Prove that the cycles(12...n)and (12)generate the symmetric group Sn

时间: 2023-09-23 10:05:25 浏览: 197
### 回答1： 证明循环（12...n）和（12）产生对称群Sn：首先，证明每一个元素在Sn中都有一个对应的元素。具体来说，如果给定一个循环（a1，a2，a3，...，an），它可以与它的逆序（an，an-1，...，a2，a1）组成一个对称的元素。同样，给定一个循环（b1，b2，...，bn），它可以与它的逆序（bn，bn-1，...，b2，b1）组成一个对称的元素。因此，所有的循环（12...n）和（12）都可以生成对称群Sn。 ### 回答2： 要证明循环置换 (12...n) 和 (12) 能够生成对称群 Sn，我们需要证明通过不断使用这两个置换操作，我们能够生成 Sn 中的所有置换。 首先，我们注意到 (12) 表示将数字 1 和 2 互换的置换。通过使用 (12)^2 = (12)(12) = (1)，我们可以看出它实际上是将位置 1 和位置 2 的数字互换的置换。 接下来，我们考虑循环置换 (12...n)。它表示将位置 1 上的数字移到位置 2，位置 2 上的数字移到位置 3，一直移动到位置 n 上的数字移到位置 1。通过将这个置换连续应用 k 次，我们可以得到 (12...n)^k 表示将位置 i 上的数字移动到位置 (i+k) mod n 上的置换。 现在，我们可以通过使用这两个置换操作生成 Sn 中的给定置换。任何一个置换可以被表示为一系列循环置换的乘积。例如，对于一个给定的置换 (a1 a2...am)，我们可以通过先将数字 a1 平移到位置 1，然后将数字 a2 平移到位置 2，一直移动到数字 am 平移到位置 m，最终得到一个以 m 结尾的循环置换。然后我们再应用 (12) 置换，将数字 1 和数字 m+1 互换，得到一个以 m+1 开头的循环置换。通过连续使用这两个操作，我们可以得到 (a1 a2...am) 这个置换。 综上所述，我们可以看出通过使用循环置换 (12...n) 和 (12)，我们能够生成 Sn 中的所有置换，因此它们能够生成对称群 Sn。 ### 回答3： 要证明循环置换 (12...n) 和 (12) 生成对称群 Sn，首先需要证明它们生成一个群。 我们知道，循环置换 (12...n) 是一个置换，将1映射到2，2映射到3，以此类推，将n映射到1。而循环置换 (12) 是将1映射到2，2映射到1，即将1和2互换了位置。那么，将这两个循环置换连续进行两次，即 (12...n)(12...n) 和 (12)(12) 分别是将1映射到3，2映射到4，...，n映射到2。以此类推，每次将两个置换连续进行两次，都会得到这样的结果。 我们令 (12...n)(12...n) 的结果为 A，(12)(12) 的结果为 B。可以发现，将 A 连续进行任意奇数次或 B 连续进行任意偶数次，都会回到起始的位置。而对于将 A 连续进行任意偶数次或 B 连续进行任意奇数次，则会得到除了起始位置以外的其他位置。 因此，我们可以得到以下结论： 1. 由 (12...n) 和 (12) 所生成的置换集合，一定是一个群，因为它满足结合律、单位元存在、每个元素都有逆元的性质。 2. 这个群中的每一个元素都可以通过连续进行 (12...n) 和 (12) 这两种置换来得到，并且这些置换可以通过不同的次数进行组合得到不同的结果。 由于 Sn 是对称群，是由 Sn 中所有可能的置换组成的，而我们已经证明了 (12...n) 和 (12) 可以生成一个群，并且这个群中的每一个元素都可以通过连续进行这两种置换得到。所以我们可以得出结论，(12...n) 和 (12) 生成的置换群是 Sn，即它们生成对称群 Sn。
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