteen points is eight transpositions, and a generating linear transvection is four transpositions, so the full affine group lies in $A_{16}$. Even the tempting product-action $S_4\wr S_2$ on $4^2$ points is contained in this affine group: identify each four-point coordinate with ${\mathbb{F}}_2^2$. Alternatively, each base transposition repeats four times and coordinate exchange makes six transpositions, both even. Thus none is an additional maximum of $S_{16}$. There are eleven conjugacy classes in the displayed list.

In degree sixty the complete list isThe notation $D(A_5,2)$ denotes the full diagonal normalizer, not just $A_5\times A_5$. It is generated on the set $A_5$ by $x\mapsto ax{b}^{-1}$, an outer automorphism induced by odd conjugation in $S_5$, and $x\mapsto x^{-1}$. Its order isIts two alternating factors form the group socle. The point stabilizer intersects them in the diagonal $A_5$, of index sixty. Odd conjugation by a transposition of $S_5$ fixes its six commuting elements in $A_5$, hence makes $(60-6)/2=27$ transpositions and is odd. Inversion fixes the identity and the fifteen involutions, so makes twenty-two transpositions and is even. This verifies that the full diagonal group is not lost inside $A_{60}$.

The almost-simple maximum $PG{L}_2(59)$ has order $59(59^2-1)=205320$ and its usual two-transitive projective-line action. Multiplication by a primitive element of ${\mathbb{F}}_{59}^{\ast }$ makes one 58-cycle and fixes zero and infinity, so it too contains odd permutations. The simple subgroup $PS{L}_2(59)$ alone is not maximal. Degree sixty is neither a prime power nor a proper power, excluding affine and product/compound types; its diagonal possibility is $T=A_5,r=2$. The almost-simple primitive possibilities, besides the natural alternating/symmetric groups, have socle $PS{L}_2(59)$. Their full normalizer is the displayed $PG{L}_2(59)$. These degree-specific classification facts supply the exhaustive primitive part of the list. Altogether there are forty-two conjugacy classes of maximal subgroups of $S_{60}$: twenty-nine intransitive, ten imprimitive, the alternating group, and two further primitive classes.

Take $g=(12\cdots n)$. Since $n\equiv 3(\mathrm{mod}4)$ is odd, its sign is $(-1{)}^{n-1}=1$, so $g\in A_n$. Define $s$ to fix $1$ and reverse the other cyclic labels: it interchanges $2$ with $n$, $3$ with $n-1$, and so on. Then $sg{s}^{-1}=g^{-1}$, whileThe centralizer of an $n$-cycle in $S_n$ is $⟨g⟩$: a commuting permutation is determined by the image of one point, and that image fixes the corresponding power of the cycle. All its elements are even, because $g$ is even. Any other conjugator taking $g$ to $g^{-1}$ differs from $s$ by a centralizer element and therefore remains odd. Thus $g$ and $g^{-1}$ are not conjugate in $A_n$. This is the parity obstruction described by inversion of an odd cycle in an alternating group.

For any complex character of a finite group, a unitary realization givesIf every $\theta \in \mathrm{Irr}(A_n)$ had real value at $g$, all these characters would take equal values on $g$ and $g^{-1}$. But irreducible characters separate conjugacy classes: they form a basis of class functions, so equality of every irreducible value would also give equality of the class indicator functions. The two distinct conjugacy classes cannot have that property. ConsequentlyIn other words, these alternating groups are not ambivalent groups, even though all symmetric-group irreducible characters are integer-valued.

Work in the finite-group setting. If $M$ is maximal in a nonabelian simple $G$, then $M\ne 1$: otherwise every nonidentity element would generate $G$, forcing a cyclic group of prime order. Choose a minimal normal subgroup $N\ne 1$ of $M$. Every characteristic subgroup of $N$ is normal in $M$, so minimality makes $N$ a characteristically simple group. Its normalizer contains $M$. By maximality it is $M$ or $G$, and the latter would make $N$ a nontrivial normal subgroup of the simple group $G$, impossible because $N\le M<G$. This proves the maximal subgroup normalizer criterion:

For $A_5$, stabilizer subgroups have order 12. Stabilizer subgroups of a two-element subset have order $|(S_2\times S_3)\cap A_5|=6$, and are isomorphic to $S_3$; for example $⟨(123),(12)(45)⟩$. The normalizer of a Sylow 5-subgroup has order 10: its normalizer in $S_5$ has order $5\cdot 4$, and exactly half its permutations are even, since a multiplier of order four on ${\mathbb{F}}_5$ is an odd four-cycle. This gives a dihedral group of order 10.

To prove completeness, an intransitive subgroup either fixes a point, or its orbits on five letters have sizes two and three. It therefore lies in one of the first two types. A transitive proper subgroup $H$ has order divisible by 5. Its number of Sylow 5-subgroups is one or six, since $A_5$ has six and the count is $1$ modulo 5. A count of six would make $30\mid |H|$, forcing an index-two subgroup and contradicting Simplicity of the alternating group A5. Therefore $H$ normalizes its unique Sylow 5-subgroup and lies in the order-ten normalizer.

Stabilizer subgroups are maximal because their index is prime. The order-ten group is transitive and cannot have a larger proper transitive overgroup by the preceding argument. The order-six group has no fixed point, cannot lie in a stabilizer subgroup, and cannot lie in the order-ten type because 3 does not divide 10. Thus all three types are maximal. $A_5$ is transitive on points and on unordered pairs, while Sylow conjugacy handles the normalizers. Consequently

Now let $G=G{L}_3(2)$. Choosing independent columns gives $|G|=7\cdot 6\cdot 4=168$. Its simplicity follows from the elementary projective-transvection proof in part 4(i), which does not use this subgroup classification.

A stabilizer subgroup and a plane stabilizer subgroup each have index seven and order 24. To construct the remaining type, identify ${\mathbb{F}}_2^3$ with ${\mathbb{F}}_8$ and take the Singer cycle $P$ of multiplications by nonzero field elements, of order seven. Any linear map commuting with a primitive multiplication commutes with all of ${\mathbb{F}}_8$, so its centralizer is precisely $P$. A normalizer sends a primitive element $a$ to another root of its minimum polynomial, namely $a,a^2,a^4$. The Frobenius map $x\mapsto x^2$ realizes those three choices. Hence

If $7\mid |H|$ for a proper $H<G$, its Sylow count is one or eight. Eight would make $56\mid |H|$, hence $|H|=56$. Its index-three coset action would inject the simple group $G$ into $S_3$: the kernel is normal and cannot be all of $G$. This is impossible. Thus $H$ normalizes a unique Sylow 7-subgroup and lies in the order-21 normalizer.