Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 (Topics in Group Theory)

For subgroups $A,B$ write $[A,B]$ for the subgroup generated by their group commutators. A nilpotent group has a central series: a finite chain $1=G_0\le G_1\le \cdots \le G_c=G$ with $[G_i,G]\le G_{i-1}$. Equivalently its upper central series, defined by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, reaches $G$. Indeed every central series term $G_i$ lies in $Z_i$ by induction, and the upper central series itself is central. A third equivalent definition is that the lower central series ${\gamma }_1=G$, ${\gamma }_{i+1}=[{\gamma }_i,G]$ terminates at ${\gamma }_{c+1}=1$: reversing this series gives a central series, and induction along any central series gives ${\gamma }_{c+1}=1$.

The least such $c$ is the nilpotency class. Nontrivial abelian groups have class one. Upper-unitriangular $3\times 3$ matrices have class two: their group commutators lie in the central top-right-entry subgroup, and further group commutators vanish. The same example works over a finite field or over the integers, so nilpotence is not a finiteness property. Subgroups and quotients inherit nilpotence by intersecting or mapping a central series; finite direct products of groups have class at most the largest factor class. Every nilpotent group is soluble, since its derived series is contained successively in the lower central series. The converse fails: $S_3$ is soluble but its center is trivial, so it cannot be a nontrivial nilpotent group. Extensions of nilpotent groups need not be nilpotent, as the extension $C_3◃S_3$ with quotient $C_2$ shows.

To prove the finite classification, first note that every nontrivial finite p-group has nontrivial center. In its class equation, every noncentral conjugacy class has size divisible by $p$, so the center's order is divisible by $p$. Induct on the group order: the smaller p-group $P/Z(P)$ is nilpotent, and lifting its central series and adjoining $Z(P)$ proves $P$ nilpotent. A finite direct product of groups of prime-power-order groups is consequently nilpotent.

Conversely suppose $G$ is finite and nilpotent. We need the normalizer condition for nilpotent groups. For any proper subgroup $H<G$, choose the first upper-central term $Z_i$ not contained in $H$. Then $Z_{i-1}\le H$. Every $z\in Z_i$ commutes with $H$ modulo $Z_{i-1}$, so it normalizes $H$. Choosing $z\notin H$ proves $H<N_G(H)$.

Let $P$ be a Sylow subgroup and put $M=N_G(P)$. Since $P$ is normal in $M$, it is the unique Sylow p-subgroup of $M$. Any element normalizing $M$ must therefore normalize $P$, so $N_G(M)=M$. If $M$ were proper, the normalizer condition would contradict this equality. Hence $N_G(P)=G$: every Sylow subgroup is normal. For Sylow subgroups $P,Q$ belonging to different primes, $[P,Q]\le P\cap Q=1$, so they commute. Their product has order the product of their orders, namely $|G|$, and has trivial intersections between each factor and the product of the others. Multiplication therefore givesThis proves finite nilpotent group decomposition in both directions without assuming the classification in the normalizer argument. It also shows that a finite group is nilpotent exactly when all its Sylow subgroups are normal. The finite hypothesis is essential to the prime-power direct-product statement; an infinite cyclic group is nilpotent but is not such a finite product.

A soluble group is one whose derived series terminates: $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, and $G^{(r)}=1$ for some finite $r$. Each factor of this series is abelian. Conversely, if a subnormal series has abelian factors, repeated commutation moves down the series, so the derived series terminates. This explains the usual alternative definition in terms of abelian factors. The least $r$ is the derived length.

Subgroups are soluble because $H^{(i)}\le G^{(i)}$; quotients are soluble because a group homomorphism maps a derived subgroup onto the derived subgroup of its image. Solubility is also closed under extensions: if $N◃G$ has derived length at most $a$ and $G/N$ has length at most $b$, then $G^{(b)}\le N$ and $G^{(a+b)}=1$. Abelian and nilpotent groups are soluble; $S_3$ has derived subgroup $A_3$ and derived length two. In a finite group, solubility is equivalent to every composition factor being cyclic of prime order: simple soluble groups are abelian and therefore cyclic of prime order, while a composition series with these factors builds a soluble group by extensions. This need not give an ambient-normal series with cyclic prime-order factors, which is the stronger property of supersolubility.

A Hall subgroup for a prime set $\pi$ is a subgroup $H$ whose order uses only primes in $\pi$ and whose index uses only primes outside $\pi$. Its order is the full $\pi$-part of $|G|$. Sylow's theorem is the singleton-prime case; solubility allows all prime sets, although Hall subgroups need not be normal, as the order-two subgroups of $S_3$ show.

Prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$. It is elementary abelian of some prime characteristic $p$: $N^{\prime }$ is characteristic in $N$ and normal in $G$, so minimality and solubility force $N^{\prime }=1$. A nontrivial Sylow subgroup of this finite abelian group is characteristic, making $N$ a p-group. The subgroup of pth powers is characteristic and proper, so minimality makes it trivial. This proves minimal normal subgroups of finite solvable groups are elementary abelian.

By induction $G/N$ has a Hall $\pi$-subgroup $\overline{H}$. Let $K$ be its inverse image. Then $[G:K]$ is a ${\pi }^{\prime }$-number. If $p\in \pi$, the whole $K$ is a $\pi$-group and is the desired Hall subgroup. If $p\notin \pi$, $K/N=Q=\overline{H}$ has order $s$ prime to $p$. We now prove the required complement existence explicitly, rather than invoking a splitting theorem as a substitute.

Write $N$ additively as an ${\mathbb{F}}_p$-vector space. Choose representatives $u(x)$ for $x\in Q$ with $u(1)=1$. Conjugation gives a well-defined action of $Q$ on $N$, because changing a representative by an element of the abelian group $N$ does not change that action. Define $f(x,y)\in N$ by $u(x)u(y)=f(x,y)u(xy)$. Associativity yieldsSince $s$ is invertible in ${\mathbb{F}}_p$, set $c(x)=s^{-1}{\sum }_{z\in Q}f(x,z)$. Sum the identity over $z$ and use the bijection $z\mapsto yz$ to obtainChanging representatives to $u^{\prime }(x)=(-c(x))u(x)$ removes the multiplication error:Thus $x\mapsto u^{\prime }(x)$ is a homomorphic section. Its image $H$ has order $|Q|$, intersects $N$ trivially, and complements $N$ in $K$. This is coprime splitting over an elementary abelian normal subgroup. Finally $|H|$ is a $\pi$-number and $[G:H]=[G:K]|N|$ is a ${\pi }^{\prime }$-number, soOnly existence was needed here; the fuller Hall theorem also gives conjugacy and containment of arbitrary $\pi$-subgroups. Solubility makes the elementary-abelian minimal-normal induction possible and cannot simply be omitted.

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.

Choose $\alpha \in \Omega$. Since $G$ acts regularly, $g\mapsto g\alpha$ identifies $\Omega$ with the underlying set of $G$, and the given action becomes left multiplication $L_g(x)=gx$.

A permutation $c$ centralizing all $L_g$ satisfiesConversely every right translation commutes with every left translation. To make the multiplication convention explicit, define $R_g(x)=x{g}^{-1}$. Then $R_g{R}_h(x)=x{h}^{-1}{g}^{-1}=x(gh{)}^{-1}=R_{gh}(x)$, using rightmost-first composition. The map $g\mapsto R_g$ is injective and its image is the entire centralizer, since any possible $c(1)$ can be written $g^{-1}$. ThusThis is centralizer of a regular permutation subgroup. If one instead writes right multiplication as $x\mapsto xg$, the resulting map is an antihomomorphism; including the inverse avoids suppressing that reversal. Choosing a different base point changes the identification but not the isomorphism type.

Retain the identification with the left regular action. A group group automorphism $\phi$ of $G$ satisfies $\phi L_g{\phi }^{-1}=L_{\phi (g)}$, so it normalizes $L(G)$ and fixes the identity point.

Conversely let a normalizing permutation $n$ fix the identity. Its conjugation on $L(G)$ induces a group group automorphism $\phi$ characterized by $n{L}_g{n}^{-1}=L_{\phi (g)}$. Evaluating at the identity gives $n(g)=\phi (g)$, so the whole identity stabilizer in the normalizer is exactly $\mathrm{Aut}(G)$. For arbitrary $n$, put $t=n(1)$. Then $L_t^{-1}n$ fixes the identity and is an group automorphism; thus every element has a unique expression $L_t\phi$.

The translation subgroup is normal in its normalizer, its intersection with the group automorphism subgroup is trivial, and the multiplication rule isConsequentlyThe action in this semidirect product is the natural evaluation of group automorphisms on $G$. This normalizer is the holomorph, realized as all permutations $x\mapsto t\phi (x)$. The decomposition is established from the regular group action itself rather than only from an order comparison.

Choose $\alpha \in \Omega$ and identify $\Omega$ with $N$ by $n\mapsto n\alpha$. Normality means that the point stabilizer $H_{\alpha }$ acts on $N$ by conjugation, andDouble transitivity therefore says that $H_{\alpha }$ is transitive on $N\setminus \{1\}$. All nonidentity elements of $N$ have the same order. By Cauchy's theorem for finite groups, some element has order a prime $p$ dividing $|N|$, so that common order is $p$. Cauchy's theorem excludes every other prime from $|N|$, hence $N$ is a finite p-group.

By the center argument from Q1, $Z(N)\ne 1$. The center is characteristic in $N$ and therefore invariant under the conjugation action of $H_{\alpha }$. Its nonidentity elements form a nonempty invariant subset of the transitive set $N\setminus \{1\}$, so $Z(N)=N$. Thus $N$ is abelian of exponent $p$:This is elementary abelian regular kernels in doubly transitive groups; knowing only that all element orders agree would not by itself prove commutativity.

Every $h\in H$ is uniquely $n{h}_0$ with $n\in N$ and $h_0\in H_{\alpha }$, because regularity supplies the unique $n$ sending $\alpha$ to $h\alpha$. Hence $H=N⋊H_{\alpha }$. The conjugation action of $H_{\alpha }$ on $N$ is faithful: an element centralizing $N$ and fixing $\alpha$ fixes every $n\alpha$, and is the identity permutation. Group automorphisms of an elementary abelian p-group are exactly invertible ${\mathbb{F}}_p$-linear maps, so $H_{\alpha }\le GL(d,p)$. Choosing a vector-space basis therefore givesUnder this embedding, $N$ is the translation subgroup and the point stabilizer is a linear subgroup transitive on nonzero vectors. The action is assumed to have at least two points, as usual for a doubly transitive group.

The orbits of a normal subgroup form a block system: $g(N\omega )=N(g\omega )$, so every group element permutes these orbits. By primitivity the orbits are either all singletons or one whole orbit. In the singleton case $N$ fixes every point, giving $N\le G_{[\Omega ]}$.

In the other case $N$ is transitive. For each $g\in G$ choose $n\in N$ with $n\alpha =g\alpha$. Then $n^{-1}g\in G_{\alpha }$, so $G=N{G}_{\alpha }$. Work in the quotient $G/N$. If $g=nh$ with $h\in G_{\alpha }$, the image of $A^g=g^{-1}Ag$ is the image of $A^h$, which equals the image $\overline{A}$ of $A$ because $A◃G_{\alpha }$. All conjugates of $A$ thus have the same image in $G/N$.

By hypothesis those conjugates generate $G$, so the quotient is generated by the single subgroup $\overline{A}$, and equals it. Since $A$ is abelian, $G/N$ is abelian. The kernel of a group homomorphism of a group homomorphism to an abelian group contains the commutator subgroup, yieldingThis is the normal-subgroup argument behind the Iwasawa simplicity lemma. Faithfulness is not needed for this version: retaining the pointwise kernel of a group homomorphism as the second alternative is essential. In a faithful primitive group action it says every nontrivial normal subgroup contains $G^{\prime }$.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Let $V=F^n$ and let $G=PSL(n,F)$ act on its one-dimensional subspaces. First identify the kernel of a group homomorphism in $SL(n,F)$. A matrix fixing every coordinate line is diagonal, and fixing every line $⟨e_i+e_j⟩$ forces all diagonal entries equal. Thus the projective kernel of a group homomorphism is $\{\lambda I:{\lambda }^n=1\}$. These scalars are central. Conversely a central matrix commutes with all $E_{ij}(1)$: comparing entries in $M{e}_{ij}=e_{ij}M$ shows all off-diagonal entries of $M$ vanish and all diagonal entries agree. Therefore the kernel of a group homomorphism is exactly $Z(SL(n,F))$. This proves scalar kernel of the projective linear action and faithfulness of the induced $G$-action.

The action is doubly transitive. Given two ordered pairs of distinct lines, choose two independent representatives of each pair and extend them to bases. A linear map taking the first basis to the second sends the desired lines correctly. If its determinant is $d\ne 1$, postcompose with a diagonal map in the target basis that scales its first basis vector by $d^{-1}$ and fixes all the others. It preserves the two target lines and corrects the determinant. Thus an element of $SL(n,F)$ sends either ordered pair to the other. Double transitivity implies primitivity: a block containing two distinct points must contain every point by the transitive point-stabilizer action.

For $\alpha =⟨e_1⟩$, considerProducts add the functionals, so $U$ is abelian. If $g$ fixes $\alpha$, write $g{e}_1=a{e}_1$. Then $g(I+e_1\phi )g^{-1}=I+e_1(a\phi g^{-1})$, and the new functional still vanishes on $e_1$. Thus the image $A$ of $U$ in $G$ is normal in the point stabilizer. Its conjugates contain all elementary transvections: changing the center line $⟨e_1⟩$ to $⟨e_i⟩$ gives every map $I+e_i\psi$ with $\psi (e_i)=0$, including $E_{ij}(t)$. Part (b) then shows that those conjugates generate $G$.

It remains to prove perfectness, not assume it. Use $[x,y]=x^{-1}{y}^{-1}xy$. For $n\ge 3$, distinct $i,j,k$ giveEvery generator is therefore a group commutator, so $SL(n,F{)}^{\prime }=SL(n,F)$. For $n=2$ and $|F|>3$, choose $a\in F^{\times }$ with $a^2\ne 1$, possible because a quadratic polynomial has at most two roots. With $D=\mathrm{diag}(a,a^{-1})$,Both coefficients are nonzero, so every upper and lower elementary transvection is again a group commutator. Part (b) proves perfectness in this case as well. Quotients of perfect groups are perfect, hence $G^{\prime }=G$ in exactly the stated range.

Apply part (a) to the faithful primitive $G$-action and its abelian normal point-stabilizer subgroup $A$. Every nontrivial normal subgroup contains $G^{\prime }=G$, so it is the whole group. A transvection is not scalar, showing $G\ne 1$. We have provedThe proof works for infinite fields too and uses no finite-order count. The excluded two-dimensional fields give the familiar soluble exceptions $PSL(2,2)\cong S_3$ and $PSL(2,3)\cong A_4$.

Let $q$ be a prime power and let $V={\mathbb{F}}_q^{2m}$ carry a nondegenerate alternating bilinear form $B$. The symplectic group over a finite field isIn a symplectic basis its matrix description is $g^{\mathsf{T}}Jg=J$, where $J=\left(\begin{array}{cc}0& I_m\\ -I_m& 0\end{array}\right)$. The definition is valid in characteristic two: alternating means $B(v,v)=0$, and then the form is also symmetric.

Count ordered symplectic bases $(e_1,f_1,\dots ,e_m,f_m)$ with $B(e_i,f_j)={\delta }_{ij}$ and all $e$-$e$ and $f$-$f$ pairings zero. There are $q^{2m}-1$ choices for $e_1\ne 0$. Nondegeneracy makes $v\mapsto B(e_1,v)$ a nonzero linear functional, so there are $q^{2m-1}$ choices for $f_1$ satisfying $B(e_1,f_1)=1$.

The span $P=⟨e_1,f_1⟩$ is nondegenerate. Thus $V=P\oplus P^{\perp }$, and its bilinear orthogonal complement is nondegenerate of dimension $2m-2$. For completeness, a vector in the radical of a bilinear form of $P^{\perp }$ is orthogonal both to $P$ and to $P^{\perp }$, hence to all $V$ and therefore zero. Recursing constructs and counts every symplectic basis. If $b_m$ is their number, $b_m=(q^{2m}-1)q^{2m-1}{b}_{m-1}$ with $b_0=1$.

The symplectic group over a finite field acts freely and transitively on symplectic bases: there is a unique linear map sending one ordered basis to another, and its equality of basis pairings ensures it preserves $B$. Its order is consequently $b_m$, givingThe exponent is $1+3+\cdots +(2m-1)=m^2$. The counting is valid for every prime power $q$, including $q=2$.

Identify a subset $A\subseteq \Omega$ with its indicator vector $a=(a_1,\dots ,a_n)\in {\mathbb{F}}_2^n$. Symmetric difference becomes coordinatewise addition modulo two, so the subsets form a vector space of dimension $n$, with the singleton subsets as a basis. The proposed form becomes the dot productDistributivity proves bilinearity, and commutativity proves symmetry. Its Gram matrix in the singleton basis is $I_n$, so it is nondegenerate. A permutation merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.

Write $u=(1,\dots ,1)$ for the full subset. Its bilinear orthogonal complement $E=u^{\perp }$ consists exactly of even-cardinality subsets. If $n\ge 2$ is even, $B(u,u)=n=0$ in ${\mathbb{F}}_2$, so $⟨u⟩\le E$. For any $v\in E$, $B(v,v)={\sum }_i{v}_i^2={\sum }_i{v}_i=0$, so the restricted form is alternating.

Nondegeneracy of the full form gives $E^{\perp }=⟨u⟩$; one can also see this directly by testing against all vectors $e_i+e_j$, which forces all coordinates of an element of $E^{\perp }$ to be equal. Hence the radical of a bilinear form of the restricted form isDefine the quotient form by $\overline{B}(v+⟨u⟩,w+⟨u⟩)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\mathrm{rad}(B{|}_E)=⟨u⟩$, hence is zero. ThereforeBoth spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the symplectic quotient of the binary subset module. The evenness assumption is essential because otherwise $u$ does not lie in its own bilinear orthogonal complement.

Set $n=6$ in part (b). The quotient $W=u^{\perp }/⟨u⟩$ has dimension four and a nondegenerate alternating form. The coordinate action therefore definesProve it is faithful before comparing orders. If a permutation lies in the kernel of a group homomorphism, it fixes the quotient class of each two-element subset $A$. That class has exactly the two representatives $A$ and $A+\Omega =\Omega \setminus A$. Their cardinalities are two and four. A coordinate permutation preserves cardinality, so it must fix $A$ itself, not exchange it with its complement. Thus it fixes every two-element subset. Intersecting the fixed sets $\{i,j\}$ and $\{i,k\}$ shows it fixes each singleton $\{i\}$, so it is the identity. Hence $\rho$ is injective.

By part (a),An injection between finite groups of equal order is surjective. Choosing a symplectic basis of $W$ identifies the form with the standard one, soThis realizes the exceptional isomorphism through a concrete binary permutation module. Equality of orders alone would not establish it; the faithful form-preserving action supplies the required group homomorphism.