Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 (Topics in Group Theory)

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the commutator subgroup. The group is soluble ifEquivalently it has a finite series with abelian factors. The trivial group is included.

If $H\le G$, induction gives $H^{(i)}\le G^{(i)}$, because commutators of elements of a subgroup are also commutators in the larger group. Thus termination of the derived series of $G$ forces termination for $H$.

For a normal subgroup $N$, the quotient map sends $[x,y]$ to the commutator of their images. Surjectivity then givesConsequently subgroups and quotient groups of a soluble group are soluble. The assertion about a quotient uses a normal subgroup; it is not a quotient by an arbitrary subgroup.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.

A Hall subgroup for a prime set $\pi$ is a subgroup $H\le G$ whose order has only prime divisors in $\pi$ and whose index has no prime divisor in $\pi$:Here one is allowed in either class, and ${\pi }^{\prime }$ is the complementary set of primes. Equivalently $|H|$ contains the complete prime-power contribution to $|G|$ for each prime in $\pi$.

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.

Use the Sylow theorems. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.

If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\ge p$ and $n_r\ge q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at leastplaces, since the excess is $(p-1)(q-1)>0$. Hence some Sylow subgroup of order $s\in \{q,r\}$ is normal.

In the quotient by this normal subgroup, the largest prime $p$ has a normal Sylow subgroup: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a normal subgroup of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that normal subgroup and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.

Let $P$ be this normal Sylow subgroup. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a normal Hall $\{p,q\}$-subgroup. The serieshas factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.

Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.

Here is the double-coset criterion for a one-point extension. Let $H=G_{\alpha }$, and suppose $x$ swaps $\omega$ and $\alpha$. Then $⟨G,x⟩$ is a one-point extension if and only ifFor necessity, in an extension $H$ fixes both $\omega$ and $\alpha$, so $x$ normalizes it and $x^2$ lies in it. Since $G$ is transitive on $\Omega$, the extension has exactly two double cosets relative to $G$: $G$ and $GxG$. For $g\notin H$, $xgx$ moves $\omega$ into $\Omega$ and is in the latter double coset.

For sufficiency, the displayed conditions make $G\cup GxG$ closed under multiplication. Products with middle element in $H$ reduce using $xhx=(xh{x}^{-1})x^2\in G$; those with middle element outside $H$ remain in $GxG$. A finite nonempty multiplication-closed set of permutations containing the identity is a group. It contains $G$ and $x$, hence equals $⟨G,x⟩$. Every element in $GxG$ moves $\omega$, while $G$ fixes it, giving the required stabilizer subgroup. The group is transitive because $G$ is transitive on $\Omega$ and $x$ moves the additional point.

For $1\le t\le |\Omega |$, a group is sharply t-transitive when any two ordered $t$-tuples of distinct points are related by exactly one group element. Equivalently its action on the set of such tuples is regular. In the finite caseand the stabilizer subgroup of an ordered $t$-tuple is trivial. The condition includes both existence and uniqueness, not just transitivity.

Put $n=|\Omega |\ge 2$, and let $H=G_{\alpha }$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ stabilizer subgroups shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a normal subgroup, rather than presuming that fixed-point-free elements are closed under multiplication.

For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex characters of a finite group. Let $\pi$ be the permutation character and $\psi =\pi -1$, the character of the permutation representation with its constant line removed. For every nontrivial irreducible character $\theta$ of $H$, form the virtual characterIts values are $\theta (1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus \{1\}$; induction gives ${\phi }_{\theta }(g)=\theta (h)$, because there is exactly one fixed coset.

The identity and the $n-1$ fixed-point-free elements together contribute $n\theta (1)\overline{\eta (1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ stabilizer subgroups. Hence character orthogonality givesA virtual character of norm one is plus or minus an irreducible character: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta (1)$ selects the plus sign. Thus each ${\phi }_{\theta }$ is an actual irreducible character.

For a group representation, $\chi (g)=\chi (1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the ${\phi }_{\theta }$. Conversely, a nonidentity element fixing a point gives $h\ne 1$ in $H$. Some nontrivial irreducible character of $H$ has $\theta (h)\ne \theta (1)$, since otherwise the regular representation of $H$ would not vanish at $h$. ConsequentlyThis proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a regular permutation subgroup.

Now $H$ acts transitively by conjugation on $N\setminus \{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is elementary abelian of order $p^d=n$.

Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every group automorphism. We have provedThe character argument supplies the regular kernel of a finite sharply two-transitive group; the final Sylow argument establishes the stronger characteristic assertion.

Iwasawa's simplicity lemma states the following. Suppose $G$ acts faithfully and primitively on a set, and a stabilizer subgroup $H$ has an abelian normal subgroup $A$ whose $G$-conjugates generate $G$. Then every nontrivial normal subgroup $N$ contains $G^{\prime }$. In particular, if $G$ is nontrivial and perfect, then $G$ is simple.

Indeed a nontrivial normal subgroup in a faithful primitive action is transitive, so $G=NH$. Since $H$ normalizes $A$, all conjugates of $A$ have the same image in $G/N$. Those images generate the quotient, which is therefore abelian. This gives $G^{\prime }\le N$ and proves the stated conclusion.

Use the field $F={\mathbb{F}}_2[\zeta ]/({\zeta }^3+\zeta +1)$. The defining polynomial has no root in ${\mathbb{F}}_2$, hence is irreducible. Label its elements bySince $F^{\ast }$ has prime order seven, $\zeta$ has order seven. Multiplication by $\zeta$ is exactly $b$. The identities ${\zeta }^3=\zeta +1$, ${\zeta }^6={\zeta }^2+1$ and ${\zeta }^5={\zeta }^4+1$ show that translation by one is exactly $a$.

Conjugating $a$ by powers of $b$ gives the translations $T_{{\zeta }^i}:z\mapsto z+{\zeta }^i$. The translations by $1,\zeta ,{\zeta }^2$ generate all eight translations. ConsequentlyFor distinct $u,v$ and distinct target points $u^{\prime },v^{\prime }$, the unique affine map has $s=(v^{\prime }-u^{\prime })/(v-u)$ and $t=u^{\prime }-su$. Thus the action is sharply two-transitive.

Its regular characteristic subgroup isTranslations act regularly, $K$ is normal, and its order eight makes it the unique Sylow two-subgroup, hence characteristic. This identifies the abstract subgroup and explicit generators in the original permutation notation.

Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, isThe label $2=1\in F$ is fixed. Set $H=G_1=\{M_s:z\mapsto sz\}$, where the subscript one denotes the original point label, namely field zero. Then $x^2=1\in H$ and $x{M}_{s}x=M_{s^{-1}}$, so $x$ normalizes $H$.

For $g(z)=sz+t$ outside $H$, we have $t\ne 0$. On the projective line,The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. ThusAll conditions of the double-coset criterion for a one-point extension hold. Therefore $G^{+}=⟨G,x⟩$ is a one-point extension, with $|G^{+}|=9\cdot 56=504$. Its stabilizer subgroup is sharply two-transitive, so its action on nine points is sharply three-transitive.

Use the projective-line realization from part (ii). The subgroup $K$ of translations is abelian and normal in the stabilizer subgroup $G$ of infinity. We verify both remaining conditions of Iwasawa's simplicity lemma, instead of concluding simplicity from transitivity alone.

Let $R$ be generated by all conjugates of $K$ in $G^{+}$. It contains the matricesHere matrices act by fractional linear transformations. For $t\ne 0$,In particular $w(1)=x\in R$, and $w(t)w(1)=\mathrm{diag}(t,t^{-1})$ acts as multiplication by $t^2$. Squaring is a bijection of $F^{\ast }$, so every $M_s$ belongs to $R$. Hence $R$ contains $G$ and $x$, and $\boxed{R=G^{+}}$.

Choose $s\ne 1$. The commutator, with convention $[u,v]=uv{u}^{-1}{v}^{-1}$, isAs $t$ varies, this gives all translations. Thus $(G^{+}{)}^{\prime }$ contains $K$, and normality makes it contain every conjugate of $K$. Since those generate $G^{+}$, the group is a perfect group.

A sharply three-transitive action on nine points is a primitive group action, and this permutation action is a faithful group action. All Iwasawa hypotheses now hold, soIt may be identified with ${\mathrm{PSL}}_2(8)={\mathrm{PGL}}_2(8)$: the generators are fractional linear transformations and $|{\mathrm{PGL}}_2(8)|=8(8^2-1)=504$, while all nonzero field elements are squares. The simplicity proof above does not rely on assuming simplicity of that named family.

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

For the block computation only, place the $f$-vectors in the order $f_1,\dots ,f_m$ after the $e$-vectors. This temporary reversal of the second block changes no transformations. The form matrix becomes $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$.

Let $T$ be upper unitriangular of size $m$, and let $S$ be symmetric. The matricessatisfy $gJ{g}^T=J$ by direct block multiplication. The generator $x_{ij}(\lambda )$ is $d(I+\lambda E_{ij})$, because the inverse transpose adds $-\lambda f_i$ to $f_j$. The generators $y_{ij}(\lambda )$ and $z_i(\lambda )$ are respectively $n(\lambda (E_{ij}+E_{ji}))$ and $n(\lambda E_{ii})$. This proves all of them preserve the form, in every characteristic.

The $x$-generators generate every upper unitriangular $T$, by elimination of off-diagonal entries. The $y,z$ generators add all elementary symmetric entries, so they generate every $n(S)$. Moreoverwhich keeps symmetry. Therefore the generated group is exactlyThe two factors are uniquely determined by its diagonal and off-diagonal blocks. Their counts are $q^{m(m+1)/2}$ and $q^{m(m-1)/2}$, givingIt is a $p$-group, and this equals the full $p$-part found in part (a); hence $U_0$ is a Sylow $p$-subgroup. In the original reversed-$f$ ordering, these matrices are upper unitriangular throughout, exactly as the prescribed generators suggest. This is consistent with the finite symplectic group order. No factor of two was divided out, so characteristic two is included.

Keep row-vector action and the block order $(W^{\prime },U,W)$. Let $R$ be the reversal matrix of size $k$, arising from the original reversed order of $f_1,\dots ,f_k$, and let $J_U$ be the alternating bilinear form matrix on $U$. ThenFor an element of $Q$, the equation $gJ{g}^T=J$ givesThus $D$, a $k\times (2m-2k)$ matrix, is arbitrary and uniquely determines $E$. The right side of the second equation is alternating, including in characteristic two: its diagonal entries vanish because $J_U$ represents an alternating bilinear form.

For any alternating matrix $K$, the equation $T-T^T=K$ has exactly $q^{k(k+1)/2}$ solutions. For each pair $i<j$, choose one entry freely and solve for the opposite entry; each diagonal entry is free. This works in characteristic two as well as odd characteristic. Taking $T=FR$ therefore givesThis is the unipotent radical count for a symplectic parabolic subgroup. It does not incorrectly replace the alternating constraint by division by two.

For a block-diagonal element, form preservation saysGiven any $C\in \mathrm{GL}(W)$, the first equation uniquely determinesWith dual bases ordered in matching rather than reversed order, the same relation is simply $A=C^{-T}$. This is the contragredient action on the paired space $W^{\prime }$.

The second equation independently allows every $B\in \mathrm{Sp}(U)$. The map taking a block-diagonal element to $(C,B)$ is a group isomorphism, with inverse $(C,B)\mapsto (R{C}^{-T}R,B,C,0,0,0)$. Thus

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.

A point is an element of $H$. A duad is an unordered two-element subset. A syntheme is a partition of $H$ into three duads, and a total of synthemes is a collection of five synthemes whose duads partition all fifteen duads. In graph terms these are vertices, edges, perfect matchings and one-factorization of $K_6$.

The first counts are $6$, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$, andsynthemes. To count totals, first observe that two edge-disjoint synthemes have union a six-cycle. Its complement in $K_6$ is a triangular prism: two triangles on alternate cycle vertices, joined by the three remaining cross edges. Its perfect matchings are the matching using all three cross edges and three matchings using one cross edge each. The all-cross matching cannot be used in a factorization, because the remaining two odd triangles cannot be matched. The other three matchings partition the prism edges. Therefore every pair of disjoint synthemes extends to a unique total.

Fix a syntheme $s$. Each of its three duads belongs to three synthemes. Inclusion-exclusion shows that $3\cdot 3-3+1=7$ synthemes share a duad with $s$, including $s$ itself. Thus eight are disjoint from $s$. A total containing $s$ uses four of these, and each disjoint syntheme determines exactly one such total. Hence $s$ belongs to $8/4=2$ totals. Counting incidences givesTwo different totals share at most one syntheme, by the unique-completion assertion. There are fifteen pairs of totals and fifteen synthemes each belonging to two totals; consequently each pair of totals has exactly one common syntheme. This incidence property drives the next construction.

Write $T_a=\Theta (a)$ for the total of $H^{\prime }$ assigned to a point $a\in H$. The duad-syntheme duality on six points is constructed entirely from incidence, as follows.

For a duad $d=\{a,b\}$, define $\Theta (d)$ to be the unique syntheme common to $T_a$ and $T_b$. This is a bijection between the fifteen duads and the fifteen synthemes of $H^{\prime }$, by the last incidence count in part (a).

For a duad $d^{\prime }$ of $H^{\prime }$, the three synthemes containing $d^{\prime }$ each belong to two totals. Each total contains exactly one of these synthemes, since its five matchings cover every duad exactly once. Their three pairs of totals therefore partition all six totals. Pulling these pairs back to $H$ gives a syntheme $s$. Different $d^{\prime }$ give different $s$, since two distinct synthemes of $H^{\prime }$ containing $d^{\prime }$ have intersection exactly that duad. There are fifteen of each, so this construction is bijective. Define $\Theta (s)=d^{\prime }$ by its inverse. Equivalently, the three synthemes $\Theta (d)$ for $d\in s$ have common duad $\Theta (s)$. In particular,

For a total $T$ of $H$, map its five synthemes to five duads of $H^{\prime }$. Any two of the original synthemes are disjoint. Their image duads must intersect: if two image duads were disjoint, the unique syntheme containing both in $H^{\prime }$ would give a common duad in the original two synthemes through the pairs-of-totals construction. Conversely intersecting duads cannot lie together in a syntheme and give disjoint original synthemes. Five distinct pairwise-intersecting edges must form the full star at one point. Indeed two edges meeting at a point either force every other edge through that point or leave only the three edges of a triangle, which cannot contain five edges. Define $\Theta (T)$ to be the star's center. Distinct totals give distinct stars; since there are six of each, this is a bijection to the points of $H^{\prime }$.

It remains to extend to unordered three-versus-three partitions. Start with a partition $B\mid B^c$ of $H^{\prime }$. Its six cross synthemes are the perfect matchings between the two triples, parametrized by permutations in $S_3$. Two of these are disjoint exactly when the quotient of their permutations is a three-cycle. Hence the six cross synthemes split into two classes of three: within a class any two are disjoint, and between classes any pair shares a duad. This unordered division into two classes is independent of the chosen orderings of the triples.

Every total has exactly two cross synthemes. To see this, any syntheme has either one or three cross duads. If a total has $h$ all-cross synthemes, it covers $3h+(5-h)$ cross duads; the whole complete graph has nine, so $h=2$. Its two cross synthemes belong to the same parity class. Conversely any pair in one class extends to a unique total. Thus the six totals split into two triples, the three totals arising from pairs in each parity class. Pulling them back through the original point-total bijection defines a partition $A\mid A^c$ of $H$.

Under the duad mapping, its six internal duads become exactly the six cross synthemes of $B\mid B^c$: a syntheme in a parity class belongs to the two totals formed by pairing it with the other two members. These incidences give the three edges of a triangle on each triple of totals. Therefore the partition is characterized byThis correspondence is injective: the six cross synthemes determine all nine cross duads of $K_{3,3}$, whose bipartition is unique up to interchange. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions on each side, so it is bijective. Define $\Theta (A\mid A^c)=B\mid B^c$ by the inverse of the construction above.

The inverse incidence rule is also useful. If a duad $d^{\prime }$ is internal to $B\mid B^c$, none of the cross synthemes contains it, so its inverse syntheme has no internal duad of $A\mid A^c$ and is entirely cross. If $d^{\prime }$ is cross, exactly two cross synthemes contain it, so the inverse syntheme has two internal duads and one cross duad. Thus internal duads and cross synthemes exchange roles in both directions. All the extensions are natural: they use intersections and incidence, with no auxiliary ordering left in the answer.

A Steiner system $S(l,m,n)$ is an $n$-point set together with $m$-element blocks such that every $l$-element subset is contained in exactly one block. On $X=H\cup H^{\prime }$, define the following six-element blocks using the incidence extensions of $\Theta$.

Take the two blocks $H$ and $H^{\prime }$. For each duad $d$ of $H$ and each duad $d^{\prime }$ in the syntheme $\Theta (d)$, takeThese give forty-five blocks of type $(2,4)$ and forty-five of type $(4,2)$. Finally, for every corresponding partition pair $A\mid A^c$ and $B\mid B^c$, take all four unionsThere are ten partition pairs and forty blocks of type $(3,3)$. Distinct indexing data give distinct blocks within each family, and different types have different intersection sizes with $H$. Thus the total is

To prove the defining property, fix a five-element subset $Y$ and set $r=|Y\cap H|$.

If $r=5$ or $0$, only $H$ or $H^{\prime }$ can contain it. If $r=1$, write $Y\cap H=\{a\}$ and $Y\cap H^{\prime }=C^{\prime }$ with $|C^{\prime }|=4$. A containing block must have type $(2,4)$ and its omitted duad is $d^{\prime }=H^{\prime }\setminus C^{\prime }$. The total $T_a$ has exactly one syntheme containing $d^{\prime }$; the other total containing that syntheme determines a unique second point $b$. Thus the unique block is $B(\{a,b\},d^{\prime })$.

If $r=4$, write $Y\cap H=C$ and $Y\cap H^{\prime }=\{a^{\prime }\}$. A containing block must have type $(4,2)$, with omitted duad $d=H\setminus C$. Exactly one duad $d^{\prime }\in \Theta (d)$ contains $a^{\prime }$, so $(H\setminus d)\cup d^{\prime }$ is the unique block.

If $r=2$, write $Y\cap H=d$ and $Y\cap H^{\prime }=C^{\prime }$, where $|C^{\prime }|=3$. There are only two possible types. A $(2,4)$ block exists exactly when the matching $\Theta (d)$ has a duad contained in $H^{\prime }\setminus C^{\prime }$. There is then exactly one such duad, since two disjoint duads cannot fit inside a triple. For a perfect matching on two triples, either all three pairs are cross, or there is one internal pair in each triple and one cross pair. Thus a $(2,4)$ block exists precisely when $\Theta (d)$ is not entirely cross for $C^{\prime }\mid (H^{\prime }\setminus C^{\prime })$.

On the other hand, a $(3,3)$ block containing $Y$ must use the unique partition of $H$ corresponding to $C^{\prime }\mid (H^{\prime }\setminus C^{\prime })$. It exists precisely when $d$ is contained in one of that partition's triples, and is then unique. By the partition incidence rule in part (b), this happens precisely when $\Theta (d)$ is entirely cross. Hence exactly one of the two possible block types exists, always uniquely.

For $r=3$, interchange the roles of $H$ and $H^{\prime }$ and use the inverse partition incidence rule proved in part (b). More explicitly, the duad $Y\cap H^{\prime }$ either has an inverse syntheme with an internal pair in the complement of the triple $Y\cap H$, yielding a unique $(4,2)$ block, or its inverse syntheme is entirely cross, yielding the unique $(3,3)$ block. These alternatives are exclusive and exhaustive for the same matching-on-two-triples reason.

Every split $r=0,\dots ,5$ therefore gives exactly one containing block. We have constructedAs an independent count, each block contains six five-subsets and $132\cdot 6=792=\left(\genfrac{}{}{0ex}{}{12}{5}\right)$. The case proof establishes uniqueness and existence; the count alone would not have done so.