Conjugacy in $S_n$, Generation, and the Simplicity of $A_n$

For each $j$ modulo $k$, $(gc{g}^{-1})(g(a_j))=g(c(a_j))=g(a_{j+1})$.

If $y$ is outside $\{g(a_1),\dots ,g(a_k)\}$, then $g^{-1}(y)$ is outside the support of $c$, so [F2] gives $(gc{g}^{-1})(y)=y$.

Steps 1.1--1.2 describe exactly the cycle on the right, including the $k=1$ case.

Suppose $\tau =g\sigma g^{-1}$. Conjugate every factor in the disjoint-cycle decomposition of $\sigma$; [F3] preserves each cycle length and relabels fixed points.

Conversely, suppose the cycle types agree. By [F1] and [F2], pair every cycle of $\sigma$ with a cycle of $\tau$ of the same length, including the fixed points, and define $g$ by sending entries positionwise in each pair.

Hence $\sigma$ and $\tau$ have the same multiplicities $c_k$ and the same cycle type.

The paired cycles partition the underlying set, so the map in step 1.2 is a bijection and hence an element of $S_n$.

Applying [F3] to each paired cycle gives $g\sigma g^{-1}=\tau$.

Assign to a conjugacy class the cycle type of any representative; [F2] makes this well-defined and injective.

Conversely, for any such tuple, partition $n$ symbols into $c_k$ blocks of size $k$ and put a $k$-cycle on each block; their product has that cycle type. For $n=0$, use the empty product on the empty set.

The orbits counted in [F1] partition the $n$ symbols, so counting their points gives ${\sum }_{k}k{c}_k=n$. Thus the image consists of tuples satisfying the displayed equation.

Thus the assignment is surjective and hence a bijection.

If $g\sigma =\sigma g$ and $O$ is a $⟨\sigma ⟩$-orbit, then $g(O)$ is another orbit of the same size.

For the $c_k$ orbits of size $k$, $g$ may permute those orbits in $c_k!$ ways by [F3]. Once a target orbit is chosen, the image of one marked point has $k$ choices, and commutation forces all other images; hence there are $k^{c_k}{c}_k!$ choices at length $k$.

Conversely, arbitrary orbit permutations and cyclic offsets from step 2.1 assemble on the disjoint orbits to a unique bijection $g$, and the forced-image rule makes $g\sigma =\sigma g$.

Choices for distinct lengths are independent, so [F4] and steps 2.1--3.1 give the displayed product. If $c_k=0$ its factor is $k^{0}0!=1$; for the identity the result is $n!$, and for $n=0$ it is the empty product $1$.

By [F1], index the conjugacy classes by the displayed tuples.

For a tuple $(c_k)$, [F2], [F3], and [F5] give class size $n!\mathrm{/}\prod k^{c_k}{c}_k!$.

In [F4], the central term counts the singleton conjugacy classes and the sum counts every remaining class. Thus summing the sizes from step 2.1 and using [F5] for $\mathrm{\mid }{S}_n\mathrm{\mid }$ gives the displayed identity.

At $n=0$, [F1] gives the one empty type and all empty products and $0!$ equal $1$, so the same formula reads $1=1$.

For $n=0,1$, the groups are trivial; $S_2$ has two elements and is cyclic. Hence all three are abelian, as is also consistent with the conjugacy and centralizer descriptions in [F3].

Let $n\ge 3$ and suppose, for contradiction, that $\sigma \mathrm{\ne }1$ lies in $Z(S_n)$. Choose $a$ with $b=\sigma (a)\mathrm{\ne }a$ and choose $c$ distinct from $a,b$.

For $t=(ac)$, [F2] gives $\sigma t{\sigma }^{-1}=(b\sigma (c))$. This cannot equal $(ac)$ because its support contains $b\mathrm{\notin }\{a,c\}$.

Thus $\sigma$ does not commute with $t$, contradicting [F1] and the assumption that it is central. Therefore only the identity is central for $n\ge 3$, while step 1.1 gives the stated exceptional centers.

For $1\le a<b\le n$, pointwise calculation using [F3] gives $(ab)=s_a{s}_{a+1}\cdots s_{b-2}{s}_{b-1}{s}_{b-2}\cdots s_{a+1}{s}_a\mathrm{.}$ When $b=a+1$, this word is just $s_a$.

Thus every transposition lies in the subgroup generated by the adjacent ones, and [F1]--[F2] show that this subgroup is $S_n$.

If $n=0$ or $1$, $S_n$ is trivial, so [F2] says the empty displayed set generates it.

For $0\le j\le n-2$, [F1] gives $c^{j}t{c}^{-j}=(j+1j+2)$.

Every element in step 1.1 belongs to $⟨c,t⟩$ by [F3], so that subgroup contains all standard adjacent transpositions.

By [F2], $⟨c,t⟩=S_n$. At $n=2$, $c=t=(12)$ and the same argument has the single adjacent transposition.

Let $\sigma \in A_n$. By [F2], write it as a product of transpositions; [F1] and [F3] imply that the number of factors is even.

Pair consecutive transpositions. An equal pair cancels; two sharing one entry multiply to a $3$-cycle; and two disjoint pairs satisfy $(ab)(cd)=(acb)(acd)$.

Thus every $\sigma \in A_n$ is a product of $3$-cycles.

Conversely, [F3] makes every $3$-cycle even, hence an element of $A_n$ by [F1]. Therefore [F4] and step 2.1 prove that the $3$-cycles generate $A_n$.

Choose $t\in G\setminus H$. By [F3], $G=H\bigsqcup tH$.

Let $C$ be the $H$-class of $x$. Using [F1] and step 1.1, the $G$-class is $C\cup tC{t}^{-1}$; these are $H$-classes and therefore are equal or disjoint.

Conjugation by $t$ is a bijection $C\to tC{t}^{-1}$, so in the disjoint case the two classes have equal cardinality.

The classes agree exactly when $tx{t}^{-1}=hx{h}^{-1}$ for some $h\in H$, which is equivalent to $h^{-1}t\in C_G(x)$. This element lies outside $H$.

Conversely, if $c\in C_G(x)\setminus H$, then $c=h^{-1}t$ for some $h\in H$, and $tx{t}^{-1}=hx{h}^{-1}\in C$; hence the two classes agree.

Steps 3.2--3.3 give the outside-centralizer criterion; negating it gives the equivalent containment criterion for splitting.

If $\sigma$ has an even-length cycle, that cycle commutes with every disjoint cycle of $\sigma$, so it lies in $C_{S_n}(\sigma )$; [F2] makes it odd.

If two cycles have the same odd length $k$, including $k=1$, the permutation swapping their entries positionwise is a product of $k$ transpositions. It centralizes $\sigma$ and is odd by [F2].

Conversely, suppose all cycle lengths are odd and distinct. Products of independent powers of the disjoint cycles form a subgroup $P\subseteq C_{S_n}(\sigma )$ of cardinality ${\prod }_{k}k$. Here each $c_k$ is $0$ or $1$, so [F5] gives $\mathrm{\mid }{C}_{S_n}(\sigma )\mathrm{\mid }={\prod }_{k}k=\mathrm{\mid }P\mathrm{\mid }$. Hence $C_{S_n}(\sigma )=P$: every centralizing permutation preserves each cycle and restricts to a power of it.

In either case the centralizer is not contained in $A_n$, so [F3] says the class does not split.

Each cycle has odd length, so [F2] makes it and all its powers even. By step 1.3 every centralizing permutation is a product of such powers and is therefore even.

Hence $C_{S_n}(\sigma )\subseteq A_n$, so [F3] gives two equal $A_n$-classes.

Every $3$-cycle has cycle type consisting of one $3$-cycle and $n-3$ fixed points, so [F1] puts all of them in one $S_n$-class.

Since $n-3\ge 2$, the length $1$ is repeated. Thus [F2] says that this class does not split in $A_n$.

Therefore all $3$-cycles form one $A_n$-conjugacy class.

By [F1], $N$ contains the entire $A_n$-conjugacy class of its given $3$-cycle.

By [F2], this means that $N$ contains every $3$-cycle.

Since those cycles generate $A_n$ by [F3], one has $N=A_n$.

Choose $1\mathrm{\ne }\sigma \in N$ having as many fixed points as possible; this is possible because $A_n$ is finite.

For any $3$-cycle $\tau$, the commutator $h=\tau \sigma {\tau }^{-1}{\sigma }^{-1}$ lies in $N$ by [F1]. Its support is contained in $\mathrm{supp}(\tau )\cup \sigma (\mathrm{supp}(\tau ))$.

If $\sigma$ itself is a $3$-cycle, there is nothing to prove. Suppose instead that $\sigma$ has a $3$-cycle $(a_1{a}_2{a}_3)$ and also moves a point $b$ outside it. The remaining disjoint cycles form a nonidentity even permutation, so [F4] shows that they move at least three points; hence $\sigma$ moves at least six points.

It remains that every nontrivial cycle of $\sigma$ is a transposition. Because $\sigma \in A_n$, [F4] makes their number $r$ even, so $r\ge 2$.

If $\sigma$ has a cycle $(a_1{a}_2{a}_3{a}_4\dots )$ of length at least $4$, take $\tau =(a_1{a}_2{a}_3)$. Direct use of [F3] gives $h=(a_1{a}_2{a}_4)$, a $3$-cycle in $N$.

Put $\tau =(a_1{a}_{2}b)$. Its support is not preserved by $\sigma$, so $h\mathrm{\ne }1$; step 1.2 shows that $h$ moves at most the five points $a_1,a_2,a_3,b,\sigma (b)$. Thus $h$ fixes more points than $\sigma$, contradicting step 1.1.

If $r\ge 3$, write two factors as $(ab)(cd)$ and take $\tau =(abc)$. Calculation gives $h=(ac)(bd)$, which is nonidentity and moves four points, fewer than the at least six moved by $\sigma$; this again contradicts step 1.1.

If $r=2$, write $\sigma =(ab)(cd)$. Since $n\ge 5$, choose a fixed point $e$ and take $\tau =(abe)$. Calculation gives $h=(aeb)$, a $3$-cycle in $N$.

The exhaustive cycle cases in [F2] show that either $\sigma$ itself, step 2.1, or step 2.4 supplies a $3$-cycle, while steps 2.2 and 2.3 exclude every other case. Therefore $N$ contains a $3$-cycle.

The cycles $(123)$ and $(345)$ have sign $(-1{)}^2=+1$ by [F4], so they belong to $A_n$; they do not commute, so $A_n$ is nontrivial.

Let $N⊴A_n$. If $N$ is nontrivial, [F2] gives a $3$-cycle in $N$, and [F3] then gives $N=A_n$.

Thus the only normal subgroups are $\{1\}$ and $A_n$; together with step 1.1, [F1] proves simplicity.

The intersection $N\cap A_n$ is normal in $A_n$ by [F5], so [F2] makes it either $\{1\}$ or $A_n$.

Suppose $N\cap A_n=A_n$.

Suppose $N\cap A_n=\{1\}$.

In the case of step 1.2, $A_n\subseteq N$. Since [F1] gives only two cosets, $N=A_n$ or $N=S_n$.

In the case of step 1.3, the restriction of sign to $N$ is injective by [F3], so $\mathrm{\mid }N\mathrm{\mid }\le 2$.

If the subgroup in step 2.2 were nontrivial, it would have a unique nonidentity element $t$. Normality [F5] would make every conjugate of $t$ the same unique element, so $t\in Z(S_n)$, contradicting [F4]. Thus this case gives $N=\{1\}$.

The alternatives in step 1.1 are exhaustive, and steps 2.1 and 3.1 give exactly $N\in \{\{1\},A_n,S_n\}$.

Since $S_n\mathrm{/}{A}_n$ is the abelian sign image, [F3] gives $[S_n,S_n]\subseteq A_n$.

Every $3$-cycle satisfies $(abc)=[(bc),(ab)]$ under the convention in [F1], so [F4] gives $A_n\subseteq [S_n,S_n]$ for $n\ge 3$.

For $n\ge 5$, the $3$-cycles $(123)$ and $(345)$ do not commute, so $[A_n,A_n]$ is nontrivial by [F1].

At $n=2$, both $A_2$ and the commutator subgroup of the abelian group $S_2$ are trivial. Thus steps 1.1--1.2 prove the first formula for all $n\ge 2$.