p -群，幂零群和可解群

6.1 $p$ -GROUPS,NILPOTENT GROUPS,AND SOLVABLE GROUPS

6.1 $p$ -群，幂零群和可解群

Let $p$ be a prime and let $G$ be a finite group of order ${p}^{a}n$ ,where $p$ does not divide $n$ . Recall that a (finite) $p$ -group is any group whose order is a power of $p$ . Sylow’s Theorem shows that $p$ -groups abound as subgroups of $G$ and in order to exploit this phenomenon to unravel the structure of finite groups it will be necessary to establish some basic properties of $p$ -groups. In the next section we shall apply these results in many specific instances.

设 $p$ 为一个素数，$G$ 为一个阶为 ${p}^{a}n$ 的有限群，其中 $p$ 不整除 $n$ 。回顾一下，一个（有限）$p$ -群是任何阶为 $p$ 的幂的群。西罗定理表明 $p$ -群作为 $G$ 的子群大量存在，为了利用这一现象来揭示有限群的结构，我们将需要建立一些 $p$ -群的基本性质。在下一节中，我们将在许多具体情况下应用这些结果。

Before giving the results on $p$ -groups we first recall a definition that has appeared in some earlier exercises.

在给出关于 $p$ -群的结果之前，我们首先回顾一个在一些早期练习中已经出现的定义。

Definition. A maximal subgroup of a group $G$ is a proper subgroup $M$ of $G$ such that there are no subgroups $H$ of $G$ with $M < H < G$ .

定义。一个群 $G$ 的极大子群 $M$ 是 $G$ 的一个真子群 $M$，使得 $G$ 中不存在其他子群 $H$ 满足 $M < H < G$ 。

By order considerations every proper subgroup of a finite group is contained in some maximal subgroup. In contrast, infinite groups may or may not have maximal subgroups. For example, $p\mathbb{Z}$ is a maximal subgroup of $\mathbb{Z}$ whereas $\mathbb{Q}$ (under +) has no maximal subgroups (cf. Exercise 16 at the end of this section).

由阶数考虑，一个有限群的每个真子群都包含在某些极大子群中。相比之下，无限群可能有极大子群，也可能没有。例如，$p\mathbb{Z}$ 是 $\mathbb{Z}$ 的一个极大子群，而 $\mathbb{Q}$（在加法下）没有极大子群（参见本节末尾的练习 16）。

We now collect all the properties of $p$ -groups we shall need into an omnibus theorem:

现在我们将所有需要的关于 $p$ -群的性质收集到一个综合定理中：

Theorem 1. Let $p$ be a prime and let $P$ be a group of order ${p}^{a},a \geq 1$ . Then

定理 1。设 $p$ 是一个素数，$P$ 是一个阶数为 ${p}^{a},a \geq 1$ 的群。那么

(1) The center of $P$ is nontrivial: $Z\left( P\right) \neq 1$ .

(1) $P$ 的中心是非平凡的：$Z\left( P\right) \neq 1$ 。

(2) If $H$ is a nontrivial normal subgroup of $P$ then $H$ intersects the center non-trivially: $H \cap Z\left( P\right) \neq 1$ . In particular,every normal subgroup of order $p$ is contained in the center.

(2) 如果 $H$ 是 $P$ 的一个非平凡正规子群，那么 $H$ 与中心非平凡相交：$H \cap Z\left( P\right) \neq 1$ 。特别是，每个阶数为 $p$ 的正规子群都包含在中心中。

(3) If $H$ is a normal subgroup of $P$ then $H$ contains a subgroup of order ${p}^{b}$ that is normal in $P$ for each divisor ${p}^{b}$ of $\left| H\right|$ . In particular, $P$ has a normal subgroup of order ${p}^{b}$ for every $b \in \{ 0,1,\ldots ,a\}$ .

(3) 如果 $H$ 是 $P$ 的一个正规子群，那么对于 $\left| H\right|$ 的每个除数 ${p}^{b}$，$H$ 包含一个在 $P$ 中正规的阶数为 ${p}^{b}$ 的子群。特别是，对于每个 $b \in \{ 0,1,\ldots ,a\}$，$P$ 都有一个阶数为 ${p}^{b}$ 的正规子群。

(4) If $H < P$ then $H < {N}_{P}\left( H\right)$ (i.e.,every proper subgroup of $P$ is a proper subgroup of its normalizer in $P$ ).

(4) 如果 $H < P$，那么 $H < {N}_{P}\left( H\right)$（即，$P$ 的每个真子群是其正规化子在 $P$ 中的真子群）。

(5) Every maximal subgroup of $P$ is of index $p$ and is normal in $P$ .

(5) $P$ 的每个极大子群的指数为 $p$ 并且在 $P$ 中是正规的。

Proof: These results rely ultimately on the class equation and it may be useful for the reader to review Section 4.3.

证明：这些结果最终依赖于类方程，读者回顾第4.3节可能是有用的。

Part 1 is Theorem 8 of Chapter 4 and is also the special case of part 2 when $H = P$ . We therefore begin by proving (2); we shall not quote Theorem 8 of Chapter 4 although the argument that follows is only a slight generalization of the one in Chapter 4. Let $H$ be a nontrivial normal subgroup of $P$ . Recall that for each conjugacy class $\mathcal{C}$ of $P$ , either $\mathcal{C} \subseteq H$ or $\mathcal{C} \cap H = \varnothing$ because $H$ is normal (this easy fact was shown in a remark preceding Theorem 4.12). Pick representatives of the conjugacy classes of $P$ :

第1部分是第4章的定理8，也是第2部分在 $H = P$ 时的特例。因此，我们首先证明（2）；尽管以下论证只是第4章中论证的轻微推广，我们不会引用第4章的定理8。设 $H$ 是 $P$ 的一个非平凡正规子群。回想对于 $P$ 的每个共轭类 $\mathcal{C}$ ，要么 $\mathcal{C} \subseteq H$ 要么 $\mathcal{C} \cap H = \varnothing$ ，因为 $H$ 是正规子群（这一简单事实在定理4.12之前的注解中已经展示）。选取 $P$ 的共轭类的代表：

with ${a}_{1},\ldots ,{a}_{k} \in H$ and ${a}_{k + 1},\ldots ,{a}_{r} \notin H$ . Let ${\mathcal{C}}_{i}$ be the conjugacy class of ${a}_{i}$ in $P$ , for all $i$ . Thus

其中 ${a}_{1},\ldots ,{a}_{k} \in H$ 和 ${a}_{k + 1},\ldots ,{a}_{r} \notin H$ 。设 ${\mathcal{C}}_{i}$ 是 ${a}_{i}$ 在 $P$ 中的共轭类，对于所有 $i$ 。因此

By renumbering ${a}_{1},\ldots ,{a}_{k}$ if necessary we may assume ${a}_{1},\ldots ,{a}_{s}$ represent classes of size 1 (i.e.,are in the center of $P$ ) and ${a}_{s + 1},\ldots ,{a}_{k}$ represent classes of size $> 1$ . Since $H$ is the disjoint union of these we have

通过必要时重新编号 ${a}_{1},\ldots ,{a}_{k}$ ，我们可以假设 ${a}_{1},\ldots ,{a}_{s}$ 代表大小为1的类（即，在 $P$ 的中心中），而 ${a}_{s + 1},\ldots ,{a}_{k}$ 代表大小为 $> 1$ 的类。由于 $H$ 是这些类的并集，我们有

Now $p$ divides $\left| H\right|$ and $p$ divides each term in the sum $\mathop{\sum }\limits_{{i = s + 1}}^{k}\left| {P : {C}_{P}\left( {a}_{i}\right) }\right|$ so $p$ divides their difference: $\left| {H \cap Z\left( P\right) }\right|$ . This proves $H \cap Z\left( P\right) \neq 1$ . If $\left| H\right| = p$ ,since $H \cap Z\left( P\right) \neq 1$ we must have $H \leq Z\left( P\right)$ . This completes the proof of (2).

现在 $p$ 整除 $\left| H\right|$ ，并且 $p$ 整除每个项的和 $\mathop{\sum }\limits_{{i = s + 1}}^{k}\left| {P : {C}_{P}\left( {a}_{i}\right) }\right|$ ，所以 $p$ 整除它们的差：$\left| {H \cap Z\left( P\right) }\right|$ 。这证明了 $H \cap Z\left( P\right) \neq 1$ 。如果 $\left| H\right| = p$ ，由于 $H \cap Z\left( P\right) \neq 1$ ，我们必须有 $H \leq Z\left( P\right)$ 。这完成了（2）的证明。

Next we prove (3) by induction on $a$ . If $a \leq 1$ or $H = 1$ ,theresult is trivial. Assume therefore that $a > 1$ and $H \neq 1$ . By part 2, $H \cap Z\left( P\right) \neq 1$ so by Cauchy’s Theorem $H \cap Z\left( P\right)$ contains a (normal) subgroup $Z$ of order $p$ . Use bar notation to denote passage to the quotient group $P/Z$ . This quotient has order ${p}^{a - 1}$ and $\bar{H} \trianglelefteq \bar{P}$ . By induction,for every nonnegative integer $b$ such that ${p}^{b}$ divides $|\overline{H}|$ there is a subgroup $\bar{K}$ of $\bar{H}$ of order ${p}^{b}$ that is normal in $\bar{P}$ . If $K$ is the complete preimage of $\bar{K}$ in $P$ then $\left| K\right| = {p}^{b + 1}$ . The set of all subgroups of $H$ obtained by this process together with the identity subgroup provides a subgroup of $H$ that is normal in $P$ for each divisor of $\left| H\right|$ . The second assertion of part 3 is the special case $H = P$ . This establishes part 3.

接下来，我们通过归纳法证明 (3) ，归纳对象是 $a$ 。如果 $a \leq 1$ 或 $H = 1$ ，结果是平凡的。因此假设 $a > 1$ 且 $H \neq 1$ 。由第2部分，$H \cap Z\left( P\right) \neq 1$ ，根据柯西定理 $H \cap Z\left( P\right)$ 包含一个（正规）子群 $Z$ ，其阶为 $p$ 。使用横杠符号表示到商群的过渡 $P/Z$ 。这个商群的阶为 ${p}^{a - 1}$ 且 $\bar{H} \trianglelefteq \bar{P}$ 。通过归纳，对于每个非负整数 $b$ ，如果 ${p}^{b}$ 整除 $|\overline{H}|$ ，则在 $\bar{H}$ 中存在一个阶为 ${p}^{b}$ 的子群 $\bar{K}$ ，且在 $\bar{P}$ 中是正规的。如果 $K$ 是 $\bar{K}$ 在 $P$ 中的完全逆像，那么 $\left| K\right| = {p}^{b + 1}$ 。通过这个过程得到的 $H$ 的所有子群集合，加上单位子群，为每个 $\left| H\right|$ 的因数提供了一个在 $P$ 中正规的 $H$ 的子群。第3部分的第二个断言是 $H = P$ 的特例。这证明了第3部分。

We prove (4) also by induction on $\left| P\right|$ . If $P$ is abelian then all subgroups of $P$ are normal in $P$ and the result is trivial. We may therefore assume $\left| P\right| > p$ (in fact, $\left| P\right| > {p}^{2}$ by Corollary 4.9). Let $H$ be a proper subgroup of $P$ . Since all elements of $Z\left( P\right)$ commute with all elements of $P,Z\left( P\right)$ normalizes every subgroup of $P$ . By part 1 we have that $Z\left( P\right) \neq 1$ . If $Z\left( P\right)$ is not contained in $H$ ,then $H$ is properly contained in $\langle H,Z\left( P\right) \rangle$ and the latter subgroup is contained in ${N}_{P}\left( H\right)$ so (4) holds. We may therefore assume $Z\left( P\right) \leq H$ . Use bar notation to denote passage to the quotient $P/Z\left( P\right)$ . Since $\bar{P}$ has smaller order than $P$ by (1),by induction $\bar{H}$ is properly contained in ${N}_{\bar{P}}\left( \bar{H}\right)$ . It follows directly from the Lattice Isomorphism Theorem that ${N}_{P}\left( H\right)$ is the complete preimage in $P$ of ${N}_{\bar{P}}\left( \bar{H}\right)$ ,hence we obtain proper containment of $H$ in its normalizer in this case as well. This completes the induction.

我们通过在 $\left| P\right|$ 上进行归纳来证明（4）。如果 $P$ 是阿贝尔群，那么 $P$ 的所有子群在 $P$ 中都是正规子群，结果是平凡的。因此，我们可以假设 $\left| P\right| > p$（实际上，根据引理4.9，是 $\left| P\right| > {p}^{2}$）。设 $H$ 是 $P$ 的一个真子群。由于 $Z\left( P\right)$ 的所有元素与 $P,Z\left( P\right)$ 的所有元素交换，$P,Z\left( P\right)$ 正规化 $P$ 的每个子群。根据第1部分，我们有 $Z\left( P\right) \neq 1$。如果 $Z\left( P\right)$ 不包含在 $H$ 中，那么 $H$ 真正包含在 $\langle H,Z\left( P\right) \rangle$ 中，后者子群包含在 ${N}_{P}\left( H\right)$ 中，因此（4）成立。因此，我们可以假设 $Z\left( P\right) \leq H$。使用横杠符号表示到商的过渡 $P/Z\left( P\right)$。由于 $\bar{P}$ 的阶小于 $P$ 的阶，根据（1），通过归纳 $\bar{H}$ 真正包含在 ${N}_{\bar{P}}\left( \bar{H}\right)$ 中。直接从格同构定理得出 ${N}_{P}\left( H\right)$ 是 $P$ 中 ${N}_{\bar{P}}\left( \bar{H}\right)$ 的完全原像，因此我们在这个情况下也得到了 $H$ 在其正规子群中的真正包含。这完成了归纳。

To prove (5) let $M$ be a maximal subgroup of $P$ . By definition, $M < P$ so by part $4,M < {N}_{P}\left( M\right)$ . By definition of maximality we must therefore have ${N}_{P}\left( M\right) = P$ , i.e., $M \trianglelefteq P$ . The Lattice Isomorphism Theorem shows that $P/M$ is a $p$ -group with no proper nontrivial subgroups because $M$ is a maximal subgroup. By part 3,however, $P/M$ has subgroups of every order dividing $\left| {P/M}\right|$ . The only possibility is $\left| {P/M}\right| = p$ . This proves (5) and completes the proof of the theorem.

为了证明（5），设 $M$ 是 $P$ 的一个极大子群。根据定义，$M < P$，所以根据部分 $4,M < {N}_{P}\left( M\right)$。由于极大性的定义，我们必须有 ${N}_{P}\left( M\right) = P$，即 $M \trianglelefteq P$。格同构定理表明 $P/M$ 是一个 $p$ -群，且没有适当的非平凡子群，因为 $M$ 是一个极大子群。然而，根据第3部分，$P/M$ 有 $\left| {P/M}\right|$ 的每个除数的子群。唯一可能的是 $\left| {P/M}\right| = p$。这证明了（5），并完成了定理的证明。

Definition.

定义。

(1) For any (finite or infinite) group $G$ define the following subgroups inductively:

（1）对于任何（有限或无限）群 $G$，以下子群是递归定义的：

and ${Z}_{i + 1}\left( G\right)$ is the subgroup of $G$ containing ${Z}_{i}\left( G\right)$ such that

并且 ${Z}_{i + 1}\left( G\right)$ 是包含 ${Z}_{i}\left( G\right)$ 的 $G$ 的子群，使得

(i.e., ${Z}_{i + 1}\left( G\right)$ is the complete preimage in $G$ of the center of $G/{Z}_{i}\left( G\right)$ under the natural projection). The chain of subgroups

（即 ${Z}_{i + 1}\left( G\right)$ 是 $G$ 中 $G/{Z}_{i}\left( G\right)$ 的中心在自然投影下的完全逆像）。子群链

is called the upper central series of $G$ . (The use of the term "upper" indicates that ${Z}_{i}\left( G\right) \leq {Z}_{i + 1}\left( G\right)$ .)

被称为 $G$ 的上中央列。（使用“上”这个术语表示 ${Z}_{i}\left( G\right) \leq {Z}_{i + 1}\left( G\right)$ 。）

(2) A group $G$ is called nilpotent if ${Z}_{c}\left( G\right) = G$ for some $c \in \mathbb{Z}$ . The smallest such $c$ is called the nilpotence class of $G$ .

（2）如果对于某个 $c \in \mathbb{Z}$ ，群 $G$ 被称为幂零群。最小的这样的 $c$ 被称为 $G$ 的幂零类。

One of the exercises at the end of this section shows that ${Z}_{i}\left( G\right)$ is a characteristic (hence normal) subgroup of $G$ for all $i$ . We use this fact freely from now on.

本节末尾的练习之一表明 ${Z}_{i}\left( G\right)$ 是 $G$ 的一个特征子群（因此是正规子群），对所有 $i$ 都成立。我们从现在开始自由地使用这个事实。

Remarks:

备注：

(1) If $G$ is abelian then $G$ is nilpotent (of class 1,provided $\left| G\right| > 1$ ),since in this case $G = Z\left( G\right) = {Z}_{1}\left( G\right) .\;$ One should think of nilpotent groups as lying between abelian and solvable groups in the hierarchy of structure (recall that solvable groups were introduced in Section 3.4; we shall discuss solvable groups further at the end of this section):

(1) 如果 $G$ 是阿贝尔群，那么 $G$ 是幂零群（在类1中，假定 $\left| G\right| > 1$ ），因为在这种情况下 $G = Z\left( G\right) = {Z}_{1}\left( G\right) .\;$ 应该将幂零群视为在阿贝尔群和解群之间的结构层次中（回想一下，解群在3.4节中引入；我们将在本节的末尾进一步讨论解群）：

$\begin{matrix} \text{cyclic}\;\text{groups} & \subset & \text{abelian}\;\text{groups} & \subset & \text{nilpotent}\;\text{groups} & \subset & \text{solvable}\;\text{groups} & \subset & \text{all}\;\text{groups} \end{matrix}$

(all of the above containments are proper, as we shall verify shortly).

（以上所有包含关系都是正确的，我们稍后会验证这一点）。

(2) For any finite group there must,by order considerations,be an integer $n$ such that

（2）对于任何有限群，由于阶数考虑，必定存在一个整数 $n$ 使得

For example, ${Z}_{n}\left( {S}_{3}\right) = 1$ for all $n \in {\mathbb{Z}}^{ + }$ . Once two terms in the upper central series are the same, the chain stabilizes at that point (i.e., all terms thereafter are equal to these two). For example,if $G = {Z}_{2} \times {S}_{3}$ ,

例如， ${Z}_{n}\left( {S}_{3}\right) = 1$ 对于所有 $n \in {\mathbb{Z}}^{ + }$ 。一旦上中央系列中的两项相同，链在该点稳定（即，之后的所有项都等于这两个）。例如，如果 $G = {Z}_{2} \times {S}_{3}$ ，

By definition, ${Z}_{n}\left( G\right)$ is a proper subgroup of $G$ for all $n$ for non-nilpotent groups. (3) For infinite groups $G$ it may happen that all ${Z}_{i}\left( G\right)$ are proper subgroups of $G$ (so $G$ is not nilpotent) but

根据定义， ${Z}_{n}\left( G\right)$ 是 $G$ 的真子群，对于所有 $n$ 对于非幂零群。 （3）对于无限群 $G$ ，可能发生所有 ${Z}_{i}\left( G\right)$ 都是 $G$ 的真子群（所以 $G$ 不是幂零群）但

Groups for which this hold are called hypernilpotent - they enjoy some (but not all) of the properties of nilpotent groups. While we shall be dealing mainly with finite nilpotent groups, results that do not involve the notion of order, Sylow subgroups etc. also hold for infinite groups. Even for infinite groups one of the main techniques for dealing with nilpotent groups is induction on the nilpotence class.

满足这种情况的群被称为超幂零群 - 它们具有一些（但不是所有）幂零群的性质。虽然我们将主要处理有限幂零群，但那些不涉及阶数、 Sylow 子群等概念的结果也适用于无限群。即使对于无限群，处理幂零群的主要技术之一也是对幂零类进行归纳。

Proposition 2. Let $p$ be a prime and let $P$ be a group of order ${p}^{a}$ . Then $P$ is nilpotent of nilpotence class at most $a - 1$ .

命题2。设 $p$ 是一个素数， $P$ 是一个阶数为 ${p}^{a}$ 的群。那么 $P$ 是幂零群，幂零类至多为 $a - 1$ 。

Proof: For each $i \geq 0,P/{Z}_{i}\left( P\right)$ is a $p$ -group,so

证明：对于每个 $i \geq 0,P/{Z}_{i}\left( P\right)$ 是一个 $p$ -群，所以

by Theorem 1(1). Thus if ${Z}_{i}\left( P\right) \neq G$ then $\left| {{Z}_{i + 1}\left( P\right) }\right| \geq p\left| {{Z}_{i}\left( P\right) }\right|$ and so $\left| {{Z}_{i + 1}\left( P\right) }\right| \geq$ ${p}^{i + 1}$ . In particular, $\left| {{Z}_{a}\left( P\right) }\right| \geq {p}^{a}$ ,so $P = {Z}_{a}\left( P\right)$ . Thus $P$ is nilpotent of class $\leq a$ . The only way $P$ could be of nilpotence class exactly equal to $a$ would be if $\left| {{Z}_{i}\left( P\right) }\right| = {p}^{i}$ for all $i$ . In this case,however, ${Z}_{a - 2}\left( P\right)$ would have index ${p}^{2}$ in $P$ ,so $P/{Z}_{a - 2}\left( P\right)$ would be abelian (by Corollary 4.9). But then $P/{Z}_{a - 2}\left( P\right)$ would equal its center and so ${Z}_{a - 1}\left( P\right)$ would equal $P$ ,a contradiction. This proves that the class of $P$ is $\leq a - 1$ .

由定理1(1)可知。因此如果 ${Z}_{i}\left( P\right) \neq G$ 那么 $\left| {{Z}_{i + 1}\left( P\right) }\right| \geq p\left| {{Z}_{i}\left( P\right) }\right|$ 并且因此 $\left| {{Z}_{i + 1}\left( P\right) }\right| \geq$ ${p}^{i + 1}$ 。特别地， $\left| {{Z}_{a}\left( P\right) }\right| \geq {p}^{a}$ ，所以 $P = {Z}_{a}\left( P\right)$ 。因此 $P$ 是阶为 $\leq a$ 的幂零群。唯一可能使 $P$ 的幂零类恰好等于 $a$ 的情况是如果对所有 $\left| {{Z}_{i}\left( P\right) }\right| = {p}^{i}$ 有 $i$ 。然而在这种情况下，${Z}_{a - 2}\left( P\right)$ 在 $P$ 中的指数为 ${p}^{2}$ ，所以 $P/{Z}_{a - 2}\left( P\right)$ 将是阿贝尔群（由推论4.9得出）。但这样 $P/{Z}_{a - 2}\left( P\right)$ 就等于它的中心，因此 ${Z}_{a - 1}\left( P\right)$ 将等于 $P$ ，这产生了矛盾。这证明了 $P$ 的类是 $\leq a - 1$ 。

Example

示例

Both ${D}_{8}$ and ${Q}_{8}$ are nilpotent of class 2 . More generally, ${D}_{{2}^{n}}$ is nilpotent of class $n - 1$ . This can be proved inductively by showing that $\left| {Z\left( {D}_{{2}^{n}}\right) }\right| = 2$ and ${D}_{{2}^{n}}/Z\left( {D}_{{2}^{n}}\right) \cong {D}_{{2}^{n - 1}}$ for $n \geq 3$ (the details are left as an exercise). If $n$ is not a power of $2,{D}_{2n}$ is not nilpotent (cf. Exercise 10).

${D}_{8}$ 和 ${Q}_{8}$ 都是2阶幂零群。更一般地，${D}_{{2}^{n}}$ 是 $n - 1$ 阶幂零群。这可以通过归纳法证明，通过展示 $\left| {Z\left( {D}_{{2}^{n}}\right) }\right| = 2$ 和 ${D}_{{2}^{n}}/Z\left( {D}_{{2}^{n}}\right) \cong {D}_{{2}^{n - 1}}$ 对 $n \geq 3$ 成立（细节留作练习）。如果 $n$ 不是 $2,{D}_{2n}$ 的幂，则 不是幂零群（参见练习10）。

We now give some equivalent (and often more workable) characterizations of nilpo-tence for finite groups:

现在我们给出一些有限群幂零性的等价（并且通常更实用）的特征：

Theorem 3. Let $G$ be a finite group,let ${p}_{1},{p}_{2},\ldots ,{p}_{s}$ be the distinct primes dividing its order and let ${P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) ,1 \leq i \leq s$ . Then the following are equivalent:

定理3。设 $G$ 是一个有限群，设 ${p}_{1},{p}_{2},\ldots ,{p}_{s}$ 是其阶的不同的素数因子，设 ${P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) ,1 \leq i \leq s$ 。那么以下条件是等价的：

(1) $G$ is nilpotent

(1) $G$ 是幂零群

(2) if $H < G$ then $H < {N}_{G}\left( H\right)$ ,i.e.,every proper subgroup of $G$ is a proper subgroup of its normalizer in $G$

(2) 如果 $H < G$ 则 $H < {N}_{G}\left( H\right)$ ，即 $G$ 的每个真子群都是其在 $G$ 中的正规化子的真子群

(3) ${P}_{i} \trianglelefteq G$ for $1 \leq i \leq s$ ,i.e.,every Sylow subgroup is normal in $G$

(3) ${P}_{i} \trianglelefteq G$ 对于 $1 \leq i \leq s$ ，即每个 Sylow 子群在 $G$ 中是正规的

(4) $G \cong {P}_{1} \times {P}_{2} \times \cdots \times {P}_{s}$ .

(4) $G \cong {P}_{1} \times {P}_{2} \times \cdots \times {P}_{s}$ 。

Proof: The proof that (1) implies (2) is the same argument as for $p$ -groups - the only fact we needed was if $G$ is nilpotent then so is $G/Z\left( G\right)$ —so the details are omitted (cf. the exercises).

证明：(1) 推导出 (2) 的证明与 $p$ -群的论证相同 - 我们唯一需要的事实是如果 $G$ 是幂零的，那么 $G/Z\left( G\right)$ 也是 - 因此细节被省略（参见练习）。

To show that (2) implies (3) let $P = {P}_{i}$ for some $i$ and let $N = {N}_{G}\left( P\right)$ . Since $P \trianglelefteq N$ ,Corollary 4.20 gives that $P$ is characteristic in $N$ . Since $P$ char $N \trianglelefteq {N}_{G}\left( N\right)$ we get that $P \trianglelefteq {N}_{G}\left( N\right)$ . This means ${N}_{G}\left( N\right) \leq N$ and hence ${N}_{G}\left( N\right) = N$ . By (2) we must therefore have $N = G$ ,which gives (3).

为了证明 (2) 推导出 (3)，设 $P = {P}_{i}$ 对于某个 $i$ ，并设 $N = {N}_{G}\left( P\right)$ 。由于 $P \trianglelefteq N$ ，根据推论 4.20 可得 $P$ 在 $N$ 中是特征子群。因为 $P$ char $N \trianglelefteq {N}_{G}\left( N\right)$ ，我们得到 $P \trianglelefteq {N}_{G}\left( N\right)$ 。这意味着 ${N}_{G}\left( N\right) \leq N$ ，因此 ${N}_{G}\left( N\right) = N$ 。根据 (2) 我们必须有 $N = G$ ，这给出了 (3)。

Next we prove (3) implies (4). For any $t,\;1 \leq t \leq s$ we show inductively that

接下来我们证明 (3) 推导出 (4)。对于任意的 $t,\;1 \leq t \leq s$ ，我们通过归纳法证明

Note first that each ${P}_{i}$ is normal in $G$ so ${P}_{1}\cdots {P}_{t}$ is a subgroup of $G$ . Let $H$ be the product ${P}_{1}\cdots {P}_{t - 1}$ and let $K = {P}_{t}$ ,so by induction $H \cong {P}_{1} \times \cdots \times {P}_{t - 1}$ . In particular, $\left| H\right| = \left| {P}_{1}\right| \cdot \left| {P}_{2}\right| \cdots \left| {P}_{t - 1}\right|$ . Since $\left| K\right| = \left| {P}_{t}\right|$ ,the orders of $H$ and $K$ are relatively prime. Lagrange’s Theorem implies $H \cap K = 1$ . By definition, ${P}_{1}\cdots {P}_{t} = {HK}$ , hence Theorem 5.9 gives

首先注意到每个 ${P}_{i}$ 在 $G$ 中是正规的，所以 ${P}_{1}\cdots {P}_{t}$ 是 $G$ 的一个子群。设 $H$ 为 ${P}_{1}\cdots {P}_{t - 1}$ 的乘积，并设 $K = {P}_{t}$ ，所以通过归纳 $H \cong {P}_{1} \times \cdots \times {P}_{t - 1}$ 。特别地， $\left| H\right| = \left| {P}_{1}\right| \cdot \left| {P}_{2}\right| \cdots \left| {P}_{t - 1}\right|$ 。由于 $\left| K\right| = \left| {P}_{t}\right|$ ，$H$ 和 $K$ 的阶互质。根据拉格朗日定理可得 $H \cap K = 1$ 。由定义， ${P}_{1}\cdots {P}_{t} = {HK}$ ，因此定理 5.9 给出

which completes the induction. Now take $t = s$ to obtain (4).

这完成了归纳。现在取 $t = s$ 以得到 (4)。

Finally, to prove (4) implies (1) use Exercise 1 of Section 5.1 to obtain

最后，为了证明（4）蕴含（1），使用5.1节的练习1来得到

By Exercise 14 in Section 5.1,

由5.1节的练习14，

Thus the hypotheses of (4) also hold for $G/Z\left( G\right)$ . By Theorem 1,if ${P}_{i} \neq 1$ then $Z\left( {P}_{i}\right) \neq 1$ ,so if $G \neq 1,\left| {G/Z\left( G\right) }\right| < \left| G\right|$ . By induction, $G/Z\left( G\right)$ is nilpotent,so by Exercise 6, $G$ is nilpotent. This completes the proof.

因此，（4）的假设也适用于 $G/Z\left( G\right)$ 。根据定理1，如果 ${P}_{i} \neq 1$ ，那么 $Z\left( {P}_{i}\right) \neq 1$ ，所以如果 $G \neq 1,\left| {G/Z\left( G\right) }\right| < \left| G\right|$ 。通过归纳，$G/Z\left( G\right)$ 是幂零的，因此根据练习6，$G$ 是幂零的。这完成了证明。

Note that the first part of the Fundamental Theorem of Finite Abelian Groups (Theorem 5 in Section 5.2) follows immediately from the above theorem (we shall give another proof later as a consequence of the Chinese Remainder Theorem):

注意，有限阿贝尔群的基本定理的第一部分（5.2节的定理5）立即从上述定理得出（我们稍后作为中国剩余定理的推论给出另一个证明）：

Corollary 4. A finite abelian group is the direct product of its Sylow subgroups.

推论4。一个有限阿贝尔群是其西洛子群的直积。

Next we prove a proposition which will be used later to show that the multiplicative group of a finite field is cyclic (without using the Fundamental Theorem of FiniteAbelian Groups).

接下来我们证明一个命题，该命题将在后面用来证明有限域的乘法群是循环群（不使用有限阿贝尔群的基本定理）。

Proposition 5. If $G$ is a finite group such that for all positive integers $n$ dividing its order, $G$ contains at most $n$ elements $x$ satisfying ${x}^{n} = 1$ ,then $G$ is cyclic.

命题5。如果 $G$ 是一个有限群，对于所有整除其阶的正整数 $n$ ，$G$ 包含至多 $n$ 个满足 ${x}^{n} = 1$ 的元素 $x$ ，那么 $G$ 是循环群。

Proof: Let $\left| G\right| = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{s}^{{\alpha }_{s}}$ and let ${P}_{i}$ be a Sylow ${p}_{i}$ -subgroup of $G$ for $i = 1,2,\ldots ,s$ . Since ${p}_{i}^{{\alpha }_{i}} \mid \left| G\right|$ and the ${p}_{i}^{{\alpha }_{i}}$ elements of ${P}_{i}$ are solutions of ${x}^{{p}_{i}^{{\alpha }_{i}}} = 1$ , by hypothesis ${P}_{i}$ must contain all solutions to this equation in $G$ . It follows that ${P}_{i}$ is the unique (hence normal) Sylow ${p}_{i}$ -subgroup of $G$ . By Theorem 3, $G$ is the direct product of its Sylow subgroups. By Theorem 1,each ${P}_{i}$ possesses a normal subgroup ${M}_{i}$ of index ${p}_{i}$ . Since $\left| {M}_{i}\right| = {p}_{i}^{{\alpha }_{i} - 1}$ and $G$ has at most ${p}_{i}^{{\alpha }_{i} - 1}$ solutions to ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} = 1$ , by Lagrange’s Theorem (Corollary 9,Section 3.2) $M$ contains all elements $x$ of $G$ satisfying ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} = 1$ . Thus any element of ${P}_{i}$ not contained in ${M}_{i}$ satisfies ${x}^{{p}_{i}^{{\alpha }_{i}}} = 1$ but ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} \neq 1$ ,i.e., $x$ is an element of order ${p}_{i}^{{\alpha }_{i}}$ . This proves ${P}_{i}$ is cyclic for all $i$ ,so $G$ is the direct product of cyclic groups of relatively prime order, hence is cyclic.

证明：设 $\left| G\right| = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{s}^{{\alpha }_{s}}$ 且 ${P}_{i}$ 是 $G$ 的一个 Sylow ${p}_{i}$ 子群，对于 $i = 1,2,\ldots ,s$ 。由于 ${p}_{i}^{{\alpha }_{i}} \mid \left| G\right|$ 且 ${P}_{i}$ 的 ${p}_{i}^{{\alpha }_{i}}$ 元素是 ${x}^{{p}_{i}^{{\alpha }_{i}}} = 1$ 的解，根据假设 ${P}_{i}$ 必须包含 $G$ 中这个方程的所有解。因此，${P}_{i}$ 是 $G$ 的唯一（因此是正规）Sylow ${p}_{i}$ 子群。根据定理3，$G$ 是其 Sylow 子群的直积。根据定理1，每个 ${P}_{i}$ 都有一个指标为 ${p}_{i}$ 的正规子群 ${M}_{i}$ 。由于 $\left| {M}_{i}\right| = {p}_{i}^{{\alpha }_{i} - 1}$ 且 $G$ 至多有 ${p}_{i}^{{\alpha }_{i} - 1}$ 个 ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} = 1$ 的解，根据拉格朗日定理（3.2节的推论9）$M$ 包含所有满足 ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} = 1$ 的 $G$ 的元素 $x$ 。因此，任何不在 ${M}_{i}$ 中的 ${P}_{i}$ 的元素都满足 ${x}^{{p}_{i}^{{\alpha }_{i}}} = 1$ 但不满足 ${x}^{{p}_{i}^{{\alpha }_{i} - 1}} \neq 1$ ，即 $x$ 是阶为 ${p}_{i}^{{\alpha }_{i}}$ 的元素。这证明了对于所有 $i$ ，${P}_{i}$ 是循环的，因此 $G$ 是互质阶数的循环群的直积，因此是循环的。

The next proposition is called Frattini's Argument. We shall apply it to give another characterization of finite nilpotent groups. It will also be a valuable tool in the next section.

下一个命题被称为弗拉蒂尼论证。我们将应用它来给出有限幂零群的另一个特征。在下一节中，它也将是一个有价值的工具。

Proposition 6. (Frattini’s Argument) Let $G$ be a finite group,let $H$ be a normal subgroup of $G$ and let $P$ be a Sylow $p$ -subgroup of $H$ . Then $G = H{N}_{G}\left( P\right)$ and $\left| {G : H}\right|$ divides $\left| {{N}_{G}\left( P\right) }\right|$

命题6（Frattini论证）设 $G$ 是一个有限群，$H$ 是 $G$ 的正规子群，$P$ 是 $H$ 的 Sylow $p$ 子群。那么 $G = H{N}_{G}\left( P\right)$ 并且 $\left| {G : H}\right|$ 整除 $\left| {{N}_{G}\left( P\right) }\right|$

Proof: By Corollary 3.15, $H{N}_{G}\left( P\right)$ is a subgroup of $G$ and $H{N}_{G}\left( P\right) = {N}_{G}\left( P\right) H$ since $H$ is a normal subgroup of $G$ . Let $g \in G$ . Since ${P}^{g} \leq {H}^{g} = H$ ,both $P$ and ${P}^{g}$ are Sylow $p$ -subgroups of $H$ . By Sylow’s Theorem applied in $H$ ,there exists $x \in H$ such that ${P}^{g} = {P}^{x}$ . Thus $g{x}^{-1} \in {N}_{G}\left( P\right)$ and so $g \in {N}_{G}\left( P\right) x$ . Since $g$ was an arbitrary element of $G$ ,this proves $G = {N}_{G}\left( P\right) H$ .

证明：根据推论3.15，$H{N}_{G}\left( P\right)$ 是 $G$ 的子群，因为 $H$ 是 $G$ 的正规子群。设 $g \in G$ 。由于 ${P}^{g} \leq {H}^{g} = H$ ，$P$ 和 ${P}^{g}$ 都是 $H$ 的 Sylow $p$ 子群。应用 Sylow 定理于 $H$ ，存在 $x \in H$ 使得 ${P}^{g} = {P}^{x}$ 。因此 $g{x}^{-1} \in {N}_{G}\left( P\right)$ 从而 $g \in {N}_{G}\left( P\right) x$ 。由于 $g$ 是 $G$ 的任意元素，这证明了 $G = {N}_{G}\left( P\right) H$ 。

Apply the Second Isomorphism Theorem to $G = {N}_{G}\left( P\right) H$ to conclude that

将第二同构定理应用于 $G = {N}_{G}\left( P\right) H$ ，得出结论

so $\left| {G : H}\right|$ divides $\left| {{N}_{G}\left( P\right) }\right|$ ,completing the proof.

因此 $\left| {G : H}\right|$ 整除 $\left| {{N}_{G}\left( P\right) }\right|$ ，完成了证明。

Proposition 7. A finite group is nilpotent if and only if every maximal subgroup is normal.

命题7。一个有限群是幂零群当且仅当它的每个极大子群都是正规子群。

Proof: Let $G$ be a finite nilpotent group and let $M$ be a maximal subgroup of $G$ . As in the proof of Theorem 1,since $M < {N}_{G}\left( M\right)$ (by Theorem 3(2)) maximality of $M$ forces ${N}_{G}\left( M\right) = G$ ,i.e., $M \trianglelefteq G$ .

证明：设 $G$ 是一个有限幂零群，$M$ 是 $G$ 的一个极大子群。如同定理1的证明中，由于 $M < {N}_{G}\left( M\right)$ （根据定理3(2)），$M$ 的极大性迫使 ${N}_{G}\left( M\right) = G$ ，即 $M \trianglelefteq G$ 。

Conversely,assume every maximal subgroup of the finite group $G$ is normal. Let $P$ be a Sylow $p$ -subgroup of $G$ . We prove $P \trianglelefteq G$ and conclude that $G$ is nilpotent by Theorem 3(3). If $P$ is not normal in $G$ let $M$ be a maximal subgroup of $G$ containing ${N}_{G}\left( P\right)$ . By hypothesis, $M \trianglelefteq G$ hence by Frattini’s Argument $G = M{N}_{G}\left( P\right)$ . Since ${N}_{G}\left( P\right) \leq M$ we have $M{N}_{G}\left( P\right) = M$ ,a contradiction. This establishes the converse.

相反地，假设有限群 $G$ 的每个极大子群都是正规子群。令 $P$ 为 $G$ 的一个 Sylow $p$ 子群。我们证明 $P \trianglelefteq G$ 并由定理3(3)得出 $G$ 是幂零群。如果 $P$ 在 $G$ 中不是正规子群，令 $M$ 为包含 ${N}_{G}\left( P\right)$ 的 $G$ 的一个极大子群。根据假设，$M \trianglelefteq G$ 因此由 Frattini 论证 $G = M{N}_{G}\left( P\right)$ 。由于 ${N}_{G}\left( P\right) \leq M$ 我们得到 $M{N}_{G}\left( P\right) = M$ ，这是一个矛盾。这证明了逆命题。

Commutators and the Lower Central Series

换位子与下中央列

For the sake of completeness we include the definition of the lower central series of a group and state its relation to the upper central series. Since we shall not be using these results in the future, the proofs are left as (straightforward) exercises.

为了完整性，我们包含群的下中央列的定义，并陈述其与上中央列的关系。由于我们将来不会使用这些结果，证明过程留给读者作为（直接的）练习。

Recall that the commutator of two elements $x,y$ in a group $G$ is defined as

回顾一下，两个元素 $x,y$ 在群 $G$ 中的换位子定义为

and the commutator of two subgroups $H$ and $K$ of $G$ is

以及两个子群 $H$ 和 $K$ 在 $G$ 中的换位子是

Basic properties of commutators and the commutator subgroup were established in Section 5.4.

换位子和换位子子群的基本性质在5.4节中已经建立。

Definition. For any (finite or infinite) group $G$ define the following subgroups inductively:

定义。对于任何（有限或无限）群 $G$ ，以下子群是递归定义的：

The chain of groups

群链

is called the lower central series of $G$ . (The term "lower" indicates that ${G}^{i} \geq {G}^{i + 1}$ .)

被称为 $G$ 的下中央列。（“下”这个术语表示 ${G}^{i} \geq {G}^{i + 1}$ 。）

As with the upper central series we include in the exercises at the end of this section the verification that ${G}^{i}$ is a characteristic subgroup of $G$ for all $i$ . The next theorem shows the relation between the upper and lower central series of a group.

与上中央列一样，我们在本节末尾的练习中包含了验证 ${G}^{i}$ 对于所有 $i$ 是 $G$ 的一个特征子群的验证。下一个定理展示了群的上中央列和下中央列之间的关系。

Theorem 8. A group $G$ is nilpotent if and only if ${G}^{n} = 1$ for some $n \geq 0$ . More precisely, $G$ is nilpotent of class $c$ if and only if $c$ is the smallest nonnegative integer such that ${G}^{c} = 1$ . If $G$ is nilpotent of class $c$ then

定理8。一个群 $G$ 是幂零的当且仅当 ${G}^{n} = 1$ 对于某个 $n \geq 0$ 。更准确地说，$G$ 是阶为 $c$ 的幂零群当且仅当 $c$ 是使得 ${G}^{c} = 1$ 的最小非负整数。如果 $G$ 是阶为 $c$ 的幂零群，那么

Proof: This is proved by a straightforward induction on the length of either the upper or lower central series.

证明：这可以通过对上或下中心列的长度进行直接归纳证明。

The terms of the upper and lower central series do not necessarily coincide in general although in some groups this does occur.

在一般情况下，上中心列和下中心列的项不一定重合，尽管在某些群中确实会发生这种情况。

Remarks:

备注：

(1) If $G$ is abelian,we have already seen that ${G}^{\prime } = {G}^{1} = 1$ so the lower central series terminates in the identity after one term.

(1) 如果 $G$ 是阿贝尔群，我们已经知道 ${G}^{\prime } = {G}^{1} = 1$ ，因此下中心列在第一项后终止于单位元。

(2) As with the upper central series, for any finite group there must, by order considerations,be an integer $n$ such that

(2) 与上中心列一样，对于任何有限群，由于阶数考虑，必须存在一个整数 $n$ ，使得

For non-nilpotent groups, ${G}^{n}$ is a nontrivial subgroup of $G$ . For example,in Section 5.4 we showed that ${S}_{3}^{\prime } = {S}_{3}^{1} = {A}_{3}$ . Since ${S}_{3}$ is not nilpotent,we must have ${S}_{3}^{2} = {A}_{3}$ . In fact

对于非幂零群，${G}^{n}$ 是 $G$ 的一个非平凡子群。例如，在第5.4节中我们证明了 ${S}_{3}^{\prime } = {S}_{3}^{1} = {A}_{3}$ 。由于 ${S}_{3}$ 不是幂零的，我们必须有 ${S}_{3}^{2} = {A}_{3}$ 。实际上

Once two terms in the lower central series are the same, the chain stabilizes at that point i.e.,all terms thereafter are equal to these two. Thus ${S}_{3}^{i} = {A}_{3}$ for all $i \geq 2$ . Note that ${S}_{3}$ is an example where the lower central series has two distinct terms whereas all terms in the upper central series are equal to the identity (in particular, for non-nilpotent groups these series need not have the same length).

一旦下中心列中的两项相同，链在该点稳定，即此后所有的项都等于这两个项。因此 ${S}_{3}^{i} = {A}_{3}$ 对于所有 $i \geq 2$ 。请注意，${S}_{3}$ 是一个例子，其中下中心列有两个不同的项，而所有上中心列的项都等于单位元（特别是，对于非幂零群，这些列的长度可能不相同）。

Solvable Groups and the Derived Series

可解群与导出列

Recall that in Section 3.4 a solvable group was defined as one possessing a series:

回顾在第3.4节中定义的可解群，它是具有以下性质的群：

such that each factor ${H}_{i + 1}/{H}_{i}$ is abelian. We now give another characterization of solvability in terms of a descending series of characteristic subgroups.

使得每个因子 ${H}_{i + 1}/{H}_{i}$ 是阿贝尔群。我们现在用特征子群的降序列来给出可解性的另一个刻画。

Definition. For any group $G$ define the following sequence of subgroups inductively:

定义。对于任何群 $G$ ，以下子群的序列可以递归地定义如下：

This series of subgroups is called the derived or commutator series of $G$ .

这个子群序列被称为导出序列或换位子序列 $G$ 。

The terms of this series are also often written as: ${G}^{\left( 1\right) } = {G}^{\prime },{G}^{\left( 2\right) } = {G}^{\prime \prime }$ ,etc. Again it is left as an exercise to show that each ${G}^{\left( i\right) }$ is characteristic in $G$ for all $i$ .

这个序列的项通常也写作：${G}^{\left( 1\right) } = {G}^{\prime },{G}^{\left( 2\right) } = {G}^{\prime \prime }$ 等等。再次留作练习来证明每个 ${G}^{\left( i\right) }$ 在 $G$ 中对所有的 $i$ 都是特征子群。

It is important to note that although ${G}^{\left( 0\right) } = {G}^{0}$ and ${G}^{\left( 1\right) } = {G}^{1}$ ,it is not in general true that ${G}^{\left( i\right) } = {G}^{i}$ . The difference is that the definition of the $i + {1}^{\text{st }}$ term in the lower central series is the commutator of the ${i}^{\text{th }}$ term with the whole group $G$ whereas the $i + {1}^{\text{st }}$ term in the derived series is the commutator of the ${i}^{\text{th }}$ term with itself. Hence

重要的是要注意，尽管 ${G}^{\left( 0\right) } = {G}^{0}$ 和 ${G}^{\left( 1\right) } = {G}^{1}$ ，但一般来说 ${G}^{\left( i\right) } = {G}^{i}$ 不成立。不同之处在于，下中央序列中 $i + {1}^{\text{st }}$ 项的定义是 ${i}^{\text{th }}$ 项与整个群 $G$ 的换位子，而导出序列中的 $i + {1}^{\text{st }}$ 项是 ${i}^{\text{th }}$ 项与自身的换位子。因此

and the containment can be proper. For example,in $G = {S}_{3}$ we have already seen that ${G}^{1} = {G}^{\prime } = {A}_{3}$ and ${G}^{2} = \left\lbrack {{S}_{3},{A}_{3}}\right\rbrack = {A}_{3}$ ,whereas ${G}^{\left( 2\right) } = \left\lbrack {{A}_{3},{A}_{3}}\right\rbrack = 1\left( {{A}_{3}\text{being}}\right.$ abelian).

包含关系可能是适当的。例如，在 $G = {S}_{3}$ 中我们已经看到 ${G}^{1} = {G}^{\prime } = {A}_{3}$ 和 ${G}^{2} = \left\lbrack {{S}_{3},{A}_{3}}\right\rbrack = {A}_{3}$ ，而 ${G}^{\left( 2\right) } = \left\lbrack {{A}_{3},{A}_{3}}\right\rbrack = 1\left( {{A}_{3}\text{being}}\right.$ （阿贝尔群）。

Theorem 9. A group $G$ is solvable if and only if ${G}^{\left( n\right) } = 1$ for some $n \geq 0$ .

定理9。一个群 $G$ 是可解的当且仅当 ${G}^{\left( n\right) } = 1$ 对于某个 $n \geq 0$ 。

Proof: Assume first that $G$ is solvable and so possesses a series

证明：首先假设 $G$ 是可解的，因此具有一个序列

such that each factor ${H}_{i + 1}/{H}_{i}$ is abelian. We prove by induction that ${G}^{\left( i\right) } \leq {H}_{s - i}$ . This is true for $i = 0$ ,so assume ${G}^{\left( i\right) } \leq {H}_{s - i}$ . Then

使得每个因子 ${H}_{i + 1}/{H}_{i}$ 是阿贝尔群。我们通过归纳法证明 ${G}^{\left( i\right) } \leq {H}_{s - i}$ 。对于 $i = 0$ 这是正确的，所以假设 ${G}^{\left( i\right) } \leq {H}_{s - i}$ 。那么

Since ${H}_{s - i}/{H}_{s - i - 1}$ is abelian,by Proposition 5.7(4), $\left\lbrack {{H}_{s - i},{H}_{s - i}}\right\rbrack \leq {H}_{s - i - 1}$ . Thus ${G}^{\left( i + 1\right) } \leq {H}_{s - i - 1}$ ,which completes the induction. Since ${H}_{0} = 1$ we have ${G}^{\left( s\right) } = 1.$

由于 ${H}_{s - i}/{H}_{s - i - 1}$ 是阿贝尔群，根据命题5.7(4)， $\left\lbrack {{H}_{s - i},{H}_{s - i}}\right\rbrack \leq {H}_{s - i - 1}$ 。因此 ${G}^{\left( i + 1\right) } \leq {H}_{s - i - 1}$ ，这完成了归纳。由于 ${H}_{0} = 1$ 我们有 ${G}^{\left( s\right) } = 1.$

Conversely,if ${G}^{\left( n\right) } = 1$ for some $n \geq 0$ ,Proposition 5.7(4) shows that if we take ${H}_{i}$ to be ${G}^{\left( n - i\right) }$ then ${H}_{i}$ is a normal subgroup of ${H}_{i + 1}$ with abelian quotient,so the derived series itself satisfies the defining condition for solvability of $G$ . This completes the proof.

相反地，如果 ${G}^{\left( n\right) } = 1$ 对于某些 $n \geq 0$ 成立，命题5.7(4)表明如果我们取 ${H}_{i}$ 为 ${G}^{\left( n - i\right) }$，那么 ${H}_{i}$ 是 ${H}_{i + 1}$ 的正规子群，其商群是阿贝尔群，因此导出列本身满足 $G$ 可解性的定义条件。这完成了证明。

If $G$ is solvable,the smallest nonnegative $n$ for which ${G}^{\left( n\right) } = 1$ is called the solvable length of $G$ . The derived series is a series of shortest length whose successive quotients are abelian and it has the additional property that it consists of subgroups that are characteristic in the whole group (as opposed to each just being normal in the next in the initial definition of solvability). Its "intrinsic" definition also makes it easier to work with in many instances, as the following proposition (which reproves some results and exercises from Section 3.4) illustrates.

如果 $G$ 是可解的，那么使得 ${G}^{\left( n\right) } = 1$ 成立的最小非负整数 $n$ 被称为 $G$ 的可解长度。导出列是一系列最短长度的列，其连续商群是阿贝尔群，并且它具有一个额外的性质，即它由在整个群中具有特征性的子群组成（与可解性初始定义中仅在每个下一个群中是正规子群不同）。它的“内在”定义也使得在许多情况下更容易处理，如下面的命题所示（它重新证明了3.4节中的一些结果和练习）。

Proposition 10. Let $G$ and $K$ be groups,let $H$ be a subgroup of $G$ and let $\varphi : G \rightarrow K$ be a surjective homomorphism.

命题10。设 $G$ 和 $K$ 是群，$H$ 是 $G$ 的子群，$\varphi : G \rightarrow K$ 是一个满同态。

(1) ${H}^{\left( i\right) } \leq {G}^{\left( i\right) }$ for all $i \geq 0$ . In particular,if $G$ is solvable,then so is $H$ ,i.e., subgroups of solvable groups are solvable (and the solvable length of $H$ is less than or equal to the solvable length of $G$ ).

(1) ${H}^{\left( i\right) } \leq {G}^{\left( i\right) }$ 对于所有 $i \geq 0$ 成立。特别地，如果 $G$ 是可解的，那么 $H$ 也是可解的，即可解群的子群是可解的（且 $H$ 的可解长度小于或等于 $G$ 的可解长度）。

(2) $\varphi \left( {G}^{\left( i\right) }\right) = {K}^{\left( i\right) }$ . In particular,homomorphic images and quotient groups of solvable groups are solvable (of solvable length less than or equal to that of the domain group).

(2) $\varphi \left( {G}^{\left( i\right) }\right) = {K}^{\left( i\right) }$ 。特别地，可解群的同态像和商群是可解的（其可解长度小于或等于定义域群的长度）。

(3) If $N$ is normal in $G$ and both $N$ and $G/N$ are solvable then so is $G$ .

如果 $N$ 在 $G$ 中是正规的，并且 $N$ 和 $G/N$ 都是可解的，那么 $G$ 也是可解的。

Proof: Part 1 follows from the observation that since $H \leq G$ ,by definition of commutator subgroups, $\left\lbrack {H,H}\right\rbrack \leq \left\lbrack {G,G}\right\rbrack$ ,i.e., ${H}^{\left( 1\right) } \leq {G}^{\left( 1\right) }$ . Then,by induction,

证明：第一部分来自于以下观察，由于 $H \leq G$ ，根据交换子子群的定义，$\left\lbrack {H,H}\right\rbrack \leq \left\lbrack {G,G}\right\rbrack$ ，即 ${H}^{\left( 1\right) } \leq {G}^{\left( 1\right) }$ 。然后，通过归纳，

In particular,if ${G}^{\left( n\right) } = 1$ for some $n$ ,then also ${H}^{\left( n\right) } = 1$ . This establishes (1).

特别地，如果 ${G}^{\left( n\right) } = 1$ 对于某个 $n$ 成立，那么 ${H}^{\left( n\right) } = 1$ 也成立。这证明了（1）。

To prove (2) note that by definition of commutators,

为了证明（2），请注意，根据交换子的定义，

so by induction $\varphi \left( {G}^{\left( i\right) }\right) \leq {K}^{\left( i\right) }$ . Since $\varphi$ is surjective,every commutator in $K$ is the image of a commutator in $G$ ,hence again by induction we obtain equality for all $i$ . Again,if ${G}^{\left( n\right) } = 1$ for some $n$ then ${K}^{\left( n\right) } = 1$ . This proves (2).

因此，通过归纳 $\varphi \left( {G}^{\left( i\right) }\right) \leq {K}^{\left( i\right) }$ 。由于 $\varphi$ 是满射，$K$ 中的每个交换子都是 $G$ 中交换子的像，因此再次通过归纳我们得到对所有 $i$ 的等式。同样，如果 ${G}^{\left( n\right) } = 1$ 对于某个 $n$ 成立，那么 ${K}^{\left( n\right) } = 1$ 。这证明了（2）。

Finally,if $G/N$ and $N$ are solvable,of lengths $n$ and $m$ respectively then by (2) applied to the natural projection $\varphi : G \rightarrow G/N$ we obtain

最后，如果 $G/N$ 和 $N$ 是可解的，长度分别为 $n$ 和 $m$ ，那么通过对自然投影 $\varphi : G \rightarrow G/N$ 应用（2），我们得到

i.e., ${G}^{\left( n\right) } \leq N$ . Thus ${G}^{\left( n + m\right) } = {\left( {G}^{\left( n\right) }\right) }^{\left( m\right) } \leq {N}^{\left( m\right) } = 1$ . Theorem 9 shows that $G$ is solvable, which completes the proof.

即 ${G}^{\left( n\right) } \leq N$ 。因此 ${G}^{\left( n + m\right) } = {\left( {G}^{\left( n\right) }\right) }^{\left( m\right) } \leq {N}^{\left( m\right) } = 1$ 。定理9表明 $G$ 是可解的，这完成了证明。

Some additional conditions under which finite groups are solvable are the following:

以下是有限群可解的一些额外条件：

Theorem 11. Let $G$ be a finite group.

定理11。设 $G$ 是一个有限群。

(1) (Burnside) If $\left| G\right| = {p}^{a}{q}^{b}$ for some primes $p$ and $q$ ,then $G$ is solvable.

（1）（Burnside）如果 $\left| G\right| = {p}^{a}{q}^{b}$ 对于某些素数 $p$ 和 $q$ 成立，那么 $G$ 是可解的。

(2) (Philip Hall) If for every prime $p$ dividing $\left| G\right|$ we factor the order of $G$ as $\left| G\right| = {p}^{a}m$ where $\left( {p,m}\right) = 1$ ,and $G$ has a subgroup of order $m$ ,then $G$ is solvable (i.e.,if for all primes $p,G$ has a subgroup whose index equals the order of a Sylow $p$ -subgroup,then $G$ is solvable — such subgroups are called Sylow $p$ -complements).

(2) (Philip Hall) 如果对于每个素数 $p$ 除以 $\left| G\right|$ 我们将 $G$ 的阶分解为 $\left| G\right| = {p}^{a}m$ 其中 $\left( {p,m}\right) = 1$ ，并且 $G$ 有一个阶为 $m$ 的子群，那么 $G$ 是可解的（即，如果对于所有素数 $p,G$ 有一个指数等于 Sylow $p$ -子群的阶的子群，那么 $G$ 是可解的 —— 这样的子群被称为 Sylow $p$ -补）。

(3) (Feit-Thompson) If $\left| G\right|$ is odd then $G$ is solvable.

(3) (Feit-Thompson) 如果 $\left| G\right|$ 是奇数，那么 $G$ 是可解的。

(4) (Thompson) If for every pair of elements $x,y \in G,\langle x,y\rangle$ is a solvable group, then $G$ is solvable.

(4) (Thompson) 如果对于每对元素 $x,y \in G,\langle x,y\rangle$ 是一个可解群，那么 $G$ 是可解的。

We shall prove Burnside’s Theorem in Chapter 19 and deduce Philip Hall’s generalization of it. As mentioned in Section 3.5, the proof of the Feit-Thompson Theorem takes 255 pages. Thompson’s Theorem was first proved as a consequence of a 475 page paper (that in turn relies ultimately on the Feit-Thompson Theorem).

我们将在第19章证明Burnside定理，并推导出Philip Hall对其的推广。如第3.5节所述，Feit-Thompson定理的证明长达255页。Thompson定理最初是作为一个475页论文的推论证明的（该论文反过来最终依赖于Feit-Thompson定理）。

A Proof of the Fundamental Theorem of Finite Abelian Groups

有限阿贝尔群基本定理的证明

We sketch a group-theoretic proof of the result that every finite abelian group is a direct product of cyclic groups (i.e., Parts 1 and 2 of Theorem 5, Section 5.2) - the Classification of Finitely Generated Abelian Groups (Theorem 3, Section 5.2) will be derived as a consequence of a more general theorem in Chapter 12.

我们概述了一个群论证明，该证明表明每个有限阿贝尔群都是循环群的直积（即，定理5的第1部分和第2部分，第5.2节）—— 第12章中一个更一般定理的推论将导出有限生成阿贝尔群的分类（定理3，第5.2节）。

By Corollary 4 it suffices to prove that for $p$ a prime,any abelian $p$ -group is a direct product of cyclic groups (the divisibility condition in Theorem 5.5 is trivially achieved by reordering factors). Let $A$ be an abelian $p$ -group. We proceed by induction on $\left| A\right|$ .

由推论4可知，只需证明对于$p$是一个素数，任何阿贝尔$p$群都是循环群的直积（定理5.5中的可整除性条件通过重新排序因子即可 trivially 实现）。设$A$是一个阿贝尔$p$群。我们对$\left| A\right|$进行归纳。

If $E$ is an elementary abelian $p$ -group (i.e., ${x}^{p} = 1$ for all $x \in E$ ),we first prove the following result:

如果$E$是一个初等阿贝尔$p$群（即，对于所有$x \in E$），我们首先证明以下结果：

If $x = 1$ ,let $M = E$ . Otherwise let $M$ be a subgroup of $E$ of maximal order subject to the condition that $x$ not be an element of $M$ . If $M$ is not of index $p$ in $E$ ,let $\bar{E} = E/M$ . Then $\bar{E}$ is elementary abelian and there exists $\bar{y} \in \bar{E} - \langle \bar{x}\rangle$ . Since $\bar{y}$ has order $p$ ,we also have $\bar{x} \notin \langle \bar{y}\rangle$ . The complete preimage of $\langle \bar{y}\rangle$ in $E$ is a subgroup of $E$ that does not contain $x$ and whose order is larger than the order of $M$ ,contrary to the choice of $M$ . This proves $\left| {E : M}\right| = p$ ,hence

如果$x = 1$，设$M = E$。否则设$M$是$E$的一个极大阶子群，满足条件$x$不是$M$的元素。如果$M$在$E$中的指数不是$p$，设$\bar{E} = E/M$。那么$\bar{E}$是初等阿贝尔群，且存在$\bar{y} \in \bar{E} - \langle \bar{x}\rangle$。由于$\bar{y}$的阶为$p$，我们还得到$\bar{x} \notin \langle \bar{y}\rangle$。$\langle \bar{y}\rangle$在$E$中的完全逆像是一个不包含$x$的$E$的子群，且其阶大于$M$的阶，与选择$M$相矛盾。这证明了$\left| {E : M}\right| = p$，因此

By the recognition theorem for direct products,Theorem 5.9, $E = M \times \langle x\rangle$ ,as asserted.

由直积的识别定理，即定理5.9，$E = M \times \langle x\rangle$，如所断言。

Now let $\varphi : A \rightarrow A$ be defined by $\varphi \left( x\right) = {x}^{p}$ (see Exercise 7,Section 5.2). Then $\varphi$ is a homomorphism since $A$ is abelian. Denote the kernel of $\varphi$ by $K$ and denote the image of $\varphi$ by $H$ . By definition $K = \left\{ {x \in A \mid {x}^{p} = 1}\right\}$ and $H$ is the subgroup of $A$ consisting of ${p}^{\text{th }}$ powers. Note that both $K$ and $A/H$ are elementary abelian. By the First Isomorphism Theorem

现令 $\varphi : A \rightarrow A$ 由 $\varphi \left( x\right) = {x}^{p}$ 定义（见练习7，第5.2节）。那么 $\varphi$ 是同态，因为 $A$ 是阿贝尔群。记 $\varphi$ 的核为 $K$ ，记 $\varphi$ 的像为 $H$ 。根据定义 $K = \left\{ {x \in A \mid {x}^{p} = 1}\right\}$ ，而 $H$ 是由 $A$ 中的 ${p}^{\text{th }}$ 次幂组成的子群。注意 $K$ 和 $A/H$ 都是初等阿贝尔群。根据第一同构定理

By induction,

通过归纳，

By definition of $\varphi$ ,there exist elements ${g}_{i} \in A$ such that ${g}_{i}^{p} = {h}_{i},1 \leq i \leq r$ . Let ${A}_{0} = \left\langle {{g}_{1},\cdots ,{g}_{r}}\right\rangle$ . It is an exercise to see that

根据 $\varphi$ 的定义，存在元素 ${g}_{i} \in A$ 使得 ${g}_{i}^{p} = {h}_{i},1 \leq i \leq r$ 。令 ${A}_{0} = \left\langle {{g}_{1},\cdots ,{g}_{r}}\right\rangle$ 。验证以下内容是一个练习

(a) ${A}_{0} = \left\langle {g}_{1}\right\rangle \times \cdots \times \left\langle {g}_{r}\right\rangle$ ,

(a) ${A}_{0} = \left\langle {g}_{1}\right\rangle \times \cdots \times \left\langle {g}_{r}\right\rangle$ ，

(b) ${A}_{0}/H = \left\langle {{g}_{1}H}\right\rangle \times \cdots \times \left\langle {{g}_{r}H}\right\rangle$ is elementary abelian of order ${p}^{r}$ ,and

(b) ${A}_{0}/H = \left\langle {{g}_{1}H}\right\rangle \times \cdots \times \left\langle {{g}_{r}H}\right\rangle$ 是阶为 ${p}^{r}$ 的初等阿贝尔群，并且

(c) $H \cap K = \left\langle {h}_{1}^{{p}^{{\alpha }_{1} - 1}}\right\rangle \times \cdots \times \left\langle {h}_{r}^{{p}^{{\alpha }_{r} - 1}}\right\rangle$ is elementary abelian of order ${p}^{r}$ .

(c) $H \cap K = \left\langle {h}_{1}^{{p}^{{\alpha }_{1} - 1}}\right\rangle \times \cdots \times \left\langle {h}_{r}^{{p}^{{\alpha }_{r} - 1}}\right\rangle$ 是阶为 ${p}^{r}$ 的初等阿贝尔群。

If $K$ is contained in $H$ ,then $\left| K\right| = \left| {K \cap H}\right| = {p}^{r} = \left| {{A}_{0} : H}\right|$ . In this case by comparing orders we see that ${A}_{0} = A$ and the theorem is proved. Assume therefore that $K$ is not a subgroup of $H$ and use the bar notation to denote passage to the quotient group $A/H$ . Let $x \in K - H$ ,so $\left| \bar{x}\right| = \left| x\right| = p$ . By the initial remark of the proof applied to the elementary abelian $p$ -group $E = \bar{A}$ ,there is a subgroup $\bar{M}$ of $\bar{A}$ such that

如果 $K$ 包含在 $H$ 中，那么 $\left| K\right| = \left| {K \cap H}\right| = {p}^{r} = \left| {{A}_{0} : H}\right|$ 。在这种情况下，通过比较阶数，我们看到 ${A}_{0} = A$ ，定理得证。因此假设 $K$ 不是 $H$ 的子群，并使用横线符号表示过渡到商群 $A/H$ 。令 $x \in K - H$ ，所以 $\left| \bar{x}\right| = \left| x\right| = p$ 。根据证明初始注释应用于初等阿贝尔 $p$ -群 $E = \bar{A}$ ，存在 $\bar{A}$ 的子群 $\bar{M}$ 使得

If $M$ is the complete preimage in $A$ of $\bar{M}$ ,then since $x$ has order $p$ and $x \notin M$ ,we have $\langle x\rangle \cap M = 1$ . By the recognition theorem for direct products,

如果 $M$ 是 $A$ 中 $\bar{M}$ 的完全逆像，那么由于 $x$ 的阶为 $p$ 并且 $x \notin M$ ，我们有 $\langle x\rangle \cap M = 1$ 。根据直积的识别定理，

By induction, $M$ is a direct product of cyclic groups,hence so is $A$ . This completes the proof.

通过归纳，$M$ 是循环群的直积，因此 $A$ 也是。这完成了证明。

The uniqueness of the decomposition of a finite abelian group into a direct product of cyclic groups (Part 3 of Theorem 5.5) can also be proved by induction using the ${p}^{\mathrm{{th}}}$ - power map (i.e., using Exercise 7, Section 5.2). This is essentially the procedure we follow in Section 12.1 for the uniqueness part of the proof of the Fundamental Theorem of Finitely Generated Abelian Groups.

有限阿贝尔群分解为循环群直积的唯一性（定理5.5的第3部分）也可以通过使用 ${p}^{\mathrm{{th}}}$ -次幂映射（即，使用练习7，第5.2节）的归纳法证明。这基本上是我们第12.1节中证明有限生成阿贝尔群基本定理唯一性部分时遵循的程序。

EXERCISES

练习

1. Prove that ${Z}_{i}\left( G\right)$ is a characteristic subgroup of $G$ for all $i$ .

2. 证明 ${Z}_{i}\left( G\right)$ 是 $G$ 的特征子群，对于所有的 $i$ 。

3. Prove Parts 2 and 4 of Theorem 1 for $G$ a finite nilpotent group,not necessarily a $p$ -group.

4. 对于有限幂零群 $G$ ，不一定是 $p$ -群，证明定理1的第2部分和第4部分。

5. If $G$ is finite prove that $G$ is nilpotent if and only if it has a normal subgroup of each order dividing $\left| G\right| ,$ and is cyclic if and only if it has a unique subgroup of each order dividing $\left| G\right| .$

6. 如果 $G$ 是有限的，证明 $G$ 是幂零的当且仅当它有一个每个除数阶的正常子群，并且当且仅当它有一个每个除数阶的唯一子群时是循环的。

7. Prove that a maximal subgroup of a finite nilpotent group has prime index.

8. 证明有限幂零群的最大子群具有素数指数。

9. Prove Parts 2 and 4 of Theorem 1 for $G$ an infinite nilpotent group.

10. 对于无限幂零群 $G$ ，证明定理1的第2部分和第4部分。

11. Show that if $G/Z\left( G\right)$ is nilpotent then $G$ is nilpotent.

12. 证明如果 $G/Z\left( G\right)$ 是幂零的，那么 $G$ 也是幂零的。

13. Prove that subgroups and quotient groups of nilpotent groups are nilpotent (your proof should work for infinite groups). Give an explicit example of a group $G$ which possesses a normal subgroup $H$ such that both $H$ and $G/H$ are nilpotent but $G$ is not nilpotent.

14. 证明幂零群的子群和商群是幂零的（你的证明应适用于无限群）。给出一个群 $G$ 的显式例子，它具有一个正常子群 $H$ ，使得 $H$ 和 $G/H$ 都是幂零的，但 $G$ 不是幂零的。

15. Prove that if $p$ is a prime and $P$ is a non-abelian group of order ${p}^{3}$ then $\left| {Z\left( P\right) }\right| = p$ and $P/Z\left( P\right) \cong {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}$

16. 证明如果 $p$ 是一个素数且 $P$ 是一个阶为 ${p}^{3}$ 的非阿贝尔群，那么 $\left| {Z\left( P\right) }\right| = p$ 和 $P/Z\left( P\right) \cong {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}$。

17. Prove that a finite group $G$ is nilpotent if and only if whenever $a,b \in G$ with $\left( {\left| a\right| ,\left| b\right| }\right) = 1$ then ${ab} = {ba}$ . [Use Part 4 of Theorem 3.]

18. 证明一个有限群 $G$ 是幂零群当且仅当每当 $a,b \in G$ 且 $\left( {\left| a\right| ,\left| b\right| }\right) = 1$ 时，则 ${ab} = {ba}$ 。[使用定理3的第4部分。]

19. Prove that ${D}_{2n}$ is nilpotent if and only if $n$ is a power of 2. [Use Exercise 9.]

20. 证明 ${D}_{2n}$ 是幂零群当且仅当 $n$ 是2的幂。[使用练习9。]

21. Give another proof of Proposition 5 under the additional assumption that $G$ is abelian by invoking the Fundamental Theorem of Finite Abelian Groups.

22. 在假设 $G$ 是阿贝尔群的基础上，通过引用有限阿贝尔群的基本定理，给出命题5的另一个证明。

23. Find the upper and lower central series for ${A}_{4}$ and ${S}_{4}$ .

24. 找出 ${A}_{4}$ 和 ${S}_{4}$ 的上中心列和下中心列。

25. Find the upper and lower central series for ${A}_{n}$ and ${S}_{n},n \geq 5$ .

26. 找出 ${A}_{n}$ 和 ${S}_{n},n \geq 5$ 的上中心列和下中心列。

27. Prove that ${G}^{i}$ is a characteristic subgroup of $G$ for all $i$ .

28. 证明 ${G}^{i}$ 是 $G$ 的一个特征子群，对所有 $i$ 成立。

29. Prove that ${Z}_{i}\left( {D}_{{2}^{n}}\right) = {D}_{{2}^{n}}^{n - 1 - i}$ .

30. 证明 ${Z}_{i}\left( {D}_{{2}^{n}}\right) = {D}_{{2}^{n}}^{n - 1 - i}$ 。

31. Prove that $\mathbb{Q}$ has no maximal subgroups. [Recall Exercise 21,Section 3.2.]

32. 证明 $\mathbb{Q}$ 没有极大子群。[回顾练习21，第3.2节。]

33. Prove that ${G}^{\left( i\right) }$ is a characteristic subgroup of $G$ for all $i$ .

34. 证明 ${G}^{\left( i\right) }$ 是 $G$ 的一个特征子群，对所有 $i$ 成立。

35. Show that if ${G}^{\prime }/{G}^{\prime \prime }$ and ${G}^{\prime \prime }/{G}^{\prime \prime \prime }$ are both cyclic then ${G}^{\prime \prime } = 1$ . [You may assume ${G}^{\prime \prime \prime } = 1$ . Then $G/{G}^{\prime \prime }$ acts by conjugation on the cyclic group ${G}^{\prime \prime }$ .]

36. 证明如果 ${G}^{\prime }/{G}^{\prime \prime }$ 和 ${G}^{\prime \prime }/{G}^{\prime \prime \prime }$ 都是循环群，那么 ${G}^{\prime \prime } = 1$ 。[你可以假设 ${G}^{\prime \prime \prime } = 1$ 。那么 $G/{G}^{\prime \prime }$ 通过共轭作用在循环群 ${G}^{\prime \prime }$ 上。]

37. Show that there is no group whose commutator subgroup is isomorphic to ${S}_{4}$ . [Use the preceding exercise.]

38. 证明不存在一个群的换位子群同构于 ${S}_{4}$ 。[使用前一个练习。]

39. Let $p$ be a prime,let $P$ be a $p$ -subgroup of the finite group $G$ ,let $N$ be a normal subgroup of $G$ whose order is relatively prime to $p$ and let $\bar{G} = G/N$ . Prove the following:

40. 设 $p$ 为一个质数，$P$ 为有限群 $G$ 的一个 $p$ -子群，$N$ 为 $G$ 的一个正规子群，其阶与 $p$ 互质，并设 $\bar{G} = G/N$ 。证明以下命题：

(a) ${N}_{\bar{G}}\left( \bar{P}\right) = \overline{{N}_{G}\left( P\right) }$ [Use Frattini’s Argument.]

(a) ${N}_{\bar{G}}\left( \bar{P}\right) = \overline{{N}_{G}\left( P\right) }$ [使用弗拉蒂尼论证。]

(b) ${C}_{\bar{G}}\left( \bar{P}\right) = \overline{{C}_{G}\left( P\right) }$ . [Use part (a).]

(b) ${C}_{\bar{G}}\left( \bar{P}\right) = \overline{{C}_{G}\left( P\right) }$ 。[使用部分 (a)。]

For any group $G$ the Frattini subgroup of $G$ (denoted by $\Phi \left( G\right)$ ) is defined to be the intersection of all the maximal subgroups of $G$ (if $G$ has no maximal subgroups,set $\Phi \left( G\right) = G$ ). The next few exercises deal with this important subgroup.

对于任意群 $G$ ，其弗拉蒂尼子群（记作 $\Phi \left( G\right)$ ）定义为所有极大子群的交集（如果 $G$ 没有极大子群，则设 $\Phi \left( G\right) = G$ ）。下面几个练习将讨论这个重要的子群。

21. Prove that $\Phi \left( G\right)$ is a characteristic subgroup of $G$ .

22. 证明 $\Phi \left( G\right)$ 是 $G$ 的一个特征子群。

23. Prove that if $N \trianglelefteq G$ then $\Phi \left( N\right) \leq \Phi \left( G\right)$ . Give an explicit example where this containment does not hold if $N$ is not normal in $G$ .

24. 证明如果 $N \trianglelefteq G$ ，则 $\Phi \left( N\right) \leq \Phi \left( G\right)$ 。给出一个明确的例子，说明如果 $N$ 在 $G$ 中不是正规子群，则这种包含关系不成立。

25. Compute $\Phi \left( {S}_{3}\right) ,\Phi \left( {A}_{4}\right) ,\Phi \left( {S}_{4}\right) ,\Phi \left( {A}_{5}\right)$ and $\Phi \left( {S}_{5}\right)$ .

26. 计算 $\Phi \left( {S}_{3}\right) ,\Phi \left( {A}_{4}\right) ,\Phi \left( {S}_{4}\right) ,\Phi \left( {A}_{5}\right)$ 和 $\Phi \left( {S}_{5}\right)$ 。

27. Say an element $x$ of $G$ is a nongenerator if for every proper subgroup $H$ of $G,\langle x,H\rangle$ is also a proper subgroup of $G$ . Prove that $\Phi \left( G\right)$ is the set of nongenerators of $G$ (here $\left| G\right| > 1)$ .

28. 设 $x$ 是 $G$ 的一个非生成元，如果对于 $G$ 的每个真子群 $H$ ，$G,\langle x,H\rangle$ 也是 $G$ 的一个真子群。证明 $\Phi \left( G\right)$ 是 $G$ 的非生成元集合（这里 $\left| G\right| > 1)$ ）。

29. Let $G$ be a finite group. Prove that $\Phi \left( G\right)$ is nilpotent. [Use Frattini’s Argument to prove that every Sylow subgroup of $\Phi \left( G\right)$ is normal in $G$ .]

30. 设 $G$ 为一个有限群。证明 $\Phi \left( G\right)$ 是幂零群。[使用弗拉蒂尼论证来证明 $\Phi \left( G\right)$ 的每个 Sylow 子群在 $G$ 中是正规子群。]

31. Let $p$ be a prime,let $P$ be a finite $p$ -group and let $\bar{P} = P/\Phi \left( P\right)$ .

32. 设 $p$ 为一个质数，$P$ 为一个有限 $p$ -群，并设 $\bar{P} = P/\Phi \left( P\right)$ 。

(a) Prove that $\bar{P}$ is an elementary abelian $p$ -group. [Show that ${P}^{\prime } \leq \Phi \left( P\right)$ and that ${x}^{p} \in \Phi \left( P\right)$ for all $x \in P$ .]

(a) 证明 $\bar{P}$ 是一个初等阿贝尔 $p$ -群。[证明 ${P}^{\prime } \leq \Phi \left( P\right)$ 且对于所有 $x \in P$ ，${x}^{p} \in \Phi \left( P\right)$ 。]

(b) Prove that if $N$ is any normal subgroup of $P$ such that $P/N$ is elementary abelian then $\Phi \left( P\right) \leq N$ . State this (universal) property in terms of homomorphisms and commutative diagrams.

(b) 证明如果 $N$ 是 $P$ 的一个正规子群，且 $P/N$ 是初等阿贝尔的，那么 $\Phi \left( P\right) \leq N$ 。用同态和交换图的形式陈述这个（通用）性质。

(c) Let $\bar{P}$ be elementary abelian of order ${p}^{r}$ (by (a)). Deduce from Exercise 24 that if $\overline{{x}_{1}},\overline{{x}_{2}},\ldots ,\overline{{x}_{r}}$ are any basis for the $r$ -dimensional vector space $\bar{P}$ over ${\mathbb{F}}_{p}$ and if ${x}_{i}$ is any element of the coset $\overline{{x}_{i}}$ ,then $P = \left\langle {{x}_{1},{x}_{2},\ldots ,{x}_{r}}\right\rangle$ . Show conversely that if ${y}_{1},{y}_{2},\ldots ,{y}_{s}$ is any set of generators for $P$ ,then $s \geq r$ (you may assume that every minimal generating set for an $r$ -dimensional vector space has $r$ elements,i.e., every basis has $r$ elements). Deduce Burnside’s Basis Theorem: a set ${y}_{1},\ldots ,{y}_{s}$ is a minimal generating set for $P$ if and only if $\overline{{y}_{1}},\ldots ,\overline{{y}_{s}}$ is a basis of $\bar{P} = P/\Phi \left( P\right)$ . Deduce that any minimal generating set for $P$ has $r$ elements.

(c) 设 $\bar{P}$ 是阶为 ${p}^{r}$ 的初等阿贝尔群（根据 (a)）。从练习 24 推出，如果 $\overline{{x}_{1}},\overline{{x}_{2}},\ldots ,\overline{{x}_{r}}$ 是 $r$ 维向量空间 $\bar{P}$ 上 ${\mathbb{F}}_{p}$ 的任意基，且如果 ${x}_{i}$ 是 $\overline{{x}_{i}}$ 的陪集中的任意元素，那么 $P = \left\langle {{x}_{1},{x}_{2},\ldots ,{x}_{r}}\right\rangle$ 。反过来证明，如果 ${y}_{1},{y}_{2},\ldots ,{y}_{s}$ 是 $P$ 的任意生成集，那么 $s \geq r$（你可以假设一个 $r$ 维向量空间的每个最小生成集都有 $r$ 个元素，即每个基都有 $r$ 个元素）。推出伯恩赛德基定理：一个集合 ${y}_{1},\ldots ,{y}_{s}$ 是 $P$ 的最小生成集当且仅当 $\overline{{y}_{1}},\ldots ,\overline{{y}_{s}}$ 是 $\bar{P} = P/\Phi \left( P\right)$ 的基。推出 $P$ 的任何最小生成集都有 $r$ 个元素。

(d) Prove that if $P/\Phi \left( P\right)$ is cyclic then $P$ is cyclic. Deduce that if $P/{P}^{\prime }$ is cyclic then so is $P$ .

(d) 证明如果 $P/\Phi \left( P\right)$ 是循环的，那么 $P$ 也是循环的。推出如果 $P/{P}^{\prime }$ 是循环的，那么 $P$ 也是循环的。

(e) Let $\sigma$ be any automorphism of $P$ of prime order $q$ with $q \neq p$ . Show that if $\sigma$ fixes the coset ${x\Phi }\left( P\right)$ then $\sigma$ fixes some element of this coset (note that since $\Phi \left( P\right)$ is characteristic in $P$ every automorphism of $P$ induces an automorphism of $P/\Phi \left( P\right)$ ). [Use the observation that $\sigma$ acts a permutation of order 1 or $q$ on the ${p}^{a}$ elements in the coset ${x\Phi }\left( P\right)$ .]

(e) 设 $\sigma$ 是 $P$ 的任意一个素数阶自同构，且满足 $q \neq p$ 。证明如果 $\sigma$ 固定了陪集 ${x\Phi }\left( P\right)$，那么 $\sigma$ 固定了该陪集的某个元素（注意由于 $\Phi \left( P\right)$ 在 $P$ 中是特征子群，$P$ 的每个自同构都会诱导出 $P/\Phi \left( P\right)$ 的一个自同构）。[利用 $\sigma$ 在陪集 ${x\Phi }\left( P\right)$ 上的 ${p}^{a}$ 元素中作用为阶数为 1 或 $q$ 的置换这一观察结果。]

(f) Use parts (e) and (c) to deduce that every nontrivial automorphism of $P$ of order prime to $p$ induces a nontrivial automorphism on $P/\Phi \left( P\right)$ . Deduce that any group of automorphisms of $P$ which has order prime to $p$ is isomorphic to a subgroup of $\operatorname{Aut}\left( \bar{P}\right) = G{L}_{r}\left( {\mathbb{F}}_{p}\right) .$

(f) 利用部分 (e) 和 (c) 推导出，每个非平凡的自同构 $P$ 的阶与 $p$ 互素时，都会在 $P/\Phi \left( P\right)$ 上诱导出一个非平凡的自同构。推导出任何阶与 $p$ 互素的 $P$ 的自同构群同构于 $\operatorname{Aut}\left( \bar{P}\right) = G{L}_{r}\left( {\mathbb{F}}_{p}\right) .$ 的一个子群。

27. Generalize part (d) of the preceding exercise as follows: let $p$ be a prime,let $P$ be a $p$ -group and let $\bar{P} = P/\Phi \left( P\right)$ be elementary abelian of order ${p}^{r}$ . Prove that $P$ has exactly $\frac{{p}^{r} - 1}{p - 1}$ maximal subgroups. [Since every maximal subgroup of $P$ contains $\Phi \left( P\right)$ ,the maximal subgroups of $P$ are,by the Lattice Isomorphism Theorem,in bijective correspondence with the maximal subgroups of the elementary abelian group $\bar{P}$ . It therefore suffices to show that the number of maximal subgroups of an elementary abelian $p$ -group of order ${p}^{r}$ is as stated above. One way of doing this is to use the result that an abelian group is isomorphic to its dual group (cf. Exercise 14 in Section 5.2) so the number of subgroups of index $p$ equals the number of subgroups of order $p$ .]

28. 将前一个练习的(d)部分推广如下：设 $p$ 为一个质数，$P$ 为一个 $p$ -群，$\bar{P} = P/\Phi \left( P\right)$ 为阶为 ${p}^{r}$ 的初等阿贝尔群。证明 $P$ 有恰好 $\frac{{p}^{r} - 1}{p - 1}$ 个极大子群。[由于 $P$ 的每个极大子群都包含 $\Phi \left( P\right)$ ，根据格同构定理，$P$ 的极大子群与初等阿贝尔群 $\bar{P}$ 的极大子群存在双射对应。因此，只需证明一个阶为 ${p}^{r}$ 的初等阿贝尔 $p$ -群的极大子群的数量如上所述。一种方法是使用这样的结果：一个阿贝尔群与其对偶群同构（参见第5.2节的练习14），因此，指数为 $p$ 的子群数量等于阶为 $p$ 的子群数量。]

29. Prove that if $p$ is a prime and $P = {Z}_{p} \times {Z}_{{p}^{2}}$ then $\left| {\Phi \left( P\right) }\right| = p$ and $P/\Phi \left( P\right) \cong {Z}_{p} \times {Z}_{p}$ . Deduce that $P$ has $p + 1$ maximal subgroups.

30. 证明如果 $p$ 是一个质数且 $P = {Z}_{p} \times {Z}_{{p}^{2}}$ ，那么 $\left| {\Phi \left( P\right) }\right| = p$ 且 $P/\Phi \left( P\right) \cong {Z}_{p} \times {Z}_{p}$ 。推导出 $P$ 有 $p + 1$ 个极大子群。

31. Prove that if $p$ is a prime and $P$ is a non-abelian group of order ${p}^{3}$ then $\Phi \left( P\right) = Z\left( P\right)$ and $P/\Phi \left( P\right) \cong {Z}_{p} \times {Z}_{p}$ . Deduce that $P$ has $p + 1$ maximal subgroups.

32. 证明如果 $p$ 是一个质数且 $P$ 是一个阶为 ${p}^{3}$ 的非阿贝尔群，那么 $\Phi \left( P\right) = Z\left( P\right)$ 且 $P/\Phi \left( P\right) \cong {Z}_{p} \times {Z}_{p}$ 。推导出 $P$ 有 $p + 1$ 个极大子群。

33. Let $p$ be an odd prime,let ${P}_{1} = {Z}_{p} \times {Z}_{{p}^{2}}$ and let ${P}_{2}$ be the non-abelian group of order ${p}^{3}$ which has an element of order ${p}^{2}$ . Prove that ${P}_{1}$ and ${P}_{2}$ have the same lattice of subgroups.

34. 设 $p$ 为一个奇质数，${P}_{1} = {Z}_{p} \times {Z}_{{p}^{2}}$ ，${P}_{2}$ 为阶为 ${p}^{3}$ 的非阿贝尔群，该群有一个阶为 ${p}^{2}$ 的元素。证明 ${P}_{1}$ 和 ${P}_{2}$ 有相同的子群格。

35. For any group $G$ a minimal normal subgroup is a normal subgroup $M$ of $G$ such that the only normal subgroups of $G$ which are contained in $M$ are 1 and $M$ . Prove that every minimal normal subgroup of a finite solvable group is an elementary abelian $p$ -group for some prime $p$ . [If $M$ is a minimal normal subgroup of $G$ ,consider its characteristic subgroups: ${M}^{\prime }$ and $\left. {\left\langle {{x}^{p} \mid x \in M}\right\rangle \text{.}}\right\rbrack$

36. 对于任何群 $G$ ，一个最小正规子群是 $G$ 的一个正规子群 $M$ ，使得 $G$ 中包含在 $M$ 中的唯一正规子群是 1 和 $M$ 。证明有限可解群的所有最小正规子群都是某个素数 $p$ 的初等阿贝尔群 $p$ 。[如果 $M$ 是 $G$ 的最小正规子群，考虑它的特征子群：${M}^{\prime }$ 和 $\left. {\left\langle {{x}^{p} \mid x \in M}\right\rangle \text{.}}\right\rbrack$ ]

37. Prove that every maximal subgroup of a finite solvable group has prime power index. [Let $H$ be a maximal subgroup of $G$ and let $M$ be a minimal normal subgroup of $G$ - cf. the preceding exercise. Apply induction to $G/M$ and consider separately the two cases: $M \leq H$ and $M \nleqslant H$ .]

38. 证明有限可解群的每个极大子群的指数是素数幂。[设 $H$ 是 $G$ 的一个极大子群，设 $M$ 是 $G$ 的最小正规子群 - 参见前一个练习。对 $G/M$ 进行归纳，并分别考虑两种情况：$M \leq H$ 和 $M \nleqslant H$ 。]

39. Let $\pi$ be any set of primes. A subgroup $H$ of a finite group is called a Hall $\pi$ -subgroup of $G$ if the only primes dividing $\left| H\right|$ are in the set $\pi$ and $\left| H\right|$ is relatively prime to $\left| {G : H}\right|$ . (Note that if $\pi = \{ p\}$ ,Hall $\pi$ -subgroups are the same as Sylow $p$ -subgroups. Hall subgroups were introduced in Exercise 10 of Section 3.3). Prove the following generalization of Sylow’s Theorem for solvable groups: if $G$ is a finite solvable group then for every set $\pi$ of primes, $G$ has a Hall $\pi$ -subgroup and any two Hall $\pi$ -subgroups (for the same set $\pi$ ) are conjugate in $G$ . [Fix $\pi$ and proceed by induction on $\left| G\right|$ ,proving both existence and conjugacy at once. Let $M$ be a minimal normal subgroup of $G$ ,so $M$ is a $p$ -group for some prime $p$ . If $p \in \pi$ ,apply induction to $G/M$ . If $p \notin \pi$ ,reduce to the case $\left| G\right| = {p}^{\alpha }n$ , where ${p}^{\alpha } = \left| M\right|$ and $n$ is the order of a Hall $\pi$ -subgroup of $G$ . In this case let $N/M$ be a minimal normal subgroup of $G/M$ ,so $N/M$ is a $q$ -group for some prime $q \neq p$ . Let $Q \in {Sy}{l}_{q}\left( N\right)$ . If $Q \trianglelefteq G$ argue as before with $Q$ in place of $M$ . If $Q$ is not normal in $G$ , use Frattini’s Argument to show ${N}_{G}\left( Q\right)$ is a Hall $\pi$ -subgroup of $G$ and establish conjugacy in this case too.]

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The following result shows how to produce normal $p$ -subgroups of some groups on which the elements of order prime to $p$ act faithfully by conjugation. Exercise 26(f) then applies to restrict these actions and give some information about the structure of the group.

以下结果展示了如何生成某些群的正则 $p$ -子群，这些群的元素对 $p$ 的阶忠实地进行共轭作用。练习 26(f) 然后应用于限制这些作用并给出有关群结构的一些信息。

34. Let $p$ be a prime dividing the order of the finite solvable group $G$ . Assume $G$ has no nontrivial normal subgroups of order prime to $p$ . Let $P$ be the largest normal $p$ -subgroup of $G$ (cf. Exercise 37,Section 4.5). Note that Exercise 31 above shows that $P \neq 1$ . Prove that ${C}_{G}\left( P\right) \leq P$ ,i.e., ${C}_{G}\left( P\right) = Z\left( P\right)$ . [Let $N = {C}_{G}\left( P\right)$ and use the preceding exercise to show $N = Z\left( P\right) \times H$ for some Hall $\pi$ -subgroup $H$ of $N$ -here $\pi$ is the set of all prime divisors of $\left| N\right|$ except for $p$ . Show $H \trianglelefteq G$ to obtain the desired conclusion: $H = 1$ .]

35. 设 $p$ 是有限可解群 $G$ 阶的因子。假设 $G$ 没有除 $p$ 外的阶为素数的非平凡正规子群。设 $P$ 是 $G$ 中最大的正规 $p$ -子群（参见练习 37，第 4.5 节）。注意上面的练习 31 显示 $P \neq 1$ 。证明 ${C}_{G}\left( P\right) \leq P$ ，即 ${C}_{G}\left( P\right) = Z\left( P\right)$ 。[设 $N = {C}_{G}\left( P\right)$ 并使用前一个练习来显示对于某个 Hall $\pi$ -子群 $H$ 的 $N$ -此处 $\pi$ 是除 $p$ 外的所有 $\left| N\right|$ 的素数因子集合，有 $N = Z\left( P\right) \times H$ 。证明 $H \trianglelefteq G$ 以得到所需的结论： $H = 1$ 。]

36. Prove that if $G$ is a finite group in which every proper subgroup is nilpotent,then $G$ is solvable. [Show that a minimal counterexample is simple. Let $M$ and $N$ be distinct maximal subgroups chosen with $\left| {M \cap N}\right|$ as large as possible and apply Part 2 of Theorem 3 to show that $M \cap N = 1$ . Now apply the methods of Exercise 53 in Section 4.5.]

37. 证明如果 $G$ 是一个有限群，其中每个真子群都是幂零的，那么 $G$ 是可解的。[展示一个最小反例是单群。设 $M$ 和 $N$ 是选择的两个不同的极大子群，使得 $\left| {M \cap N}\right|$ 尽可能大，并应用定理 3 的第二部分来显示 $M \cap N = 1$ 。现在应用第 4.5 节练习 53 的方法。]

38. Let $p$ be a prime,let $V$ be a nonzero finite dimensional vector space over the field of $p$ elements and let $\varphi$ be an element of ${GL}\left( V\right)$ of order a power of $p$ (i.e., $V$ is a nontrivial elementary abelian $p$ -group and $\varphi$ is an automorphism of $V$ of $p$ -power order). Prove that there is some nonzero element $v \in V$ such that $\varphi \left( v\right) = v$ ,i.e., $\varphi$ has a nonzero fixed point on $V$ .

39. 设 $p$ 为一个质数，$V$ 为定义在 $p$ 元素域上的非零有限维向量空间，$\varphi$ 为 ${GL}\left( V\right)$ 中的一个元素，其阶为 $p$ 的幂次（即，$V$ 是一个非平凡的基本阿贝尔 $p$ -群，$\varphi$ 是 $V$ 的一个阶为 $p$ 幂次的自同构）。证明存在一个非零元素 $v \in V$ 使得 $\varphi \left( v\right) = v$ ，即 $\varphi$ 在 $V$ 上有一个非零固定点。

40. Let $V$ be a finite dimensional vector space over the field of 2 elements and let $\varphi$ be an element of ${GL}\left( V\right)$ of order 2 . (i.e., $V$ is a nontrivial elementary abelian 2-group and $\varphi$ is an automorphism of $V$ of order 2). Prove that the map $v \mapsto v + \varphi \left( v\right)$ is a homomorphism from $V$ to itself. Show that every element in the image of this map is fixed by $\varphi$ . Deduce that the subspace of elements of $V$ which are fixed by $\varphi$ has dimension $\geq \frac{1}{2}$ (dimension $V$ ). (Note that if $G$ is the semidirect product of $V$ with $\langle \varphi \rangle$ ,where $V \trianglelefteq G$ and $\varphi$ acts by conjugation on $V$ by sending each $v \in V$ to $\varphi \left( v\right)$ ,then the fixed points of $\varphi$ on $V$ are ${C}_{V}\left( \varphi \right)$ and the above map is simply the commutator map: $v \mapsto \left\lbrack {v,\varphi }\right\rbrack$ . In this terminology the problem is to show that ${\left| {C}_{V}\left( \varphi \right) \right| }^{2} \geq \left| V\right|$ .)

41. 设 $V$ 为定义在2元素域上的有限维向量空间，$\varphi$ 为 ${GL}\left( V\right)$ 中的一个元素，其阶为2（即，$V$ 是一个非平凡的基本阿贝尔2-群，$\varphi$ 是 $V$ 的一个阶为2的自同构）。证明映射 $v \mapsto v + \varphi \left( v\right)$ 是从 $V$ 到其自身的同态。证明这个映射的像中的每个元素都被 $\varphi$ 固定。推导出由 $V$ 中被 $\varphi$ 固定的元素组成的子空间的维数为 $\geq \frac{1}{2}$（$V$ 的维数）。（注意，如果 $G$ 是 $V$ 与 $\langle \varphi \rangle$ 的半直积，其中 $V \trianglelefteq G$ 并且 $\varphi$ 通过共轭作用在 $V$ 上，将每个 $v \in V$ 映射到 $\varphi \left( v\right)$ ，那么 $\varphi$ 在 $V$ 上的固定点是 ${C}_{V}\left( \varphi \right)$ ，上述映射仅仅是交换子映射：$v \mapsto \left\lbrack {v,\varphi }\right\rbrack$。在这个术语中，问题是要证明 ${\left| {C}_{V}\left( \varphi \right) \right| }^{2} \geq \left| V\right|$。）

42. Use the preceding exercise to prove that if $P$ is a 2-group which has a cyclic center and $M$ is a subgroup of index 2 in $P$ ,then the center of $M$ has rank $\leq 2$ . [The group $G/M$ of order 2 acts by conjugation on the ${\mathbb{F}}_{2}$ vector space: $\left\{ {z \in Z\left( M\right) \mid {z}^{2} = 1}\right\}$ and the fixed points of this action are in the center of $P$ .]

43. 使用前一个练习来证明，如果 $P$ 是一个具有循环中心的2-群，并且 $M$ 是 $P$ 的一个指数为2的子群，那么 $M$ 的中心是 $\leq 2$ 阶的。 [阶为2的群 $G/M$ 通过共轭作用于 ${\mathbb{F}}_{2}$ 向量空间：$\left\{ {z \in Z\left( M\right) \mid {z}^{2} = 1}\right\}$，这种作用的固定点在 $P$ 的中心中。]