置换群的自同构群

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文章是以前的老文章 (以前的风格是半角写作. 同时 dollar , 西文字母, 及汉字交接处没有空格)

置换群的自同构群

对$n\neq 2$而言, $\mbox{inn}(S_n)\cong S_n/C(S_n)\cong S_n$. 对$n=2$, $\mbox{inn}(S_2)\cong \{e\}$. 欲探究$\mbox{aut} (S_n)$与$\mbox{inn}(S_n)$之关联, 下先证明引理一则.

引理: $\tau\in\mbox{aut}(S_n)$将对换映至对换, 若且仅若$\tau$为内自同构.

证明: 充分性显然, 因为对任意同构映射$f$皆有$f(a\,b)f^{-1}=(f(a),f(b))$. 下证明其必要性:

可以观察到$\tau$将不交对换映至不交对换, 将相交对换映射至相交对换, 亦将两两相交的三组对换映至两两相交的三组对换. 今设$\tau((ij)(il))=\tau((ilj))=(i'l'j')$, 则由$\tau((ij)(ir))=(i''r''j'')$及$|\{i'',r'',j''\}\cap\{i',l',j'\}|=2$知$\tau$将$(ij)$映至的像$(\phi(i)\phi(j))$. 同理考虑$\tau((is)(il))$知$\tau$将$(il)$映至的像$(\phi(i)\phi(l))$, 从而求出$\phi(i)$. 如此递推可知$\tau:(a,b)\mapsto(\phi(a),\phi(b))$, 因此$\tau=f_\phi, f_\phi(\sigma)=\phi\sigma\phi^{-1}$为内自同构. 必要性证毕.

记$T^{(n)}_k$由$S_n$中所有由$k$组不交对换组成的置换组成. 因此$\mbox{aut}(S_n)=\mbox{inn}(S_n)$当且仅当$|T_k^{(n)}|=|T^{(n)}_1|\Leftrightarrow k=1$. 不幸的是$|T_3^{(6)}|=|T^{(6)}_1|=15$恰为唯一的异类.

一般地, $|T^{(n)}_k|=\dfrac{\binom{n}{2}\binom{n-2}{2}\cdots\binom{n-2k+2}{2}}{k!}=\dfrac{n!}{2^k k!(n-2k)!}$ , $|T_1^{(n)}|=\dfrac{n(n-1)}{2}$.
$$
|T_k^{(n)}|/|T^{(n)}_1|=\dfrac{(n-2)\cdots(n-2k+1)}{2^{k-1}k!}
$$
其中$k\leq\dfrac{n}{2}$. $n\leq 6$时逐项验证知$(k,n)=(3,6)$为有价值之$1$解. 当$n=7,8$时,分子含有$5$因子而分母不含之, 故比值非$1$. $n\geq 9$时,
$$
|T_k^{(n)}|/|T^{(n)}_1|=\dfrac{(n-2)\cdots(n-2k+1)}{2^{k-1}k!}=\\\dfrac{n-3}{k}\cdot\dfrac{n-2}{2(k-1)}\cdot\dfrac{n-4}{2(k-2)}\cdots\dfrac{n-k-1}{2}\cdot (n-k-2)!>1
$$
故$n\neq 2,6$时, 有
$$
\mbox{aut}(S_n)=\mbox{inn}(S_n)\cong S_n\cong\mbox{aut}(A_n)=\mbox{inn}(A_n)
$$
$S_6$实则存在外自同构, 且$[\mbox{aut}(S_6):\mbox{inn}(S_6)]=2$, 今以篇幅故暂不展示. 外自同构之展现方式诸多. 图论角度上, 构造双射$GQ(2,2)\overset{\sim}{\to} K_6$即可; 代数角度上, 考虑$PGL_2(\mathbb F_5)\curvearrowright P^1(\mathbb F_5)$即可.

附上Generalized Quadrangle graph $GQ(2,2)$, 该图将$S_6$中的三对两两不交对换之积作为元素, 对应$GQ(2,2)$中的点. 两点相邻若且仅若有一对对换相同.

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The Outer Automorphism of S6('PRIMES' stands for Program for Research In Mathematics, Engineering and Science for high school students, see here) $G$ is a group, $Z(G)$ is the center of $G$, $\text{Inn}(G)$ is the inner automorphism group of $G$, $\text{Aut}(G)$ is the automorphism group of $G$. As a consequence of the first isomorphism theorem, we have

Theorem

The inner automorphisms $\text{Inn}(G)$ form a normal subgroup of $\text{Aut}(G)$ and $G / Z(G) \cong \text{Inn}(G).$

Definition

A group $G$ is complete if $Z(G)$ is trivial and every automorphism of $G$ is an inner automorphism.

$G$ complete ⇒ $\operatorname{Aut}(G) ≅ G$

Theorem

$S_n$ is complete for $n≠ 2, 6$.

The center of $S_2$ is itself. $S_6$ is a genuine exception:

Theorem (Hölder)

There exists exactly 1 outer automorphism of $S_6$ (up to composition with an inner automorphism), so that $|\operatorname{Aut}(S_6)|= 2×6! = 1440$.

Definition

A transposition is a permutation $π ∈ S_n$ that fixes exactly $n − 2$ elements (and flips the remaining two elements).

Lemma

An automorphism of $S_n$ preserves transpositions if and only if it is an inner automorphism.

Proof Overview: Completeness of $S_n$ for $n≠2, 6$

Let $T_{k}$ be the conjugacy class in $S_{n}$ consisting of products of $k$ disjoint transpositions.

• A permutation $\pi$ is an involution if and only if it lies in some $T_{k}$.
• If $f \in \operatorname{Aut}\left(S_{n}\right)$, then $f\left(T_{1}\right)=T_{k}$ for some $k$.
• It suffices to show $\left|T_{k}\right| \neq\left|T_{1}\right|$ for $k \neq 1$.
• This is true for $n \neq 6$.
• For $n=6$, it turns out that $\left|T_{1}\right|=\left|T_{3}\right|$ is the only exception.

Key Step: Construct a 120-element subgroup $H$ of $S_{6}$ that acts transitively on $\{1,2,3,4,5,6\}$.

• This subgroup cannot be $S_{5}$ or any of its conjugates (none are transitive subgroups).
• Consider the action of $S_{6}$ on the 6-element coset space $S_{6} / H$.
• Let $f$ the corresponding mapping from $S_{6}$ to $S_{6}$.
• Note that $f: H \mapsto S_{5}$. ( $H$ fixes coset consisting of $H$ and permutes all other cosets. $H$ has order 120 , the same as $S_{5}$.)
• $H$ (the preimage of $S_{5}$ in $f$ ) is transitive.
• The preimage of $S_{5}$ is not conjugate to $S_{5}$.
• $f$ cannot be inner.

Methods of Construction

$\bbox[#262686,2pt,color:#fff]{\small1}$ Simply 3-transitive action of $PGL_2(\Bbb F_5)$ on the six-element set $P^1(\Bbb F_5)$ $\bbox[#262686,2pt,color:#fff]{\small2}$ Transitive action of $S_5$ on its six 5-Sylow subgroups

Construction 1: Properties of $PGL_2(K)$

Let $K$ be a field.

Definition

$G L_{2}(K)$ is the set of $2 \times 2$ invertible matrices, whose elements are in the field $K$. $P G L_{2}(K)$ is the quotient of the group $G L_{2}(K)$ by the scalar matrices $K^{\times}$ (nonzero elements of $K$ ). $\boldsymbol{P}^{1}(K)$ is the set of one-dimensional vector spaces (lines) in $K^{2}$.

• There is a natural action of $G L_{2}(K)$ on $\boldsymbol{P}^{\mathbf{1}}(K)$
• Permutation of the lines through the origin in $K^{2}$
• Matrices of the form$$ \left(\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right) \text {, } $$where $a \in K$, fix lines, so we have an action of $P G L_{2}(K)$ on $\boldsymbol{P}^{1}(K)$

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groupprops.subwiki.org/wiki/Transposition-pre … etric_group_is_inner

Given: A finite set $S$. $G$ is the symmetric group on $S$. $\sigma$ is an automorphism of $G$ that sends transpositions to transpositions.

To prove: $\sigma$ is inner.

Proof: By fact (1), it suffices to find $g \in G$ such that conjugation by $g$ agrees with $\sigma$ on the set of transpositions. Further, we can assume that $S$ has at least three elements (The statement is obviously true for $S$ having zero, one, or two elements). We now describe how the permutation $g$ can be constructed explicitly.

Also, we use fact (2) at many steps, without explicitly acknowledging it.

1. Pick distinct elements $a,b,c \in S$. Then, both $\sigma((a,b))$ and $\sigma((a,c))$ are transpositions. We claim that these transpositions have exactly one element in common: If $\sigma((a,b))$ and $\sigma((a,c))$ have no element in common, then they commute, and hence, $(a,b)$ commutes with $(a,c)$, which is not true. Thus, $\sigma((a,b))$ and $\sigma((a,c))$ have exactly one element in commmon.
2. Pick distinct elements $a,b,c \in S$. Then, there are elements $a',b',c' \in S$ such that $\sigma((a,b)) = (a',b')$, $\sigma((b,c)) = (b',c')$ and $\sigma((c,a)) = (c',a')$: This follows directly from the previous step.
3. For any element $a \in S$, there is a unique element $g(a)$ that is involved in the transposition $\sigma((a,b))$ for every $b \ne a$: For any $b \ne c$ distinct from $a$, there exist $a',b',c'$ as described in the previous step. We now show that $\sigma((a,d))$ is also a transposition involving $a'$.
   • Since $(a,d)$ commutes with $(b,c)$, $\sigma((a,d))$ commutes with $(b',c')$. In particular, it cannot be a transposition involving $b'$ or $c'$.
   • Since $(a,b)$ does not commute with $(a,d)$, $\sigma((a,b))$ does not commute with $\sigma((a,d))$, so they must have an element in common. Thus, either $a'$ or $b'$ is involved in $\sigma((a,d))$.
   • These together force $a'$ to be involved in $\sigma((a,d))$.
   • We can thus define $g(a)$ as the unique element involved in $\sigma((a,x))$ for every transposition involving $a$.
4. The previous step yields a map $g:S \to S$ such that for any transposition $(a,b)$, $\sigma((a,b))$ is a transposition involving both $g(a)$ and $g(b)$. Step (2) makes it clear that $g(a) \ne g(b)$, so $\sigma((a,b)) = (g(a),g(b))$. Thus, $\sigma$ is induced by conjugation by the permutation $g$.