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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.

Use integral coefficients throughout, allowing a nontrivial orientation character. A dimension-$d$ Poincare duality group has a finite resolution by finitely generated projective group-ring modules andas abelian groups. The action on the top copy of $\mathbb{Z}$ may be by signs. This group-ring criterion is particularly suited to the Lyndon–Hochschild–Serre spectral sequence.

Apply that sequence to the extension with coefficient module $\mathbb{Z}G$:As an $N$-module, $\mathbb{Z}G$ is a direct sum of copies of $\mathbb{Z}N$ indexed by the cosets in $Q$. Cohomology of $N$ commutes with this direct sum because its projective resolution is finitely generated. Hence the inner cohomology vanishes except when $b=n$, where it is a direct sum of rank-one groups indexed by $Q$. The $Q$-action shifts these summands freely and transitively, with possible signs from the orientation module. Choosing a generator in one summand and transporting it by $Q$ identifies the resulting left $Q$-module with its regular module $\mathbb{Z}Q$. The residual right group action records the orientation twist; it need not be trivial.

Now the duality of $Q$ says that the only nonzero $E_2$ term is $(a,b)=(q,n)$ and that this term is $\mathbb{Z}$ as an abelian group. No differential can enter or leave it. Therefore $H^{\ast }(G;\mathbb{Z}G)$ has exactly the duality pattern in degree $n+q$. The usual extension-resolution construction also supplies a resolution by finitely generated projective modules: combine the finite kernel and quotient resolutions, lift the quotient differentials and add the necessary homotopy corrections. Its total length is at most $n+q$. Alternatively, the same spectral sequence for arbitrary coefficient modules gives the dimension bound, while the extension finiteness lemma supplies finite generation of the resolution. The surviving top term makes the bound sharp. This proves the extension rule for Poincare duality groups:Requiring a trivial top orientation action would need an additional orientation condition; the untwisted conclusion is not implicit in the hypotheses.

We next need two precise auxiliary dimension results. The Strebel infinite-index subgroup theorem says that an infinite-index subgroup of an integral $P{D}^d$ group has integral cohomological dimension at most $d-1$. The Stallings-Swan theorem says that a group of integral cohomological dimension at most one is free, including the trivial group. Since $G/N$ contains an element of infinite order, $N$ has infinite index in $G$, so $\mathrm{cd}N\le 2$. If its dimension is at most one, the required free-group conclusion follows, with finite rank because $N$ is finitely presented.

Suppose instead that $\mathrm{cd}N=2$. Finite presentation gives a partial free resolution with finitely generated modules through degree two. The kernel in degree one is finitely generated and, because the integral cohomological dimension is two, projective. Truncating at this kernel gives a finite resolution by finitely generated projective modules. Thus $N$ has the finiteness needed for group-ring cohomology to commute with direct sums.

Choose an infinite-order element $\bar{t}\in G/N$, lift it to $t\in G$, and let $H$ be the preimage of $⟨\bar{t}⟩$. Then $H=N⋊⟨t⟩$. Restriction of $\mathbb{Z}H$ to $N$ yields copies of $\mathbb{Z}N$ indexed by all powers of $t$. For every $j$, the quotient generator acts on $H^j(N;\mathbb{Z}H)$ by shifting this direct sum, possibly with an automorphism on each copy. Its invariants vanish, since an invariant finite-support sequence must be zero. Its coinvariants are one copy of $H^j(N;\mathbb{Z}N)$: the shift relations identify successive copies, and any twists can be removed by transporting the identifications along the integer index.

The infinite cyclic group has integral cohomological dimension one, so only columns zero and one occur in the spectral sequence. Column zero vanishes and column one consists of these coinvariants. We obtain the cyclic-extension shift of group-ring cohomologyas abelian groups. Moreover $H^2(N;\mathbb{Z}N)\ne 0$. Here is a useful proof of that last point: if top cohomology of a finite projective resolution were zero, its last dual differential would be onto. Its target is projective, so that differential splits. Dualizing back splits the last injection of the original resolution, shortening it and contradicting integral cohomological dimension two.

It follows that $H^3(H;\mathbb{Z}H)\ne 0$, and hence $\mathrm{cd}H=3$. Strebel's theorem forces $H$ to have finite index in $G$. It is therefore itself a $P{D}^3$ group, by finite-index invariance of Poincare duality for torsion-free groups. The displayed shift now makes $H^j(N;\mathbb{Z}N)$ zero except for $j=2$, where it is $\mathbb{Z}$. Together with its finite projective resolution, this provesThe classification of two-dimensional Poincare duality groups identifies the latter with the fundamental group of a closed aspherical surface. This classification includes orientation-twisted surface groups; it does not include the two-sphere or real projective plane, whose fundamental groups do not have integral cohomological dimension two.

In this latter case we have also proved that $G/N$ is virtually infinite cyclic, since $H/N=⟨\bar{t}⟩$ has finite index. If necessary replace the surface group by its characteristic index-two orientable subgroup and take the subgroup generated by it and $t$. This still has finite index in $G$ and is a semidirect product of an orientable closed aspherical surface group by $\mathbb{Z}$. The Dehn-Nielsen-Baer theorem for closed orientable surfaces realizes the outer automorphism defined by conjugation by $t$ as a surface homeomorphism $f$. For the torus this is just realization of $G{L}_2(\mathbb{Z})$ by linear homeomorphisms. Changing the chosen lift accounts for an inner automorphism. The mapping torusis a surface bundle over $S^1$ with fundamental group ${\pi }_1(S){⋊}_{f_{\ast }}\mathbb{Z}$. Thus a finite-index subgroup of $G$ is the fundamental group of a surface fibration over the circle. Taking a further double cover of the base if required makes the monodromy orientation-preserving.

A group automorphism is a bijective group homomorphism from $G$ to itself. The multiplication in $\mathrm{Aut}(G)$ is composition: $(\alpha \beta )(g)=\alpha (\beta (g))$. Composition is associative, the identity map is the identity element, and each inverse bijection is again a homomorphism, so these maps form the automorphism group.

For conjugation by $h$, multiplication is preserved becauseConjugation by $h^{-1}$ is its inverse, so $\psi (h)$ is an group automorphism. MoreoverHence $\psi :G\to \mathrm{Aut}(G)$ is a homomorphism, with image the inner automorphisms. Its group homomorphism kernel is the group center of $G$, since $\psi (h)$ is the identity precisely when $hg=gh$ for every $g$.

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

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

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.

The commutator subgroup $[G,G]$ is characteristic: every automorphism sends commutators to commutators and therefore preserves the subgroup they generate. Thus an automorphism $A$ inducesThis is independent of the coset representative $g$. If $A^{\prime }=c_h\circ A$ differs from $A$ by an inner automorphism, thenbecause the abelianization is abelian. Hence $q([A])=\overline{A}$ is independent of the representative of the outer class. Finally $\overline{AB}=\overline{A}\overline{B}$, sois a well-defined group homomorphism.