Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 7 (Topics in Group Theory)

A central series is a finite chain of normal subgroupswith $G_i/G_{i-1}\le Z(G/G_{i-1})$ for every $i\ge 1$. Equivalently, writing $[x,y]=x^{-1}{y}^{-1}xy$ and $[A,B]$ for the subgroup generated by these group commutators, the condition is $[G_i,G]\le G_{i-1}$. Centrality is relative to the whole quotient group, not just to the next subgroup in the chain.

A group is nilpotent if it admits a finite central series. The trivial group admits a series of length zero. For a nontrivial nilpotent group, the minimum possible length is its nilpotency class.

Use induction on the order of the finite p-group. The trivial group is nilpotent. For $G\ne 1$, the permitted center theorem gives $Z=Z(G)\ne 1$, so the quotient group $G/Z$ is a smaller p-group. By induction it has a central seriesLet $H_i$ be the inverse image of ${\overline{H}}_i$ under the quotient map, so $H_0=Z$ and $H_s=G$. Centrality in the quotient gives $[H_i,G]\le H_{i-1}$ for $i\ge 1$: the image of every such group commutator lies in ${\overline{H}}_{i-1}$. Also $[Z,G]=1$. Thereforeis a central series of $G$. This proves every finite p-group is nilpotent, including the case $Z=G$, when the quotient is trivial and the series is simply $1\le G$.

Assume $[H,G]\le K$. For $k\in K\le H$ and $g\in G$,Thus $g^{-1}Kg\le K$ for every $g$; applying this also to $g^{-1}$ gives equality. Hence $K$ is a normal subgroup of $G$, even though initially it was assumed normal only in $H$. Now the quotient $G/K$ is defined, and for $h\in H$, $g\in G$,Every element of $H/K$ consequently lies in the center of a group $Z(G/K)$.

Conversely, if $K⊴G$ and $H/K\le Z(G/K)$, each displayed quotient group commutator is trivial, so $[h,g]\in K$. Taking the subgroup generated by these elements gives $[H,G]\le K$. ThereforeThe forward argument also implies $H⊴G$, since $g^{-1}hg=h[h,g]\in H$.

Let $1=A_0\le \cdots \le A_r=A$ and $1=B_0\le \cdots \le B_s=B$ be central series. Pad the shorter chain by repeating its terminal group, so both have length $t=\max (r,s)$. In the direct product of groups, group commutators are coordinatewise, henceThe equality holds because commutators supported in either coordinate generate the corresponding factor independently. Thereforeis a central series, proving that a direct product of two nilpotent groups is nilpotent. More precisely, the coordinatewise recurrence for the lower central series gives ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$, so the nilpotency class of a direct product is the maximum of the two factor classes.

Define the lower central series by ${\Gamma }_1(G)=G$ and ${\Gamma }_{j+1}(G)=[{\Gamma }_j(G),G]$. Its terms are characteristic normal subgroups and form a descending chain. Define the upper central series by $Z_0(G)=1$ andEquivalently, $Z_{i+1}(G)=\{x\in G:[x,G]\subseteq Z_i(G)\}$. These are characteristic normal subgroups in an ascending chain, and successive factors are central in the appropriate quotient.

For any length-$r$ central series $1=G_0\le \cdots \le G_r=G$, prove the two comparisons independently. Starting at ${\Gamma }_1(G)=G_r$, if ${\Gamma }_{r-i+1}(G)\le G_i$, thenDescending induction gives the lower containment. Starting at $G_0=Z_0(G)=1$, if $G_{i-1}\le Z_{i-1}(G)$, then $[G_i,G]\le G_{i-1}\le Z_{i-1}(G)$, which by definition gives $G_i\le Z_i(G)$. Thus the central-series comparison theorem yieldsIn particular, a nilpotent group satisfies ${\Gamma }_{r+1}(G)=1$ and $Z_r(G)=G$.

Conversely, if ${\Gamma }_n(G)=1$, reverse the finite lower chain to obtain $1={\Gamma }_n\le {\Gamma }_{n-1}\le \cdots \le {\Gamma }_1=G$. The defining commutator recurrence makes it a central series. If $Z_n(G)=G$, the finite upper chain is already a central series. This provesFinally, for an integer $c>0$, termination at ${\Gamma }_{c+1}=1$ gives a central series of length $c$ and hence $Z_c=G$ by the comparison. Conversely, $Z_c=G$ gives a central series of length $c$ and hence ${\Gamma }_{c+1}=1$. ThereforeRepeated terms cause no problem; the least such $c$ for a nontrivial group is its nilpotency class.

Choose a representative Sylow subgroup with generatorsOn the first four points, $r^4=s^2=1$ and $srs=r^{-1}$, so $⟨r,s⟩$ is the dihedral group of order eight. The disjoint transposition $t$ commutes with it and generates an independent factor of order two. Thus $G=⟨r,s,t⟩\cong D_8\times C_2$ has order 16. Since the power of two dividing $7!$ is $2^{\lfloor 7/2\rfloor +\lfloor 7/4\rfloor }=2^4$, this is a Sylow 2-subgroup of $S_7$. Any other choice is conjugate to it and has conjugate series.

The dihedral relation gives $[r,s]=r^2$, while $t$ commutes with both generators. Modulo $⟨r^2⟩$ all three generators commute, so the commutator subgroup is exactly $⟨r^2⟩$, and $r^2$ is central. Hence the lower central series isWithin the dihedral factor only $1,r^2$ commute with both $r$ and $s$. The center of $G$ is therefore $⟨r^2,t⟩\cong C_2\times C_2$. Its quotient has order four and is abelian, so the upper central series isBoth series show that the nilpotency class is two.

Identify $G$ with a subgroup of $\mathrm{Sym}(\Omega )$ and adjoin one new point $\infty$, fixed by every element of $G$. A one-point extension of a permutation group is a transitive subgroup $G^{+}\le \mathrm{Sym}(\Omega \cup \{\infty \})$ whose point stabilizer at $\infty$ is exactly this embedded copy of $G$. Equivalently, one may use an isomorphic permutation representation with a distinguished point whose stabilizer acts on the remaining points as the original group.

Since $G$ is transitive on $\Omega$, the extension is doubly transitive group action. The orbit-stabilizer theorem gives the useful necessary order condition

Suppose an extension $H$ existed. Its degree would be $p$, and its order would be $2p(p-1)$ by the orbit-stabilizer theorem. A Sylow $p$-subgroup $P$ consequently has order $p$, and the Sylow theorems imply that its number $n_p$ divides $2(p-1)$ and is congruent to one modulo $p$. Since $2(p-1)<2p$, the only possibilities are $1$ and $p+1$. The latter would require $p+1\mid 2(p-1)$ and hence $p+1\mid 4$, impossible for $p\ge 5$. Thus $P$ is a normal subgroup of $H$.

A nonidentity element of $P$ is a $p$-cycle, so $P$ acts regularly on the $p$ points. Label them by ${\mathbb{F}}_p$ so that $P$ consists of translations. A permutation normalizing $P$ must have the form $x\mapsto ax+b$: its conjugation sends translation by one to translation by some $a\ne 0$, and its value at zero determines $b$. Therefore the normalizer of $P$ in $S_p$ has order $p(p-1)$. Normality would place $H$ inside this normalizer, contradicting $|H|=2p(p-1)$. Hence the specified dihedral permutation group has no one-point extension.

Here is an involution form of the double-coset criterion for a one-point extension. Embed $G$ in $\mathrm{Sym}(\Omega \cup \{\infty \})$ by fixing $\infty$, choose $a\in \Omega$, and put $H=G_a$. The necessary and sufficient condition is the existence of an involution $t$ such thatProducts in this part are compositions of maps, with the rightmost map applied first. The criterion is a finite test on permutations of the enlarged set; it does not presuppose that a larger group has already been found.

For sufficiency, put $K=G\cup GtG$. Multiplying a double coset by an element of $G$ on either side leaves it unchanged. A product of two elements of $GtG$ reduces to $g_{1}tgt{g}_2$. If $g\in H$, then $tgt\in H\le G$; otherwise the last condition puts it in $GtG$. Hence $K$ is closed under multiplication. It contains the identity and is closed under inverses, since $t^{-1}=t$, so it is a group.

Every element $g_{1}t{g}_2$ sends $\infty$ to $g_1(a)\in \Omega$, whereas elements of $G$ fix $\infty$. Thus $K_{\infty }=G$. Moreover the orbit of $\infty$ contains $a$ and then all of $\Omega$, so $K$ is transitive and is the desired one-point extension of a permutation group. For the order check, $G\cap tGt=H$: an element of the intersection fixes both $\infty$ and $a$, and the reverse containment follows from $tHt=H$. Consequently $|GtG|=|G{|}^2/|H|=|\Omega ||G|$, as required.

For necessity, let $K$ be an extension. It is doubly transitive group action, of order $|\Omega |(|\Omega |+1)|G_a|$, which is even. By Cauchy theorem for groups it contains a nontrivial involution, and a pair interchanged by this involution can be carried to $(\infty ,a)$ by double transitivity. Conjugating gives an involution $t$ interchanging these two points. Their common point stabilizer is $H$, so $tHt=H$. Finally $G$ has just two orbits on the enlarged set, namely $\{\infty \}$ and $\Omega$, giving $K=G\cup GtG$. More explicitly, for $k\in K\setminus G$, choose $g\in G$ with $g(a)=k(\infty )$; then $t{g}^{-1}k$ fixes $\infty$ and belongs to $G$. If $g\notin H$, the element $tgt$ moves $\infty$, so it belongs to $GtG$, completing necessity.

Adjoin $\infty$ to $\Omega =\{1,\dots ,n\}$ and take $G=A_n$ fixing it. Choose three distinct points $a,b,c\in \Omega$ and put $t=(\infty a)(bc)$. This is an even involution interchanging $\infty$ and $a$. Conjugation by it preserves $H=G_a$, because it permutes the remaining $n-1$ points and preserves parity. Thus $tHt=H$.

If $g\in G\setminus H$, the even permutation $k=tgt$ moves $\infty$ into $\Omega$. The alternating group $A_n$ is transitive for $n\ge 3$: a suitable 3-cycle carries any prescribed point to another. Choose $u\in G$ with $u(a)=k(\infty )$. Then $v=t{u}^{-1}k$ is even and fixes $\infty$, so $v\in G$ and $k=utv\in GtG$. All conditions of the double-coset criterion for a one-point extension hold, including when $n=3$ and $H$ is trivial.

The constructed extension lies in $A_{n+1}$ and has order $(n+1)|A_n|=(n+1)!/2$. Therefore

A nonidentity transvection is a linear automorphism $\tau =I+N$ with $\mathrm{rank}N=1$ and $N^2=0$. Equivalently, it fixes a hyperplane pointwise and its displacement has image in a line lying in that hyperplane. The notation $T\setminus \{1\}$ allows the identity as a degenerate transvection; use that convention here, with $N=0$ also allowed.

For a nonidentity transvection, choose $d\ne 0$ spanning the image of $N$. There is a unique nonzero linear functional $f$ such that $N(v)=f(v)d$. Since $N^2(v)=f(v)f(d)d$, we have $f(d)=0$, so $d\in \ker f$. Conversely, this condition makes $(df{)}^2=0$ and gives inverse $I-df$. The identity is represented by $f=0$ and any $d\ne 0$. ThusWhen $f(d)=f^{\prime }(d)=0$, the two displacement maps have zero product, givingFor clarity use left actions of linear maps. Applying $g^{-1}$, then ${\tau }_{f,d}$, then $g$ givesWith the alternative conjugation notation ${\tau }^g=g^{-1}\tau g$, the answer is ${\tau }_{f,d}^g={\tau }_{f\circ g,g^{-1}(d)}$. In either convention the transformed direction lies in the kernel of the transformed functional.

For a nonidentity transvection ${\tau }_{f,d}$, take $e_1=d$ and choose $e_2$ with $f(e_2)=1$. Complete $e_1$ to a basis of $\ker f$ using $e_3,\dots ,e_n$. Then $e_1,e_2,e_3,\dots ,e_n$ is a basis of $V$, and the matrix of the transvection is $I+E_{12}$: it sends $e_2$ to $e_2+e_1$ and fixes the other basis vectors. Consequently all nonidentity transvections form one conjugacy class in $\mathrm{GL}(V)$. This matrix has determinant one, and conjugation preserves determinant, soFor $n\ge 3$, let $k$ be any GL-conjugator carrying one transvection to another. In an adapted basis for the first, the matrix $c=\mathrm{diag}(1,1,s,1,\dots ,1)$ centralizes it and has determinant $s$. Set $s=(\det k{)}^{-1}$. Then $kc\in \mathrm{SL}(V)$ gives the same conjugation as $k$. This proves the stronger transvection conjugacy over a finite field resultThe extra coordinate is essential to this determinant adjustment; no such claim is required in dimension two.

Write $x_{ij}(a)=I+a{E}_{ij}$ for $i\ne j$. For $a\ne 0$ these are transvections, and for $a=0$ they are the identity. Left multiplication by $x_{ij}(a)$ adds $a$ times row $j$ to row $i$. We show that these elementary transvection matrices generate the special linear group.

Apply Gaussian elimination using only row additions. If a pivot is zero, invertibility ensures some lower entry in its column is nonzero; adding that row produces a nonzero pivot without interchanging or scaling rows. Clear entries below each pivot, and then above pivots in reverse order. This reduces any invertible matrix to a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with all $d_i\ne 0$, preserving determinant throughout. For a determinant-one matrix, ${\prod }_i{d}_i=1$.

A two-coordinate diagonal block is itself a product of elementary transvection matrices. On those two coordinates putThis formula is valid in characteristic two as well. For $n\ge 2$, multiply the diagonal blocks with entries $d_i,d_i^{-1}$ on coordinates $(i,n)$ for $1\le i<n$. Their product is $D$, because its last entry is ${\prod }_{i<n}{d}_i^{-1}=d_n$. Thus $D$ and the original matrix lie in the subgroup generated by transvections. Conversely every transvection has determinant one by part ii. HenceFor $n=1$, both the special linear group and the subgroup generated by the degenerate identity transvection are trivial. The argument also establishes transvections generate the special linear group over any field.

Use the group commutator convention $[x,y]=x^{-1}{y}^{-1}xy$. If $n\ge 3$, choose distinct $i,j,k$. Multiplying elementary matrices, using $E_{ik}{E}_{kj}=E_{ij}$ and all reverse or repeated products here equal to zero, givesEvery elementary generator is therefore in the commutator subgroup. Part iii shows these generators generate $\mathrm{SL}(V)$, proving it is a perfect group.

It remains to handle $n=2$ and $q>3$. The polynomial $s^2-1$ has at most two roots, whereas ${\mathbb{F}}_q^{\times }$ has more than two elements, so choose $s\ne 0$ with $s^2\ne 1$. For $h=\mathrm{diag}(s,s^{-1})$, direct conjugation givesBoth coefficients are nonzero. As $t$ varies, these group commutators give every upper and lower elementary transvection, which generate ${\mathrm{SL}}_2({\mathbb{F}}_q)$. Thus the perfectness of the special linear group is established in all requested cases:The two pairs in the PDF are exclusions, not a literal logical disjunction. They really are exceptions: ${\mathrm{SL}}_2({\mathbb{F}}_2)\cong S_3$ has a nontrivial sign quotient. Also ${\mathrm{SL}}_2({\mathbb{F}}_3)$ maps onto ${\mathrm{PSL}}_2({\mathbb{F}}_3)\cong A_4$, which has the nontrivial abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action has degree four and image of order 12, hence the index-two subgroup $A_4$ of $S_4$. Neither exceptional group is perfect.

First determine when a group automorphism can carry a transposition to another type of involution. An involution in $S_n$ is a product of $k$ disjoint transpositions, and its conjugacy class has sizeAn automorphism preserves element orders and conjugacy-class sizes. Equality with the transposition-class size $n(n-1)/2$ requiresFor $k=2$ this says $(n-2)(n-3)=4$, which has no integral solution $n\ge 4$. For $k=3$, the minimum value of the left side, at $n=6$, is $4!=24=2^2\cdot 3!$; it increases strictly for larger $n$. For $k\ge 4$, even its minimum $(2k-2)!$ exceeds $2^{k-1}k!$: at $k=4$ this is $720>192$, and the ratio of these two quantities increases at the next step by $(2k)(2k-1)/(2(k+1))>1$. Thus only in degree six can the transposition class be exchanged with a different involution class, namely the triple transpositions.

We next prove that any automorphism preserving transpositions is an inner automorphism. Regard a transposition as an edge of the complete graph on the $n$ letters. Two distinct transpositions fail to commute exactly when their edges meet. Any family of pairwise intersecting edges is contained in a vertex star or a triangle: after choosing $\{a,b\}$ and $\{a,c\}$, an edge not containing $a$ must be $\{b,c\}$, which precludes any fourth edge. For $n\ge 5$, the stars are therefore distinguished as the maximal such families of size $n-1\ge 4$. An automorphism permutes stars, inducing a permutation $\pi$ of the letters; their pairwise intersections show that $(ij)$ is sent to $(\pi (i)\pi (j))$. Since transpositions generate $S_n$, this is conjugation by a permutation.

For $n=4$, stars and triangles both have three edges, but a star generates $S_4$ while a triangle generates a subgroup $S_3$. An automorphism preserves the order of the generated subgroup and hence still distinguishes stars. For $n=3$, every permutation of the three transpositions is induced by conjugation in $S_3$, since its conjugation action on them is faithful and has order six. The groups $S_1,S_2$ have trivial automorphism groups. This completes the small cases.

It follows that all automorphisms for $n\ne 6$ are inner. In degree six, an automorphism permutes the two involution classes of size 15; the kernel of this action consists precisely of the inner automorphisms. Therefore the automorphisms of the symmetric group satisfy

A duad is an unordered pair of letters, a syntheme is a partition of the six letters into three duads, and a total of synthemes is a set of five synthemes partitioning the 15 duads. There are $6!/(2^3\cdot 3!)=15$ synthemes. Fix one, $M$. Each of its three duads lies in three synthemes, and two distinct duads of $M$ can occur together only in $M$ itself. Thus precisely $1+3\cdot 2=7$ synthemes meet $M$ in a duad, leaving eight disjoint from it.

For any syntheme $M^{\prime }$ disjoint from $M$, their union as edges in $K_6$ is a six-cycle: the two perfect matchings alternate, and no two-cycle is possible in a simple graph. Its complement is a triangular prism, namely two triangles joined by three corresponding edges. The prism has four perfect matchings. One consists of all three joining edges; each of the other three consists of one joining edge and the remaining edge in each triangle. To partition the prism into three perfect matchings, one cannot use the all-joining matching, since what remains is two odd triangles. The other three do partition it. Hence two disjoint synthemes extend to a unique synthematic total.

There are eight choices for $M^{\prime }$ and four choices of $M^{\prime }$ within each synthematic total containing $M$. Thus $M$ lies in exactly two synthematic totals. Counting incidences of synthemes and synthematic totals now gives the counting synthematic totals formulaTwo distinct synthematic totals share at most one syntheme, since sharing two would force them to be the unique synthematic total extending that pair. Each syntheme gives a pair of synthematic totals containing it, and the 15 synthemes account for all $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ pairs. Therefore every pair of synthematic totals shares exactly one syntheme.

This also explains why the action on the six synthematic totals defines an automorphism of $S_6$. A permutation fixing all synthematic totals fixes their pairwise intersections, hence every syntheme. Each duad is the intersection of two distinct synthemes containing it, so every duad is then fixed. Intersecting two duads through a point shows that every point is fixed. The action is faithful and, by equality of orders, maps $S_6$ onto the full symmetric group of the synthematic totals. Part iii shows that a transposition acts as a triple transposition, establishing that this is an outer automorphism.

Relabel each duad of each synthematic total by the chosen transposition and compare the resulting synthematic total with the six labelled synthematic totals. For example, $(12)$ interchanges $T_1,T_2$, interchanges $T_3,T_4$, and interchanges $T_5,T_6$. Applying the same calculation to the other adjacent transpositions givesThese formulae use the PDF's numbering of the synthematic totals, which is essential for the particular fixed elements below. Compose permutations from left to right, consistently with the right action on synthematic totals. Multiplying the second and fourth images gives $(23)(45)$, while multiplying the third and fifth gives $(34)(56)$. Hence the two involutionsare fixed by the outer automorphism $\alpha$. Since $\alpha$ is a group automorphism, their product is fixed as well. Applying $u$ and then $v$ yieldsUnder right-to-left composition the product is the inverse cycle, which is also fixed; the displayed convention removes that ambiguity.

A Steiner system $S(5,6,12)$ consists of 12 points and a collection of six-element blocks such that every five-element subset lies in exactly one block. The blocks are often called hexads. Necessarily their number is $\left(\genfrac{}{}{0ex}{}{12}{5}\right)/\left(\genfrac{}{}{0ex}{}{6}{5}\right)=132$. We give the synthematic construction of the small Witt design explicitly and prove this defining property.

Write $X=\{1,\dots ,6\}$ and $Y=\{T_1,\dots ,T_6\}$. For an unordered pair $P=\{T_i,T_j\}\subset Y$, let $S(P)$ denote the unique syntheme shared by its two synthematic totals. A duad $D\subset X$ belongs to three synthemes. The corresponding three pairs $P$ partition $Y$: every synthematic total contains $D$ in exactly one of its synthemes, and each syntheme lies in exactly two synthematic totals. Thus the incidence $D\in S(P)$ gives a perfect matching of the six synthematic totals for each duad.

To describe the blocks with three points in each half, take an unordered partition $X=A\bigsqcup A^c$ into triples. A syntheme is cross if each of its three duads joins $A$ to $A^c$. Form a graph on $Y$ by declaring $P$ an edge exactly when $S(P)$ is cross. This graph is the union of two disjoint triangles. Indeed a synthematic total partitions the nine cross duads; every syntheme has either one or three cross duads, so if $k$ of its five synthemes are cross, then $3k+(5-k)=9$, giving $k=2$. Each vertex therefore has degree two. The stronger assertion about its components follows directly for $A=\{1,2,3\}$, when the two triangles areany other triple partition is obtained by relabelling $X$, which also permutes the synthematic totals. Denote the two components by $B,B^c$.

The complete block collection is given by the following three rules:

There are $2+15\cdot 3\cdot 2+10\cdot 4=132$ distinct blocks. The first two rules have different intersection sizes with $X$ from the third; within the third, a block determines its triple $A$ and its component. We next verify the Steiner system property, rather than relying solely on this count.

We need one further consequence of the triangle construction. The three synthemes attached to the three pairs within $B$ are pairwise disjoint, since any two belong to their common synthematic total, and together they are all nine cross duads of the complete bipartite graph between $A$ and $A^c$. Thus $B$ determines the unordered partition $A\bigsqcup A^c$ uniquely: it is the bipartition of this connected graph. There are 10 such partitions, each giving two different components, hence 20 different triples in $Y$. Since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, each triple in $Y$ corresponds to exactly one triple partition of $X$. The same reasoning applies to $B^c$.

Let $F$ be any five-point set and put $r=|F\cap X|$. If $r=5$, only $X$ can contain it; if $r=0$, only $Y$ can contain it, because no other block has five points in either half.

If $r=4$, set $D=X\setminus F$ and write $F\cap Y=\{T\}$. The three pairs attached to $D$ partition $Y$, so exactly one contains $T$; its block $(X\setminus D)\cup P$ is the unique block through $F$. If $r=1$, put $P=Y\setminus F$ and write $F\cap X=\{x\}$. The three duads of $S(P)$ partition $X$, so exactly one contains $x$; this gives the unique block $D\cup (Y\setminus P)$.

If $r=3$, write $A=F\cap X$ and $P=F\cap Y$. A syntheme has either one or three duads crossing $A\bigsqcup A^c$. In the one-cross case, $S(P)$ has exactly one duad $D$ internal to $A^c$, and the unique block is $(X\setminus D)\cup P$. No block with three points in each half is possible, since $P$ is then not an edge of either triangle. In the three-cross case, $P$ is an edge of exactly one triangle $B$, giving the unique block $A\cup B$; no duad of $S(P)$ is internal to $A^c$, so a block with four points in $X$ is impossible.

Finally let $r=2$, write $D=F\cap X$ and $B=F\cap Y$, and use the uniquely associated partition $X=A\bigsqcup A^c$ just proved. If $D$ crosses this partition, the three synthemes belonging to the pairs inside $B$ contain each cross duad exactly once. Thus the matching of $Y$ attached to $D$ has exactly one pair internal to $B$. It has exactly one pair $P$ internal to $B^c$ as well, since a matching must leave equal numbers of vertices in the two halves for cross pairs. The unique block containing $F$ is then $D\cup (Y\setminus P)$; $D$ is not contained in either $A$ or $A^c$, so no block of the third rule applies. If $D$ is internal to $A$ or $A^c$, none of its attached pairs lies inside $B$ or $B^c$, since the synthemes attached to these internal pairs are cross. Thus the second rule gives no block. The third rule gives exactly one, namely $A\cup B$ or $A^c\cup B$, according to which part contains $D$.

All possibilities have now been exhausted, provingThis is the small Witt design; the construction uses only duads, synthemes, their synthematic totals, and the induced action on those synthematic totals.