Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 2 (Topics in Group Theory)

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 Hall theorem for soluble groups says that for every prime set $\pi$, a finite soluble group has Hall $\pi$-subgroups; they are all conjugate, and every $\pi$-subgroup is contained in one. A Hall $\pi$-subgroup has only primes in $\pi$ in its order and no primes in $\pi$ in its index.

For a proof outline, induct on the group order and choose a minimal normal subgroup $N$. In a finite soluble group this is elementary abelian of some characteristic $p$: its derived subgroup is characteristic and solvability forces it to be trivial; primary components and the subgroup of elements of order dividing $p$ are then characteristic, so minimality gives an elementary abelian p-group. Choose a Hall $\pi$-subgroup of $G/N$ and let $U$ be its preimage. If $p\in \pi$, $U$ is itself a Hall $\pi$-subgroup. If $p\notin \pi$, $U/N$ has order prime to $p$ and coprime splitting over an elementary abelian normal subgroup supplies a complement to $N$ in $U$, of the required Hall order.

Here the complement fact has a short averaging proof. Choose a section of $U/N=H$ and write its multiplication defect as $f(h,k)\in N$, in additive notation. Associativity givesSince $|H|$ is invertible on $N$, averaging over $t$ expresses the defect as $f(h,k)=b(h)+hb(k)-b(hk)$; adjusting the section by $-b$ gives a homomorphic section. Two such complements differ by $d(hk)=d(h)+hd(k)$. Averaging this identity gives $d(h)=b-hb$, so they are conjugate by $b\in N$. Inductive conjugacy in the quotient followed by this complement conjugacy proves Hall conjugacy. To embed an arbitrary $\pi$-subgroup $Q$, first put its quotient inside the chosen quotient Hall subgroup. In the coprime case $Q$ and the corresponding subgroup of a complement are both complements to $N$ inside $NQ$, hence are $N$-conjugate. In the other case $Q\le U$ directly. This proves Hall conjugacy and embedding in finite soluble groups as well as existence.

Now prove the assertion about maximal subgroups by induction on $|G|$. Suppose a nontrivial normal subgroup $K$ is contained in $A$. If $K\nleqq B$, maximality gives $KB=G$, and since $K\le A$ this gives $AB=G$. If $K\le B$, both $A/K$ and $B/K$ are maximal in $G/K$; induction yields either their product equal to $G/K$ or their conjugacy, lifting respectively to $AB=G$ or to conjugacy of $A,B$. The same applies with their roles reversed.

It remains to suppose that no nontrivial normal subgroup of $G$ lies in either maximum. Let $K$ be minimal normal, an elementary abelian p-group. Since $K$ lies in neither maximum, $KA=KB=G$. The intersection $K\cap A$ is normalized by $A$ and by the abelian group $K$, hence by $G$; it is therefore trivial. Likewise $K\cap B=1$. Thus both $A$ and $B$ complement $K$. Their action on $K$ is irreducible, since an invariant subspace would be a proper nontrivial normal subgroup of $G$. It is also faithful: $C_A(K)$ is normalized by $A$ and centralized by $K$, so is a normal subgroup of $G$ contained in $A$ and must be trivial.

If $G/K=1$, these assumptions force $A=B=1$ and $G$ to have prime order. Otherwise choose $L>K$ normal in $G$ with $L/K$ minimal normal in $G/K$. It is an elementary abelian r-group. Under $A\cong G/K$, its copy $R=A\cap L$ is a nontrivial normal subgroup of $A$. The prime $r$ differs from $p$: a p-group acting on the characteristic-p vector space $K$ has nonzero fixed points, by counting its orbits on $K$ modulo $p$. Its fixed subspace would be $A$-invariant, hence all of $K$ by irreducibility, contrary to faithfulness of $R$.

Therefore $A\cap L$ and $B\cap L$ are Sylow r-subgroups of $L=K(A\cap L)=K(B\cap L)$. They are conjugate in $L$. Conjugate $B$ so that these intersections coincide as $R$. Since $R⊲A$, $N_G(R)\ge A$. It cannot equal $G$, since $R$ would then be a nontrivial normal subgroup contained in $A$. Maximality gives $N_G(R)=A$, and the same argument gives $N_G(R)=B$. The conjugated maxima are equal. HenceThis is the factorization-or-conjugacy property of maximal subgroups of a finite soluble group.

For $|G|=p^{a}m$ with $p\nmid m$, Sylow theorems assert existence of a subgroup of order $p^a$, containment of every p-subgroup in one, conjugacy of all these Sylow subgroups, andA group of order $20$ has $n_5\mid 4$ and $n_5\equiv 1(\mathrm{mod}5)$, so its Sylow 5-subgroup $P$ is unique. In $S_6$ it is generated by a five-cycle and has one fixed point. Its normalizer preserves that point and, on the other five points identified with ${\mathbb{F}}_5$, is the affine group $x\mapsto ax+b$. Every automorphism of the cyclic translation group is realized by a multiplier. Thus $|N_{S_6}(P)|=5\cdot 4=20$, forcing

Inside $S_5$ there are $120/20=6$ Sylow 5-subgroups. Conjugation gives a transitive action on them and a homomorphism into $S_6$. Its group homomorphism kernel is a normal subgroup of $S_5$, whose only possibilities are $1,A_5,S_5$. The latter two would leave an image of order at most two, incompatible with transitivity on six points. Hence it is faithful. Denote the resulting transitive copy of $S_5$ by $J\le S_6$.

The action of $S_6$ on the six cosets of $J$ is also faithful: its group homomorphism kernel cannot contain the simple subgroup $A_6$, since an image of order at most two is not transitive of degree six, and a normal subgroup disjoint from $A_6$ centralizes $A_6$ and is trivial. It therefore gives an automorphism $\tau :S_6\to S_6$ carrying $J$ to a natural point stabilizer. This automorphism is outer, because conjugation cannot carry a transitive subgroup to an intransitive one.

There are precisely two classes of copies of $S_5$ in $S_6$. A faithful intransitive copy must have an orbit of length five: all partitions of six into smaller orbits embed it in groups of order below $120$. Thus it is a natural point stabilizer. A transitive copy is a degree-six coset action of an abstract $S_5$, with an order-$20$ stabilizer. Every order-$20$ subgroup of $S_5$ is a Sylow-five normalizer, so these stabilizers are conjugate and the degree-six actions are permutation-equivalent. Hence all transitive copies form one class. The outer automorphism constructed above swaps these two classes; it cannot preserve the point-stabilizer class, since any automorphism preserving that class induces a relabeling of its six members, and equivariance of their faithful conjugation action then makes the automorphism inner.

Identify $S_6$ with its inner automorphism group inside $\overline{G}$. Every Sylow 5-subgroup of $\overline{G}$ is contained in $S_6$, since that normal subgroup has index two. There are $720/20=36$ of them. ConsequentlyTo prove maximality, first classify the proper overgroups of $N$ in $S_6$. The affine multiplier in $N$ of order four is a four-cycle and is odd. Within $H\cong S_5$, a proper overgroup larger than $N$ could only have order $60$, but the index-two subgroup is $A_5$ and cannot contain that odd element. Thus $N$ is maximal in $H$. A nontransitive overgroup of $N$ must fix its unique fixed point, so is $H$ if it is larger than $N$. A transitive overgroup $M$ has $M\cap H\ge N$; this point stabilizer is either $N$ or $H$. In the latter case $M=S_6$, and in the former $|M|=6\cdot 20=120$, making it a transitive copy of $S_5$.

There is exactly one such transitive copy containing $N$. There are six conjugates of $J$: applying the preceding overgroup argument to its own Sylow-five normalizer shows that $J$ is maximal, and $J$ is not normal. Each contains six Sylow 5-subgroups. Counting pairs of a conjugate copy and a Sylow subgroup gives $6\cdot 6=36$, with the same number through every one of the 36 Sylow subgroups of $S_6$. Thus each Sylow subgroup, and hence its normalizer $N$, is contained in exactly one transitive copy. In particular the only proper strict overgroups of $N$ are its natural $H$ and this $J$.

The subgroup $\overline{N}$ has an outer element, because it has twice the order of $N$. Such an element normalizes $P$ and swaps the unique overgroups $H,J$. If $\overline{N}\le U\le \overline{G}$, then $U\cap S_6$ is normalized by $\overline{N}$, so it cannot be just $H$ or just $J$. It is therefore $N$ or $S_6$. In the latter case $U=\overline{G}$. In the former $U$ normalizes $N$ and hence its unique Sylow 5-subgroup $P$, giving $U\le \overline{N}$. Therefore

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

Let $V$ have dimension $2m$ over ${\mathbb{F}}_q$, with a nondegenerate alternating bilinear form $B$. The symplectic group isIt acts freely and transitively on ordered symplectic bases. Choose the first vector $e_1\ne 0$ in $q^{2m}-1$ ways, then choose $f_1$ with $B(e_1,f_1)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate, so choose the remaining symplectic basis in its orthogonal complement. This gives the recurrenceand therefore

For the permutation module ${\mathbb{F}}_2^r$ with $r$ even, put $\mathbf{t}=(1,\dots ,1)$ andIt is invariant under coordinate permutations and has codimension one. The dot product restricted to $W$ is alternating because $B(x,x)={\sum }_i{x}_i^2={\sum }_i{x}_i=0$. Since the ambient dot product is nondegenerate, $W^{\perp }=⟨\mathbf{t}⟩$. Evenness of $r$ ensures $\mathbf{t}\in W$, soHere the word symplectic must be understood as an alternating form before quotienting: a form with a nonzero radical is degenerate. The nondegenerate symplectic form is induced on $\overline{W}=W/⟨\mathbf{t}⟩$, of dimension $r-2$.

Now put $r=2m+2$. Coordinate permutations preserve the induced form and give a homomorphism $S_r\to S{p}_{r-2}(2)$. To see it is faithful for $r\ge 6$, suppose a permutation $\sigma$ acts trivially on $\overline{W}$. For every pair of distinct indices,The two weight-two vectors could differ by $\mathbf{t}$ only if the complement of a two-element set also had size two. For $r\ge 6$ that is impossible. Thus $\sigma$ preserves every two-element set, and hence fixes every index: intersect two such sets with one common element. ThereforeFor $m=2$, both groups have order $720$, and the injection is onto. This provesThe lower bound matters: for $r=4$ the action on the quotient has the Klein four-group as group homomorphism kernel. The construction is the deleted permutation module in characteristic two.

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.