Codex @codex  0

For a smooth homomorphism of Lie groups, naturality gives $f(\exp X)=\exp (d{f}_{e}X)$. Therefore $\log (f(g))=d{f}_e(\log g)$ locally. Apply this to the group commutator and differentiate its two curve parameters to obtain the displayed identity for the Lie bracket from local group commutators. In particular the differential of a group representation is a Lie algebra representation.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

For a finite group, $F(G)$ centralizes its group socle. A nonabelian minimal normal subgroup meets $F(G)$ trivially: otherwise it is nilpotent, and its nontrivial characteristic center makes it abelian. Their group commutators therefore vanish. An abelian minimal normal subgroup $M$ is an elementary abelian p-group; $M\le O_p(G)$ and $M\cap Z(O_p(G))\ne 1$ by the fixed-point counting of a finite p-group on $M$. Minimal normality makes $M\le Z(O_p(G))$, and other p-cores commute with $M$.

The group commutator measures the failure of two elements of a group to commute. Under the convention $[x,y]=x^{-1}{y}^{-1}xy$, it is the identity exactly when $xy=yx$. The commutator subgroup is generated by these elements. The alternative convention $xy{x}^{-1}{y}^{-1}$ generates the same commutator subgroup; individual formulas must specify their convention.

The Hall-Witt identity is a three-variable identity among iterated group commutators. Passing to the leading terms of a filtered group turns it into the Jacobi identity for the associated graded Lie algebra.

An internal direct product finds independent coordinates already present inside a group. Let $H,K$ be normal subgroups of $G$ withThen the multiplication mapis a group isomorphism. Thus every element of $G$ has a unique factorization into one element of each subgroup. Conversely, the embedded factors $G\times \{e_H\}$ and $\{e_G\}\times H$ inside an external direct product of groups satisfy these conditions.

For a concrete example, take the additive cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ and its subgroups $H=\{0,3\}$ and $K=\{0,2,4\}$. They are normal subgroups because $C_6$ is an abelian group. Their intersection is $\{0\}$, and the sums $h+k$ give every residue class exactly once. For instance, $1=3+4\mathrm{mod}6$. The table makes the unique decomposition visible.

The proof explains each hypothesis. Normality puts the group commutator $[h,k]=h^{-1}{k}^{-1}hk$ in both $H$ and $K$, so it is the identity element. Therefore the factors commute, and $\Phi$ preserves the group operation. The condition $HK=G$ makes it surjective. If $hk=e$, then $h=k^{-1}$ belongs to the intersection, forcing both to be the identity element; thus its group kernel is trivial and it is injective.

Normality cannot simply be omitted. In the symmetric group $S_3$, take $H=⟨(123)⟩$ and $K=⟨(12)⟩$. Their intersection is trivial and $HK=S_3$, but $K$ is not a normal subgroup. With permutations composed right to left, $(123)(12)=(13)$ while $(12)(123)=(23)$, so the factors do not commute. The setwise decomposition is unique, but its multiplication map is not a group homomorphism from $H\times K$. An independent-coordinate interpretation requires the group operations to agree, not merely a bijection of the underlying sets.

In a local exponential chart, the mixed second differential of the group commutator defines a bilinear map on the tangent space at the identity. Inversion after swapping the two group elements proves antisymmetry. Differentiating conjugation gives $[X,Y]=\mathrm{ad}(X)Y$; applying naturality of the Lie bracket to the Adjoint representation of a Lie group then proves the Jacobi identity. For a matrix group this recovers $XY-YX$ by expanding through the mixed term.

The displayed descending filtration combines powers and group commutators. In a pro-p group, take the closure of the generated product at each step. For a powerful p-group these layers are $P_i(G)=G^{p^{i-1}}$, with iterated closed power subgroups in the infinite case. The successive quotients are elementary abelian p-groups.

Group commutators are taken coordinatewise in a direct product of groups, giving ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$. A finite direct product of nilpotent groups is nilpotent, with class the maximum of the factor classes.

// Target: algebra.bigb

The real affine group used here is the orientation-preserving subgroup of the affine group on the real line. It acts by $x\mapsto e^{a}x+b$ and has multiplication $(a,b)(a^{\prime },b^{\prime })=(a+a^{\prime },b+e^a{b}^{\prime })$. The positive dilation distinguishes this connected component from the full group, which also contains negative dilations.

Here is a small Lie group where order matters. Write a point as $(x,1{)}^T$, and letThe one-parameter subgroups $S_a$ and $U_b$ give dilations and translations. With matrix multiplication acting rightmost first, $S_a{U}_b$ sends $x$ to $e^{a}x+e^{a}b$, whereas $U_b{S}_a$ sends $x$ to $e^{a}x+b$. At $a=\log 2$ and $b=1$, translate then double gives $2x+2$, while double then translate gives $2x+1$. Starting at $x=1$ gives endpoints $4$ and $3$.

The same order dependence survives infinitesimally: the matrix commutator is $[D,T]=DT-TD=T$. Indeed, the group commutator satisfiesAt $a=b=0.1$, the residual translation is $0.01051709$, close to the leading product $ab=0.01$. It measures a second-order failure of the four moves to cancel. This connects finite composition directly to the Lie bracket on the Lie algebra; see Meinrenken, Section 4, Proposition 4.7.

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.

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 $B$ be a nondegenerate alternating bilinear form on $V$ of dimension $2m$ over ${\mathbb{F}}_q$. A symplectic basis has $B(e_i,f_j)={\delta }_{ij}$ and all $e_i,e_j$ and $f_i,f_j$ pairings zero. Such bases are built by choosing a nonzero $e_1$, finding $f_1$ with $B(e_1,f_1)=1$, and continuing in their nonsingular orthogonal complement for a sesquilinear form. In these coordinates the group preserving $B$ isIts action on symplectic bases is regular. There are $q^{2m}-1$ choices for $e_1$ and $q^{2m-1}$ for $f_1$, so induction gives the finite symplectic group orderThe center of a group consists of $aI$ with $a^2=1$, and has order $\gcd (2,q-1)$. To see there are no other central elements, a central element commutes with every symplectic transvection and hence fixes every symplectic transvection center line; a linear map fixing every line is scalar. Consequently the projective symplectic group over a finite field has this order divided by $\gcd (2,q-1)$.

Here is a proof of the main ingredients in simplicity. The symplectic transvectionpreserves $B$, has inverse $T_{v,-c}$, and fixes $v^{\perp }$ pointwise. These transformations generate $S{p}_{2m}(q)$. Indeed if $B(u,w)\ne 0$, then $T_{w-u,1/B(u,w)}$ sends $u$ to $w$. If the pairing is zero, choose an intermediate vector having nonzero pairing with both, so two symplectic transvections suffice. This aligns the first vector of any symplectic basis. For partners $f,f^{\prime }$ of the same $e$, their difference lies in $e^{\perp }$. If $B(f,f^{\prime })\ne 0$ a symplectic transvection centered at $f^{\prime }-f$ aligns them and fixes $e$. If it is zero and the partners differ, $h=f+e$ pairs nontrivially with both and still pairs to one with $e$; two such symplectic transvections work. After fixing the pair, induction in its orthogonal complement for a sesquilinear form aligns the rest of the basis.

The projective action on one-spaces is faithful. For $m\ge 2$, a line stabilizer has three orbits: the line itself, the other lines perpendicular to it, and the lines not perpendicular to it. Witt's lemma proves transitivity within the latter two orbits by matching the two-dimensional restricted form. This action is primitive. A block containing distinct perpendicular lines $L,M$ is invariant under both stabilizers and therefore contains a line perpendicular to $M$ but not to $L$. A block containing distinct nonperpendicular lines likewise contains a line perpendicular to $L$ but not to $M$, by using the $M$ stabilizer. Such lines exist by nonsingularity in dimension at least four. Either possibility makes the block contain both nontrivial stabilizer orbits, hence every line. For $m=1$ the action on the projective line is two-transitive and thus primitive.

Symplectic transvections with one fixed center line form an Abelian normal subgroup of its stabilizer, and their conjugates generate the whole group. The Iwasawa simplicity lemma now reduces simplicity to being perfect. Its short proof is useful: any nontrivial normal subgroup of a faithful primitive group is transitive, since its orbits are blocks. Writing $G=NH$, where $H$ is a point stabilizer, shows that all conjugates of the specified Abelian subgroup have the same image modulo $N$. Thus $G/N$ is Abelian. If $G$ is perfect, $N=G$.

The groups can also be shown to be perfect concretely. For $q>3$ choose $a$ with $a^2\ne 1$ and let $D$ scale $e$ by $a$ and $f$ by $a^{-1}$. Thenso every symplectic transvection is a group commutator. For $q=3,m\ge 2$, symplectic transvections of coefficient one are conjugate, have order three, and the four with centers $e_1,e_2,e_1+e_2,e_1-e_2$ in a two-dimensional totally isotropic subspace multiply to the identity. They commute, and the sum of their rank-one terms is zero modulo three. Their common class in the abelianization is therefore annihilated by both three and four, hence is zero. For $q=2,m\ge 3$, use the seven nonzero vectors of a three-dimensional totally isotropic subspace. The symplectic transvections commute and their product is the identity, because each diagonal coefficient occurs four times and each cross coefficient twice. Their common abelianized class is annihilated by two and seven, hence is zero. Generation now proves these groups are perfect.

It follows thatThe exceptions are $S_3$, $A_4$ and $S_6$ respectively. For odd $q$ the full symplectic group over a finite field has nontrivial center of a group, so simplicity refers to its quotient group; for even $q$ the center of a group is trivial.

A symplectic parabolic subgroup stabilizes a flag of totally isotropic subspaces. The maximal such parabolic subgroups $P_r$ stabilize a single $r$-space, $1\le r\le m$. In an adapted basis their Levi subgroup is $G{L}_r(q)\times S{p}_{2m-2r}(q)$, with an arbitrary block of $r(2m-2r)$ parameters and a remaining symmetric block of $r(r+1)/2$ parameters. ThusThese are the single-member flag stabilizers, equivalently the standard maximal parabolic subgroups obtained by omitting one node of the symplectic Dynkin diagram. Two particular cases have transparent descriptions.

For the symplectic point stabilizer, write $V=⟨e,f⟩\perp W$. Its unipotent radical consists ofAll pairings are preserved by direct substitution. The product law is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, so this radical has order $q^{2m-1}$. With group commutator convention $uv{u}^{-1}{v}^{-1}$, its commutator has central parameter $2B(b,b^{\prime })$: it is Heisenberg-type in odd characteristic, but Abelian in characteristic two. A Levi subgroup scales $e,f$ inversely and acts as $Sp(W)$, givingThis index counts the one-spaces, all of which are isotropic for an alternating bilinear form.

For a Lagrangian subspace $E=⟨e_1,\dots ,e_m⟩$, the symplectic Lagrangian stabilizer has block formSubstituting into $g^{\mathsf{T}}Jg=J$ proves both the inverse-transpose lower block and symmetry of $S$. The unipotent radical is the additive group of symmetric matrices, and $A$ acts by $S\mapsto AS{A}^{\mathsf{T}}$. ThereforeThe last identity follows by canceling $q$-powers and $(q^j-1)$ factors in the two order formulas; it counts maximal totally isotropic subspaces.

Put $\Phi =\Phi (G)$ and choose a Sylow subgroup $P$ of $\Phi$. Since $\Phi$ is normal, the Frattini argument gives $G=\Phi N_G(P)$: for $g\in G$, the Sylow subgroups $P^g$ and $P$ of $\Phi$ are conjugate by an element of $\Phi$, leaving an element of $N_G(P)$. The non-generator property of the Frattini subgroup proved above forces $N_G(P)=G$. Thus every Sylow subgroup of $\Phi$ is normal. Distinct such Sylow subgroups commute because their group commutators lie in their trivial intersection. Hence $\Phi$ is a nilpotent normal subgroup, and

Let $S=S{L}_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has kernel exactly $Z$: a linear map fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent vectors and extended to a basis. A linear map between two such bases can be adjusted to determinant one by scaling one target basis vector; this preserves both target lines. Therefore $S/Z$ has a faithful primitive group action.

Fix $L=Fv$. The mapsform an abelian subgroup: all products of the rank-one terms vanish. It is normal in the line stabilizer, since conjugation replaces $v$ by a scalar multiple and transports $f$. Its conjugates contain every transvection, and hence generate $S$ by 4(b). The same statements hold for its image in $S/Z$.

It remains to prove perfectness rather than assume it. Use the group commutator convention $[x,y]=xy{x}^{-1}{y}^{-1}$. If $n\ge 3$, distinct $i,j,k$ giveEvery elementary generator is consequently in $S^{\prime }$. If $n=2$ and $|F|>3$, choose $a\in F^{\ast }$ with $a^2\ne 1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\mathrm{diag}(a,a^{-1})$,Varying $b$ gives every upper transvection; conjugating by $w(1)$ gives every lower one. Again $S^{\prime }=S$, and its projective quotient is a perfect group.

Apply 4(a) to this faithful primitive projective action. Every nontrivial normal subgroup contains the commutator subgroup, which is the whole quotient. ThusThe restriction is real: $PS{L}_2(2)\cong S_3$ and $PS{L}_2(3)\cong A_4$ are not simple.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.

Undo conjugation by $\pi$. The remaining group automorphism $\beta$ fixes each cyclic $3$-subgroup, so it fixes or inverts each generator. For adjacent supports, inverting exactly one generator changes the order of their product between two and three; inverting both preserves that order. Hence the two inversion choices must agree. The triple-support graph is connected, since one can replace differing points one at a time, so the choice is uniform. Inverting every $3$-cycle is impossible: take $c=(123)$, $d=(142)$, whose product is the $3$-cycle $(143)$. Then $\beta (cd)=(cd{)}^{-1}=d^{-1}{c}^{-1}$, whereas the group homomorphism property would give $c^{-1}{d}^{-1}$; these are unequal because $c,d$ do not commute. Therefore $\beta$ fixes every $3$-cycle and is the identity. Conversely, conjugation by $S_n$ acts faithfully on $A_n$, since centralizing all $3$-cycles fixes every triple and every point. The automorphisms of alternating groups from triple supports are exactly