$n$元交错群是单群($n = 3$ 或 $n ≥ 5$)

hbghlyj

people.reed.edu/~jerry/332/08ansim.pdf

1. Some Properties Of The Symmetric Group

Let $n \in \mathbb{Z}_{>0}$ be a a positive integer, and let $S_{n}$ denote the group of permutations of $n$ letters.

The fact that every element of $S_{n}$ can be written (nonuniquely) as a product of transpositions (2-cycles) is self-evident.

The fact that every element of $S_{n}$ can be written as a product of an even number of transpositions, or as an odd number of transpositions, but not both is not at all self-evident. See Gallian for a short elementary proof. However, granting the fact, the identity (composing from right to left, as seems to be the practice in introductory texts)
$$
(123 \cdots k)=(1 k) \cdots(13)(12)
$$
shows that that a $k$-cycle is even if and only if $k$ is odd.
The subgroup of even permutations in $S_{n}$ is the alternating group $A_{n}$. Its name comes from the fact that if its elements are viewed as even permutations of the symbols $r_{1}, \ldots, r_{n}$ then they preserve the polynomial expression
$$
\prod_{i<j}\left(r_{i}-r_{j}\right)
$$
whereas the odd permutations negate the expression.
An element of $S_{n}$ can also be written
$$
\tau=\left(\begin{array}{lll}
x_{1} & \cdots & x_{n} \\
y_{1} & \cdots & y_{n}
\end{array}\right)
$$
where $\left\{x_{1}, \ldots, x_{n}\right\}=\left\{y_{1}, \ldots, y_{n}\right\}=\{1, \ldots, n\}$ and typically $x_{1}=1, \ldots, x_{n}=n$ or similarly for the $y$-values. And an element of $S_{n}$ can be written as a product of disjoint cycles,$$
\sigma=\left(a_{1} \cdots a_{k}\right)\left(b_{1} \cdots b_{\ell}\right) \cdots, \quad\left\{a_{i}\right\} \sqcup\left\{b_{j}\right\} \sqcup \cdots=\{1, \ldots, n\} .
$$
Clearly the order of $\sigma$ is the least common multiple of the lengths of its cycles.
Proposition 1.1. The conjugation-action of $S_{n}$ on itself restricts to the subset of elements having each particular cycle-structure, and the action is transitive on each such subset.
Proof. If
$$
\sigma=\left(a_{1} \cdots a_{k}\right)\left(b_{1} \cdots b_{\ell}\right) \cdots
$$
and
$$
\tau=\left(\begin{array}{lllllll}
\alpha_{1} & \cdots & \alpha_{k} & \beta_{1} & \cdots & \beta_{\ell} & \cdots \\
a_{1} & \cdots & a_{k} & b_{1} & \cdots & b_{\ell} & \cdots
\end{array}\right)
$$
then (composing right-to-left)
$$
\tau^{-1} \sigma \tau=\left(\alpha_{1} \cdots \alpha_{k}\right)\left(\beta_{1} \cdots \beta_{\ell}\right) \cdots .
$$

2. Simplicity Of The Alternating Group

Let $G$ be a group. Recall that a subgroup of $G$ is normal if its normalizer is all of $G$, or equivalently, if it is the kernel of a homomorphism out of $G$. Every subgroup of an abelian group is normal.
Since the alternating group $A_{n}$ is the kernel of the homomorphism
$$
S_{n} \longrightarrow \mathbb{Z} / 2 \mathbb{Z}, \quad \sigma \longmapsto \text { parity of } \sigma,
$$
it follows that $A_{n}$ is a normal subgroup of $S_{n}$. We wonder about normal subgroups of $A_{n}$ in turn. For $n=4$, the Klein four-group
$$
V=\{e,(12)(34),(13)(24),(14)(23)\}
$$
is normal in $A_{4}$ because conjugation preserves cycle-structure. However, the Klein four-group turns out to be the only example.

Definition 2.1. A group $G$ is simple if it has no nontrivial normal subgroups.
As discussed in Gallian, the problem of classifying all finite simple groups has been a long, titanic mathematical endeavor. Our aim here is more modest.
Proposition 2.2. The alternating group $A_{n}$ is simple for $n \neq 4$.
Proof. Since $A_{1} \approx A_{2} \approx\{\cdot\}$ and $A_{3} \approx \mathbb{Z} / 3 \mathbb{Z}$, we may take $n \geq 5$. The equalities
$$
(a b)(c d)=(a c b)(a c d), \quad(a b)(b c)=(a b c)
$$
show that $A_{n}$ is generated by the set of its 3-cycles. Also, beyond Proposition 1.1, all 3-cycles are conjugate in $A_{n}$. Indeed, each $\left(a_{1} a_{2} a_{3}\right)$ takes the form $\tau^{-1}(123) \tau$ for some $\tau \in S_{n}$; if $\tau$ is even then we are done, and if $\tau$ is odd then also $\left(a_{1} a_{2} a_{3}\right)=$ $\tau^{-1}(45)(123)(45) \tau$ with $(45) \tau$ even. This argument uses the condition $n \geq 5$, but it does not assume that $(45)$ and $\tau$ are disjoint. The upshot is that given a normal subgroup $N \neq\{e\}$ of $A_{n}$ where $n \geq 5$, we need only to show that $N$ contains a 3-cycle. Now reason as follows.
• If $N$ contains an element $\sigma=(1 \cdots r) \tau$ where $r \geq 4$, then let $\delta=(123)$ and compute
$$
\sigma^{-1} \cdot \delta^{-1} \sigma \delta=(r \cdots 1)(132)(1 \cdots r)(123)=(23 r)
$$
Thus $N$ contains a 3-cycle as desired. The same argument holds for any $\sigma=\left(a_{1} \cdots a_{r}\right) \tau$, using $\delta=\left(a_{1} a_{2} a_{3}\right)$. We use this same idea for notational ease without comment from now on.
• If $N$ contains an element $\sigma=(123)(456) \tau$, then let $\delta=(124)$ and compute that we are done by the previous bullet,
$$
\sigma^{-1} \cdot \delta^{-1} \sigma \delta=(132)(465)(142)(123)(456)(124)=(12436)
$$
• If $N$ contains an element $\sigma=(123) \tau$ where $\tau$ is a product of 2-cycles, then $\sigma^{2}=(132)$ and we are done.
• If $N$ contains an element $\sigma=(12)(34) \tau$ where $\tau$ is a product of 2-cycles, then let $\delta=(123)$ so that $\sigma^{-1} \delta^{-1} \sigma \delta=(14)(23)$. Next (using $n \geq 5$ here),
$$
(152)(14)(23)(125)=(13)(45)
$$
and then we are done by the first bullet,
$$
(14)(23)(13)(45)=(12345) \text {. }
$$
This completes the proof.$□$

hbghlyj

总结一下: 设$N⊲A_n,n≥5$ 分类讨论⇒$N$包含一个3-cycle $N$对conjugation封闭,而且(Proposition 1.1.)conjugation在3-cycle的集合上的作用是transitive的⇒$N$包含所有3-cycle 3-cycles可以生成整个$A_n$⇒$N=A_n$

hbghlyj

GroupProps上的证明:
The proof is by induction on $n$.
Base case

The case $n = 5$ is dealt with separately, by a direct argument. For full proof, refer: A5 is simple

Induction step

We prove that for $n \ge 5$, if $A_n$ is simple, then $A_{n+1}$ is simple.

Given: $A_{n+1}$ is the group of even permutations on $\{ 1,2,3,\dots,n+1 \}$. $N$ is a normal subgroup of $A_{n+1}$.

To prove: $N = A_{n+1}$ or $N$ is trivial.

Proof: Let $H_i$ denote the subgroup of $A_{n+1}$ that stabilizes the letter $i$. Then, each $H_i$ consists of the even permutations on $n$ letters (the letters excluding $i$) and is hence isomorphic to $A_n$. Thus, each $H_i$ is simple.

Now, since normality satisfies transfer condition, $N \cap H_i$ is normal in $H_i$ for every $i$. By simplicity of $H_i$, either $N$ contains $H_i$, or $N$ intersects $H_i$ trivially.

Suppose there exists $i$ for which $N$ contains $H_i$. Then, by fact (3) stated above, $H_i$ is contranormal inside $A_{n+1}$, i.e., its normal closure is $A_{n+1}$. Since $N$ is normal, this forces $N = A_{n+1}$, and we are done.

Otherwise, $N \cap H_i$ is trivial for every $i$. Thus, no nontrivial element of $N$ fixes any letter. Let's use this to show that $N$ can have no nontrivial elements.

Suppose $\sigma \in N$ is nontrivial. Then, first observe that in the cycle decomposition of $\sigma$, every element must be in a cycle of the same length $k$ (otherwise, some power of $\sigma$ would fix a letter). Thus, $\sigma$ has $r$ cycles each of size $k$, where $kr = n + 1$.

Now, if $n \ge 5$, then $n + 1 \ge 6$. Choose an $i$ and a double transposition $\tau \in A_{n+1}$ such that $\tau$ fixes both $i$ and $\sigma(i)$, but such that $\tau$ does not commute with $\sigma$ (this is possible because $n+1 \ge 6$). Then, $\sigma^{-1}\tau\sigma\tau^{-1} \in N$ is a nontrivial element of $N$ fixing both $i$ and $\sigma(i)$, giving the required contradiction.