对称群

1.3 SYMMETRIC GROUPS

1.3 对称群

Let $\Omega$ be any nonempty set and let ${S}_{\Omega }$ be the set of all bijections from $\Omega$ to itself (i.e., the set of all permutations of $\Omega$ ). The set ${S}_{\Omega }$ is a group under function composition: $\circ$ . Note that $\circ$ is a binary operation on ${S}_{\Omega }$ since if $\sigma : \Omega \rightarrow \Omega$ and $\tau : \Omega \rightarrow \Omega$ are both bijections,then $\sigma \circ \tau$ is also a bijection from $\Omega$ to $\Omega$ . Since function composition is associative in general, $\circ$ is associative. The identity of ${S}_{\Omega }$ is the permutation 1 defined by $1\left( a\right) = a$ ,for all $a \in \Omega$ . For every permutation $\sigma$ there is a (2-sided) inverse function, ${\sigma }^{-1} : \Omega \rightarrow \Omega$ satisfying $\sigma \circ {\sigma }^{-1} = {\sigma }^{-1} \circ \sigma = 1$ . Thus,all the group axioms hold for $\left( {{S}_{\Omega }, \circ }\right)$ . This group is called the symmetric group on the set $\Omega$ . It is important to recognize that the elements of ${S}_{\Omega }$ are the permutations of $\Omega$ ,not the elements of $\Omega$ itself.

设 $\Omega$ 为任何非空集合，令 ${S}_{\Omega }$ 为所有从 $\Omega$ 到其自身的双射（即 $\Omega$ 的所有排列）的集合。集合 ${S}_{\Omega }$ 在函数复合下构成一个群：$\circ$。注意 $\circ$ 是 ${S}_{\Omega }$ 上的一个二元运算，因为如果 $\sigma : \Omega \rightarrow \Omega$ 和 $\tau : \Omega \rightarrow \Omega$ 都是双射，那么 $\sigma \circ \tau$ 也是一个从 $\Omega$ 到 $\Omega$ 的双射。由于函数复合通常具有结合性，$\circ$ 也是结合的。${S}_{\Omega }$ 的单位元是排列 1，定义为 $1\left( a\right) = a$，对于所有 $a \in \Omega$。对于每一个排列 $\sigma$，都有一个（双边）逆函数 ${\sigma }^{-1} : \Omega \rightarrow \Omega$，满足 $\sigma \circ {\sigma }^{-1} = {\sigma }^{-1} \circ \sigma = 1$。因此，所有群公理对于 $\left( {{S}_{\Omega }, \circ }\right)$ 都成立。这个群被称为集合 $\Omega$ 上的对称群。重要的是要认识到 ${S}_{\Omega }$ 的元素是 $\Omega$ 的排列，而不是 $\Omega$ 本身的元素。

In the special case when $\Omega = \{ 1,2,3,\ldots ,n\}$ ,the symmetric group on $\Omega$ is denoted ${S}_{n}$ ,the symmetric group of degree $n{.}^{1}$ The group ${S}_{n}$ will play an important role throughout the text both as a group of considerable interest in its own right and as a means of illustrating and motivating the general theory.

在 $\Omega = \{ 1,2,3,\ldots ,n\}$ 的特殊情况下，$\Omega$ 上的对称群记作 ${S}_{n}$，即 $n{.}^{1}$ 阶的对称群。群 ${S}_{n}$ 将在本文中扮演重要角色，既作为一个自身具有相当大兴趣的群，也作为说明和激发一般理论的手段。

First we show that the order of ${S}_{n}$ is $n$ !. The permutations of $\{ 1,2,3,\ldots ,n\}$ are precisely the injective functions of this set to itself because it is finite (Proposition 0.1) and we can count the number of injective functions. An injective function $\sigma$ can send the number 1 to any of the $n$ elements of $\{ 1,2,3,\ldots ,n\} ;\sigma \left( 2\right)$ can then be any one of the elements of this set except $\sigma \left( 1\right)$ (so there are $n - 1$ choices for $\sigma \left( 2\right)$ ); $\sigma \left( 3\right)$ can be any element except $\sigma \left( 1\right)$ or $\sigma \left( 2\right)$ (so there are $n - 2$ choices for $\sigma \left( 3\right)$ ),and so on. Thus there are precisely $n \cdot \left( {n - 1}\right) \cdot \left( {n - 2}\right) \ldots 2 \cdot 1 = n$ ! possible injective functions from $\{ 1,2,3,\ldots ,n\}$ to itself. Hence there are precisely $n$ ! permutations of $\{ 1,2,3,\ldots ,n\}$ so there are precisely $n$ ! elements in ${S}_{n}$ .

首先我们证明 ${S}_{n}$ 的阶是 $n$ !。因为它是有限的（命题 0.1），我们可以计算这个集合到自身的注入函数的数量。$\sigma$ 的一个注入函数可以将数字 1 映射到 $n$ 元素中的任何一个，$\{ 1,2,3,\ldots ,n\} ;\sigma \left( 2\right)$ 可以是这个集合中除了 $\sigma \left( 1\right)$ 之外的任何一个元素（因此对于 $\sigma \left( 2\right)$ 有 $n - 1$ 种选择）；$\sigma \left( 3\right)$ 可以是除了 $\sigma \left( 1\right)$ 或 $\sigma \left( 2\right)$ 之外的任何一个元素（因此对于 $n - 2$ 有 $n - 2$ 种选择），依此类推。因此，从 $\{ 1,2,3,\ldots ,n\}$ 到其自身的确切 $n \cdot \left( {n - 1}\right) \cdot \left( {n - 2}\right) \ldots 2 \cdot 1 = n$ ! 个可能的注入函数。因此，确切地有 $n$ ! 个 $\{ 1,2,3,\ldots ,n\}$ 的排列，所以 ${S}_{n}$ 中确切地有 $n$ ! 个元素。

We now describe an efficient notation for writing elements $\sigma$ of ${S}_{n}$ which we shall use throughout the text and which is called the cycle decomposition.

接下来我们描述一种高效的表示 $\sigma$ 中元素的方法，我们将在全文中使用这种表示法，它被称为循环分解。

A cycle is a string of integers which represents the element of ${S}_{n}$ which cyclically permutes these integers (and fixes all other integers). The cycle $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ is the permutation which sends ${a}_{i}$ to ${a}_{i + 1},1 \leq i \leq m - 1$ and sends ${a}_{m}$ to ${a}_{1}$ . For example (213) is the permutation which maps 2 to 1,1 to 3 and 3 to 2. In general, for each $\sigma \in {S}_{n}$ the numbers from 1 to $n$ will be rearranged and grouped into $k$ cycles of the form

循环是一串整数，它代表 ${S}_{n}$ 中的元素，这个元素将这些整数循环置换（并固定所有其他整数）。循环 $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ 是这样一个置换，它将 ${a}_{i}$ 映射到 ${a}_{i + 1},1 \leq i \leq m - 1$，并将 ${a}_{m}$ 映射到 ${a}_{1}$。例如，（213）是这样的置换，它将 2 映射到 1，1 映射到 3，3 映射到 2。一般来说，对于每个 $\sigma \in {S}_{n}$，从 1 到 $n$ 的数字将被重新排列并分组为形式为 $k$ 的循环。

from which the action of $\sigma$ on any number from 1 to $n$ can easily be read,as follows. For any $x \in \{ 1,2,3,\ldots ,n\}$ first locate $x$ in the above expression. If $x$ is not followed immediately by a right parenthesis ( i.e., $x$ is not at the right end of one of the $k$ cycles), then $\sigma \left( x\right)$ is the integer appearing immediately to the right of $x$ . If $x$ is followed by a right parenthesis,then $\sigma \left( x\right)$ is the number which is at the start of the cycle ending with $x$ (i.e.,if $x = {a}_{{m}_{i}}$ ,for some $i$ ,then $\sigma \left( x\right) = {a}_{{m}_{i - 1} + 1}$ (where ${m}_{0}$ is taken to be 0 )). We can represent this description of $\sigma$ by

从中可以很容易地读出 $\sigma$ 对从 1 到 $n$ 的任何数字的作用，如下所示。对于任何 $x \in \{ 1,2,3,\ldots ,n\}$，首先在上面的表达式中找到 $x$。如果 $x$ 后面没有紧跟右括号（即 $x$ 不是 $k$ 循环之一的右端），那么 $\sigma \left( x\right)$ 是紧在 $x$ 右边的整数。如果 $x$ 后面跟有右括号，那么 $\sigma \left( x\right)$ 是以 $x$ 结尾的循环开始的那个数（即，如果 $x = {a}_{{m}_{i}}$ 对于某个 $i$ 成立，那么 $\sigma \left( x\right) = {a}_{{m}_{i - 1} + 1}$（其中 ${m}_{0}$ 视为 0））。我们可以用以下方式表示对 $\sigma$ 的这种描述：

---

${}^{1}$ We shall see in Section 6 that the structure of ${S}_{\Omega }$ depends only on the cardinality of $\Omega$ ,not on the particular elements of $\Omega$ itself,so if $\Omega$ is any finite set with $n$ elements,then ${S}_{\Omega }$ "looks like" ${S}_{n}$ .

${}^{1}$ 我们将在第6节中看到，${S}_{\Omega }$ 的结构仅取决于 $\Omega$ 的基数，而不是 $\Omega$ 本身的特定元素，所以如果 $\Omega$ 是具有 $n$ 元素的任何有限集，那么 ${S}_{\Omega }$ "看起来像" ${S}_{n}$。

---

The product of all the cycles is called the cycle decomposition of $\sigma$ .

所有循环的乘积称为 $\sigma$ 的循环分解。

We now give an algorithm for computing the cycle decomposition of an element $\sigma$ of ${S}_{n}$ and work through the algorithm with a specific permutation. We defer the proof of this algorithm and full analysis of the uniqueness aspects of the cycle decomposition until Chapter 4.

我们现在给出一个计算元素 $\sigma$ 在 ${S}_{n}$ 中的循环分解的算法，并用一个特定的排列来演示这个算法。我们将这个算法的证明以及循环分解唯一性方面的完整分析推迟到第4章。

Let $n = {13}$ and let $\sigma \in {S}_{13}$ be defined by

设 $n = {13}$ 并且定义 $\sigma \in {S}_{13}$ 为

Cycle Decomposition Algorithm

循环分解算法

Method	Example
To start a new cycle pick the smallest element of $\{ 1,\;2,\ldots .\;n\}$ which has not yet appeared in a previous cycle - call it $a$ (if you are just starting, $a = 1$ ); begin the new cycle: (a	(1
Read off $\sigma \left( a\right)$ from the given description of $\sigma$ - call it $b$ . If $b = a$ ,close the cycle with a right parenthesis (without writing $b$ down); this completes a cycle - return to step 1. If $b \neq a$ , write $b$ next to $a$ in this cycle: ( ${ab}$	$\sigma \left( 1\right) = \frac{1}{2} = b,\frac{1}{2} \neq 1$ so write: (112
Read off $\sigma \left( b\right)$ from the given description of $\sigma$ - call it $c$ . If $c = a$ ,close the cycle with a right parenthesis to complete the cycle - return to step 1. If $c \neq a$ ,write $c$ next to $b$ in this cycle: ( ${abc}$ Repeat this step using the number $c$ as the new value for $b$ until the cycle closes.	$\sigma \left( {12}\right) = 8,8 \neq 1$ so continue the cycle as: (1128

Naturally this process stops when all the numbers from $\{ 1,2,\ldots ,n\}$ have appeared in some cycle. For the particular $\sigma$ in the example this gives

显然，当 $\{ 1,2,\ldots ,n\}$ 中的所有数字都出现在某个循环中时，这个过程自然停止。对于示例中的特定 $\sigma$，这给出

The length of a cycle is the number of integers which appear in it. A cycle of length $t$ is called a $t$ -cycle. Two cycles are called ${disjoint}$ if they have no numbers in common. Thus the element $\sigma$ above is the product of 5 (pairwise) disjoint cycles: a 5-cycle,a 2-cycle, a 1-cycle, a 3-cycle, and another 2-cycle.

循环的长度是出现在其中的整数的数量。长度为 $t$ 的循环被称为 $t$ -循环。如果两个循环没有共同的数字，则称这两个循环为 ${disjoint}$ 。因此上面的元素 $\sigma$ 是5个（两两）不相交循环的乘积：一个5-循环，一个2-循环，一个1-循环，一个3-循环，以及另一个2-循环。

Henceforth we adopt the convention that 1-cycles will not be written. Thus if some integer, $i$ ,does not appear in the cycle decomposition of a permutation $\tau$ it is understood that $\tau \left( i\right) = i$ ,i.e.,that $\tau$ fixes $i$ . The identity permutation of ${S}_{n}$ has cycle decomposition $\left( 1\right) \left( 2\right) \ldots \left( n\right)$ and will be written simply as 1 . Hence the final step of the algorithm is:

从此我们约定1-循环不写出。因此，如果某个整数 $i$ 不出现在排列 $\tau$ 的循环分解中，则理解为 $\tau \left( i\right) = i$ ，即 $\tau$ 固定 $i$ 。${S}_{n}$ 的恒等排列的循环分解为 $\left( 1\right) \left( 2\right) \ldots \left( n\right)$ ，并且将简单地写作1。因此，算法的最后一步是：

Cycle Decomposition Algorithm (cont.)

循环分解算法（续）

Final Step: Remove all cycles of length 1

The cycle decomposition for the particular $\sigma$ in the example is therefore

因此，示例中特定 $\sigma$ 的循环分解为

This convention has the advantage that the cycle decomposition of an element $\tau$ of ${S}_{n}$ is also the cycle decomposition of the permutation in ${S}_{m}$ for $m \geq n$ which acts as $\tau$ on $\{ 1,2,3,\ldots ,n\}$ and fixes each element of $\{ n + 1,n + 2,\ldots ,m\}$ . Thus,for example, (12)is the permutation which interchanges1and2and fixes all larger integers whether viewed in ${S}_{2},{S}_{3}$ or ${S}_{4}$ ,etc.

这个约定的优点是，元素 $\tau$ 的 ${S}_{n}$ 的循环分解也是排列在 ${S}_{m}$ 中对于 $m \geq n$ 的循环分解，它作为 $\tau$ 在 $\{ 1,2,3,\ldots ,n\}$ 上作用并固定 $\{ n + 1,n + 2,\ldots ,m\}$ 的每个元素。因此，例如，（12）是交换1和2的排列，并且在 ${S}_{2},{S}_{3}$ 或 ${S}_{4}$ 中固定所有较大的整数，等等。

As another example, the 6 elements of ${S}_{3}$ have the following cycle decompositions:

另一个例子是，${S}_{3}$ 的6个元素有以下循环分解：

The group ${S}_{3}$

群 ${S}_{3}$

Values of ${\sigma }_{i}$	Cycle Decomposition of ${\sigma }_{i}$
${\sigma }_{1}\left( 1\right) = 1,{\sigma }_{1}\left( 2\right) = 2,{\sigma }_{1}\left( 3\right) = 3$	1
${\sigma }_{2}\left( 1\right) = 1,{\sigma }_{2}\left( 2\right) = 3,{\sigma }_{2}\left( 3\right) = 2$	(23)
${\sigma }_{3}\left( 1\right) = 3,{\sigma }_{3}\left( 2\right) = 2,{\sigma }_{3}\left( 3\right) = 1$	(13)
${\sigma }_{4}\left( 1\right) = 2,{\sigma }_{4}\left( 2\right) = 1,{\sigma }_{4}\left( 3\right) = 3$	(12)
${\sigma }_{5}\left( 1\right) = 2,{\sigma }_{5}\left( 2\right) = 3,{\sigma }_{5}\left( 3\right) = 1$	(123)

For any $\sigma \in {S}_{n}$ ,the cycle decomposition of ${\sigma }^{-1}$ is obtained by writing the numbers in each cycle of the cycle decomposition of $\sigma$ in reverse order. For example,if $\sigma = \left( {1128104}\right) \left( {213}\right) \left( {5117}\right) \left( {69}\right)$ is the element of ${S}_{13}$ described before then

对于任何 $\sigma \in {S}_{n}$ ，${\sigma }^{-1}$ 的循环分解是通过将 $\sigma$ 的循环分解中的每个循环中的数字按相反顺序写出获得的。例如，如果 $\sigma = \left( {1128104}\right) \left( {213}\right) \left( {5117}\right) \left( {69}\right)$ 是之前描述的 ${S}_{13}$ 的元素，那么

Computing products in ${S}_{n}$ is straightforward,keeping in mind that when computing $\sigma \circ \tau$ in ${S}_{n}$ one reads the permutations from right to left. One simply "follows" the elements under the successive permutations. For example,in the product $\left( {123}\right) \circ$ $\left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \end{array}\right)$ the number 1 is sent to 2 by the first permutation,then 2 is sent to 3 by the second permutation, hence the composite maps 1 to 3. To compute the cycle decomposition of the product we need next to see what happens to 3 . It is sent first to 4 , then 4 is fixed, so 3 is mapped to 4 by the composite map. Similarly, 4 is first mapped to 3 then 3 is mapped to 1 , completing this cycle in the product: (134). Finally, 2 is sent to 1,then 1 is sent to 2 so 2 is fixed by this product and so $\left( {1 \geq 3}\right) \circ \left( {1 \geq 2}\right) \left( {3 \leq 4}\right) = \left( {1 \geq 4}\right)$ is the cycle decomposition of the product.

在 ${S}_{n}$ 中计算乘积是直接的，考虑到在 ${S}_{n}$ 中计算 $\sigma \circ \tau$ 时是从右到左读取排列。只需“跟随”连续排列下的元素。例如，在乘积 $\left( {123}\right) \circ$ $\left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \end{array}\right)$ 中，数字 1 通过第一个排列被送到 2，然后 2 通过第二个排列被送到 3，因此复合映射将 1 映射到 3。为了计算乘积的循环分解，我们需要接下来观察 3 发生了什么。它首先被送到 4，然后 4 被固定，所以 3 通过复合映射被映射到 4。类似地，4 首先被映射到 3，然后 3 被映射到 1，完成了乘积中的这个循环：(134)。最后，2 被送到 1，然后 1 被送到 2，所以 2 被这个乘积固定，因此 $\left( {1 \geq 3}\right) \circ \left( {1 \geq 2}\right) \left( {3 \leq 4}\right) = \left( {1 \geq 4}\right)$ 是乘积的循环分解。

As additional examples,

作为额外的例子，

In particular this shows that

特别是这表明

Each cycle $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ in a cycle decomposition can be viewed as the permutation which cyclically permutes ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ and fixes all other integers. Since disjoint cycles permute numbers which lie in disjoint sets it follows that

在循环分解中的每个循环 $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ 可以被视为循环排列 ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ 并固定所有其他整数的排列。由于不相交的循环排列的是位于不相交集合中的数字，因此

disjoint cycles commute.

不相交的循环可以交换。

Thus rearranging the cycles in any product of disjoint cycles (in particular, in a cycle decomposition) does not change the permutation.

因此，重新排列任何不相交循环的乘积中的循环（特别是在循环分解中）不会改变排列。

Also,since a given cycle, $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ ,permutes $\left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{m}}\right\}$ cyclically,the numbers in the cycle itself can be cyclically permuted without altering the permutation, i.e.,

此外，由于给定的循环 $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ 循环排列 $\left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{m}}\right\}$，循环中的数字本身可以循环排列而不改变排列，即，

Thus,for instance, $\left( {12}\right) = \left( {21}\right)$ and $\left( {1234}\right) = \left( {3412}\right)$ . By convention,the smallest number appearing in the cycle is usually written first.

因此，例如，$\left( {12}\right) = \left( {21}\right)$ 和 $\left( {1234}\right) = \left( {3412}\right)$。按照惯例，循环中出现的最小数字通常首先写出。

One must exercise some care working with cycles since a permutation may be written in many ways as an arbitrary product of cycles. For instance,in ${S}_{3},\left( \begin{array}{lll} 1 & 2 & 3 \end{array}\right) =$ $\left( {1\;2}\right) \left( {2\;3}\right) = \left( {1\;3}\right) \left( {1\;3\;2}\right) \left( {1\;3}\right) \;$ etc. But,(as we shall prove) the cycle decomposition of each permutation is the unique way of expressing a permutation as a product of disjoint cycles (up to rearranging its cycles and cyclically permuting the numbers within each cycle). Reducing an arbitrary product of cycles to a product of disjoint cycles allows us to determine at a glance whether or not two permutations are the same. Another advantage to this notation is that it is an exercise (outlined below) to prove that the order of a permutation is the l.c.m. of the lengths of the cycles in its cycle decomposition.

在处理循环时必须小心，因为一个置换可以以多种方式写成循环的任意乘积。例如，在 ${S}_{3},\left( \begin{array}{lll} 1 & 2 & 3 \end{array}\right) =$ $\left( {1\;2}\right) \left( {2\;3}\right) = \left( {1\;3}\right) \left( {1\;3\;2}\right) \left( {1\;3}\right) \;$ 等等中。但是（正如我们将证明的），每个置换的循环分解是将置换表示为不相交循环乘积的唯一方式（循环的重新排列和每个循环内数字的循环置换除外）。将任意的循环乘积简化为不相交循环的乘积，可以让我们一眼就确定两个置换是否相同。这种表示法的另一个优点是，下面的练习是证明置换的阶是其循环分解中各循环长度的最小公倍数。

EXERCISES

练习

1. Let $\sigma$ be the permutation

2. 设 $\sigma$ 为置换

and let $\tau$ be the permutation

并设 $\tau$ 为置换

Find the cycle decompositions of each of the following permutations: $\sigma ,\tau ,{\sigma }^{2},{\sigma \tau },{\tau \sigma }$ , and ${\tau }^{2}\sigma$ .

找出以下置换的循环分解：$\sigma ,\tau ,{\sigma }^{2},{\sigma \tau },{\tau \sigma }$ 和 ${\tau }^{2}\sigma$。

2. Let $\sigma$ be the permutation

3. 设 $\sigma$ 为置换

and let $\tau$ be the permutation

并设 $\tau$ 为置换

Find the cycle decompositions of the following permutations: $\sigma ,\tau ,{\sigma }^{2},{\sigma \tau },{\tau \sigma }$ ,and ${\tau }^{2}\sigma$ .

找出以下置换的循环分解：$\sigma ,\tau ,{\sigma }^{2},{\sigma \tau },{\tau \sigma }$ 和 ${\tau }^{2}\sigma$。

3. For each of the permutations whose cycle decompositions were computed in the preceding two exercises compute its order.

4. 对于前两个练习中计算出的循环分解的每个置换，计算其阶。

5. Compute the order of each of the elements in the following groups: (a) ${S}_{3}$ (b) ${S}_{4}$ .

6. 计算以下群中每个元素的阶：(a) ${S}_{3}$ (b) ${S}_{4}$。

7. Find the order of $\left( \begin{array}{llllll} 1 & 1 & 2 & 8 & {10} & 4 \end{array}\right) \left( \begin{array}{lll} 2 & 1 & 3 \end{array}\right) \left( \begin{array}{llll} 5 & 1 & 1 & 7 \end{array}\right) \left( \begin{array}{ll} 6 & 9 \end{array}\right)$ .

8. 找出 $\left( \begin{array}{llllll} 1 & 1 & 2 & 8 & {10} & 4 \end{array}\right) \left( \begin{array}{lll} 2 & 1 & 3 \end{array}\right) \left( \begin{array}{llll} 5 & 1 & 1 & 7 \end{array}\right) \left( \begin{array}{ll} 6 & 9 \end{array}\right)$ 的阶。

9. Write out the cycle decomposition of each element of order 4 in ${S}_{4}$ .

10. 写出 ${S}_{4}$ 中阶为4的每个元素的循环分解。

11. Write out the cycle decomposition of each element of order 2 in ${S}_{4}$ .

12. 写出 ${S}_{4}$ 中阶为2的每个元素的循环分解。

13. Prove that if $\Omega = \{ 1,2,3,\ldots \}$ then ${S}_{\Omega }$ is an infinite group (do not say $\infty ! = \infty$ ).

14. 证明如果 $\Omega = \{ 1,2,3,\ldots \}$ 则 ${S}_{\Omega }$ 是一个无限群（不要说 $\infty ! = \infty$）。

15. (a) Let $\sigma$ be the 12-cycle (123456789101112). For which positive integers $i$ is ${\sigma }^{i}$ also a 12-cycle?

16. (a) 设 $\sigma$ 为12-循环 (123456789101112)。对于哪些正整数 $i$ ，${\sigma }^{i}$ 也是12-循环？

(b) Let $\tau$ be the 8-cycle $\left( \begin{array}{llllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \end{array}\right)$ . For which positive integers $i$ is ${\tau }^{i}$ also an 8-cycle?

(b) 设 $\tau$ 为8-循环 $\left( \begin{array}{llllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \end{array}\right)$ 。对于哪些正整数 $i$ ，${\tau }^{i}$ 也是8-循环？

(c) Let $\omega$ be the 14-cycle (1234567891011121314). For which positive integers $i$ is ${\omega }^{i}$ also a 14-cycle?

(c) 设 $\omega$ 为14-循环 (1234567891011121314)。对于哪些正整数 $i$ ，${\omega }^{i}$ 也是14-循环？

10. Prove that if $\sigma$ is the $m$ -cycle $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ ,then for all $i \in \{ 1,2,\ldots ,m\} ,{\sigma }^{i}\left( {a}_{k}\right) = {a}_{k + i}$ , where $k + i$ is replaced by its least residue mod $m$ when $k + i > m$ . Deduce that $\left| \sigma \right| = m$ .

11. 证明如果 $\sigma$ 是 $m$ -循环 $\left( {{a}_{1}{a}_{2}\ldots {a}_{m}}\right)$ ，那么对于所有 $i \in \{ 1,2,\ldots ,m\} ,{\sigma }^{i}\left( {a}_{k}\right) = {a}_{k + i}$ ，其中 $k + i$ 被替换为其在模 $m$ 下的最小余数当 $k + i > m$ 。推导出 $\left| \sigma \right| = m$ 。

12. Let $\sigma$ be the $m$ -cycle $\left( {{12}\ldots m}\right)$ . Show that ${\sigma }^{i}$ is also an $m$ -cycle if and only if $i$ is relatively prime to $m$ .

13. 设 $\sigma$ 为 $m$ -循环 $\left( {{12}\ldots m}\right)$ 。证明 ${\sigma }^{i}$ 也是 $m$ -循环当且仅当 $i$ 与 $m$ 互质。

14. (a) If $\tau = \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \end{array}\right) \left( \begin{array}{ll} 5 & 6 \end{array}\right) \left( \begin{array}{ll} 7 & 8 \end{array}\right) \left( \begin{array}{ll} 9 & {10} \end{array}\right)$ determine whether there is a $n$ -cycle $\sigma \left( {n \geq {10}}\right)$ with $\tau = {\sigma }^{k}$ for some integer $k$ .

15. (a) 如果 $\tau = \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \end{array}\right) \left( \begin{array}{ll} 5 & 6 \end{array}\right) \left( \begin{array}{ll} 7 & 8 \end{array}\right) \left( \begin{array}{ll} 9 & {10} \end{array}\right)$ 确定是否存在一个 $n$ -循环 $\sigma \left( {n \geq {10}}\right)$ ，使得 $\tau = {\sigma }^{k}$ 对于某个整数 $k$ 。

(b) If $\tau = \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{lll} 3 & 4 & 5 \end{array}\right)$ determine whether there is an $n$ -cycle $\sigma \left( {n \geq 5}\right)$ with $\tau = {\sigma }^{k}$ for some integer $k$ .

(b) 如果 $\tau = \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{lll} 3 & 4 & 5 \end{array}\right)$ 确定是否存在一个 $n$ -循环 $\sigma \left( {n \geq 5}\right)$ ，使得 $\tau = {\sigma }^{k}$ 对于某个整数 $k$ 。

13. Show that an element has order 2 in ${S}_{n}$ if and only if its cycle decomposition is a product of commuting 2-cycles.

14. 证明一个元素在 ${S}_{n}$ 中的阶为2当且仅当其循环分解是可交换的2-循环的乘积。

15. Let $p$ be a prime. Show that an element has order $p$ in ${S}_{n}$ if and only if its cycle decomposition is a product of commuting $p$ -cycles. Show by an explicit example that this need not be the case if $p$ is not prime.

16. 设 $p$ 为一个质数。证明一个元素在 ${S}_{n}$ 中的阶等于 $p$ 当且仅当它的循环分解是可交换的 $p$ -循环的乘积。通过一个具体例子说明，如果 $p$ 不是质数，则情况可能并非如此。

17. Prove that the order of an element in ${S}_{n}$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. [Use Exercise 10 and Exercise 24 of Section 1.]

18. 证明一个元素在 ${S}_{n}$ 中的阶等于它的循环分解中各循环长度的最小公倍数。 [使用第1节的练习10和练习24。]

19. Show that if $n \geq m$ then the number of $m$ -cycles in ${S}_{n}$ is given by

20. 证明如果 $n \geq m$ ，那么 ${S}_{n}$ 中 $m$ -循环的数量由以下给出

[Count the number of ways of forming an $m$ -cycle and divide by the number of representations of a particular $m$ -cycle.]

[计算形成一个 $m$ -循环的方法数，然后除以特定 $m$ -循环的表现形式数。]

17. Show that if $n \geq 4$ then the number of permutations in ${S}_{n}$ which are the product of two disjoint 2-cycles is $n\left( {n - 1}\right) \left( {n - 2}\right) \left( {n - 3}\right) /8$ .

18. 证明如果 $n \geq 4$ ，那么 ${S}_{n}$ 中由两个不相交的2-循环乘积构成的排列的数量是 $n\left( {n - 1}\right) \left( {n - 2}\right) \left( {n - 3}\right) /8$ 。

19. Find all numbers $n$ such that ${S}_{5}$ contains an element of order $n$ . [Use Exercise 15.]

20. 找出所有满足 ${S}_{5}$ 包含一个阶为 $n$ 的元素的数 $n$ 。 [使用练习15。]

21. Find all numbers $n$ such that ${S}_{7}$ contains an element of order $n$ . [Use Exercise 15.]

22. 找出所有满足 ${S}_{7}$ 包含一个阶为 $n$ 的元素的数 $n$ 。 [使用练习15。]

23. Find a set of generators and relations for ${S}_{3}$ .

24. 找出 ${S}_{3}$ 的一组生成元和关系。