5.5 半直积

5.5 SEMIDIRECT PRODUCTS

5.5 半直积

In this section we study the "semidirect product" of two groups $H$ and $K$ ,which is a generalization of the notion of the direct product of $H$ and $K$ obtained by relaxing the requirement that both $H$ and $K$ be normal. This construction will enable us (in certain circumstances) to build a "larger" group from the groups $H$ and $K$ in such a way that $G$ contains subgroups isomorphic to $H$ and $K$ ,respectively,as in the case of direct products. In this case the subgroup $H$ will be normal in $G$ but the subgroup $K$ will not necessarily be normal (as it is for direct products). Thus, for instance, we shall be able to construct non-abelian groups even if $H$ and $K$ are abelian. This construction will allow us to enlarge considerably the set of examples of groups at our disposal. As in the preceding section, we shall then prove a recognition theorem that will enable us to decompose some familiar groups into smaller "factors," from which we shall be able to derive some classification theorems.

在这一节中，我们研究两个群 $H$ 和 $K$ 的“半直积”，这是通过放宽要求 $H$ 和 $K$ 都必须是正规群的条件，对 $H$ 和 $K$ 的直积概念的推广。这种构造将使我们（在某些情况下）能够从群 $H$ 和 $K$ 构造出一个“更大”的群，使得 $G$ 分别包含与 $H$ 和 $K$ 同构的子群，就像直积的情况一样。在这种情况下，子群 $H$ 在 $G$ 中将是正规子群，但子群 $K$ 不一定是正规子群（这与直积的情况不同）。因此，例如，即使 $H$ 和 $K$ 是阿贝尔群，我们也能够构造出非阿贝尔群。这种构造将使我们能够大大扩展可用的群例子集合。与前一节一样，我们将证明一个识别定理，这将使我们能够将一些熟悉的群分解为较小的“因子”，从而推导出一些分类定理。

By way of motivation suppose we already have a group $G$ containing subgroups $H$ and $K$ such that

出于动机的考虑，假设我们已经有一个包含子群 $G$ 和 $H$ 以及 $K$ 的群，使得

(a) $H \trianglelefteq G$ (but $K$ is not necessarily normal in $G$ ),and

(a) $H \trianglelefteq G$（但 $K$ 在 $G$ 中不一定是正规子群），并且

(b) $H \cap K = 1$ .

(b) $H \cap K = 1$。

It is still true that ${HK}$ is a subgroup of $G$ (Corollary 3.15) and,by Proposition 8, every element of ${HK}$ can be written uniquely as a product ${hk}$ ,for some $h \in H$ and $k \in K$ ,i.e.,there is a bijection between ${HK}$ and the collection of ordered pairs(h,k), given by ${hk} \mapsto \left( {h,k}\right)$ (so the group $H$ appears as the set of elements(h,1)and $K$ appears as the set of elements $\left( {1,k}\right) )$ . Given two elements ${h}_{1}{k}_{1}$ and ${h}_{2}{k}_{2}$ of ${HK}$ ,we first see how to write their product (in $G$ ) in the same form:

仍然是真的，${HK}$ 是 $G$ 的子群（推论3.15），并且根据命题8，${HK}$ 的每个元素都可以唯一地写成乘积 ${hk}$ 的形式，对于某些 $h \in H$ 和 $k \in K$，即存在一个双射 ${HK}$ 和有序对 (h,k) 的集合之间，由 ${hk} \mapsto \left( {h,k}\right)$ 给出（因此群 $H$ 出现在 (h,1) 的元素集合中，而 $K$ 出现在 $\left( {1,k}\right) )$ 的元素集合中）。给定 ${HK}$ 中的两个元素 ${h}_{1}{k}_{1}$ 和 ${h}_{2}{k}_{2}$，我们首先看看如何将它们的乘积（在 $G$ 中）写成相同的形式：

where ${h}_{3} = {h}_{1}\left( {{k}_{1}{h}_{2}{k}_{1}^{-1}}\right)$ and ${k}_{3} = {k}_{1}{k}_{2}$ . Note that since $H \trianglelefteq G,{k}_{1}{h}_{2}{k}_{1}^{-1} \in H$ ,so ${h}_{3} \in H$ and ${k}_{3} \in K$ .

其中 ${h}_{3} = {h}_{1}\left( {{k}_{1}{h}_{2}{k}_{1}^{-1}}\right)$ 和 ${k}_{3} = {k}_{1}{k}_{2}$。注意由于 $H \trianglelefteq G,{k}_{1}{h}_{2}{k}_{1}^{-1} \in H$，所以 ${h}_{3} \in H$ 和 ${k}_{3} \in K$。

These calculations were predicated on the assumption that there already existed a group $G$ containing subgroups $H$ and $K$ with $H \trianglelefteq G$ and $H \cap K = 1$ . The basic idea of the semidirect product is to turn this construction around, namely start with two (abstract) groups $H$ and $K$ and try to define a group containing (an isomorphic copy of) them in such a way that (a) and (b) above hold. To do this, we write equation (1), which defines the multiplication of elements in our group, in a way that makes sense even if we do not already know there is a group containing $H$ and $K$ as above. The point is that ${k}_{3}$ in equation (1) is obtained only from multiplication in $K$ (namely ${k}_{1}{k}_{2}$ ) and ${h}_{3}$ is obtained from multiplying ${h}_{1}$ and ${k}_{1}{h}_{2}{k}_{1}^{-1}$ in $H$ . If we can understand where the element ${k}_{1}{h}_{2}{k}_{1}^{-1}$ arises (in terms of $H$ and $K$ and without reference to $G$ ),then the group ${HK}$ will have been described entirely in terms of $H$ and $K$ . We can then use this description to define the group ${HK}$ using equation (1) to define the multiplication.

这些计算是基于这样一个假设：已经存在一个包含子群 $G$ 和 $H$ 的群 $K$ ，并且具有 $H \trianglelefteq G$ 和 $H \cap K = 1$ 。半直积的基本思想是将这个构造反过来，即从两个（抽象）群 $H$ 和 $K$ 开始，尝试定义一个包含（它们的同构副本）的群，以便上述的（a）和（b）成立。为此，我们写出方程（1），该方程定义了群中元素乘法，即使我们不知道是否存在一个包含上述 $H$ 和 $K$ 的群，这种方式也是有意义的。关键在于方程（1）中的 ${k}_{3}$ 只是通过 $K$ 中的乘法（即 ${k}_{1}{k}_{2}$ ）获得的，而 ${h}_{3}$ 是通过在 $H$ 中乘以 ${h}_{1}$ 和 ${k}_{1}{h}_{2}{k}_{1}^{-1}$ 获得的。如果我们能够理解元素 ${k}_{1}{h}_{2}{k}_{1}^{-1}$ 的来源（即 $H$ 和 $K$ 的关系，而不参考 $G$ ），那么群 ${HK}$ 将完全用 $H$ 和 $K$ 来描述。然后我们可以使用这个描述来定义使用方程（1）来定义乘法的群 ${HK}$ 。

Since $H$ is normal in $G$ ,the group $K$ acts on $H$ by conjugation:

由于 $H$ 在 $G$ 中是正规群，所以群 $K$ 通过共轭作用于 $H$ ：

(we use the symbol - to emphasize the action) so that (1) can be written

（我们使用符号 - 来强调这种作用），因此（1）可以写成

The action of $K$ on $H$ by conjugation gives a homomorphism $\varphi$ of $K$ into $\operatorname{Aut}\left( H\right)$ ,so (2) shows that the multiplication in ${HK}$ depends only on the multiplication in $H$ ,the multiplication in $K$ and the homomorphism $\varphi$ ,hence is defined intrinsically in terms of $H$ and $K$ .

由 $K$ 通过共轭作用在 $H$ 上给出一个同态 $\varphi$ 到 $\operatorname{Aut}\left( H\right)$ 的映射，因此 (2) 表明 ${HK}$ 中的乘法仅依赖于 $H$ 中的乘法、$K$ 中的乘法以及同态 $\varphi$，因此可以内在地用 $H$ 和 $K$ 来定义。

We now use this interpretation to define a group given two groups $H$ and $K$ and a homomorphism $\varphi$ from $K$ to $\operatorname{Aut}\left( H\right)$ (which will turn out to define conjugation in the resulting group).

我们现在使用这种解释来定义一个群，给定两个群 $H$ 和 $K$ 以及从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的同态 $\varphi$（这将最终定义结果群中的共轭）。

Theorem 10. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into $\operatorname{Aut}\left( H\right)$ . Let $\cdot$ denote the (left) action of $K$ on $H$ determined by $\varphi$ . Let $G$ be the set of ordered pairs(h,k)with $h \in H$ and $k \in K$ and define the following multiplication on $G$ :

定理 10。设 $H$ 和 $K$ 是群，$\varphi$ 是从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的同态。令 $\cdot$ 表示由 $\varphi$ 决定的 $K$ 在 $H$ 上的（左）作用。令 $G$ 表示由满足 $h \in H$ 和 $k \in K$ 的有序对 (h,k) 组成的集合，并在 $G$ 上定义以下乘法：

(1) This multiplication makes $G$ into a group of order $\left| G\right| = \left| H\right| \left| K\right|$ .

(1) 这种乘法使得 $G$ 成为一个阶数为 $\left| G\right| = \left| H\right| \left| K\right|$ 的群。

(2) The sets $\{ \left( {h,1}\right) \mid h \in H\}$ and $\{ \left( {1,k}\right) \mid k \in K\}$ are subgroups of $G$ and the maps $h \mapsto \left( {h,1}\right)$ for $h \in H$ and $k \mapsto \left( {1,k}\right)$ for $k \in K$ are isomorphisms of these subgroups with the groups $H$ and $K$ respectively:

(2) 集合 $\{ \left( {h,1}\right) \mid h \in H\}$ 和 $\{ \left( {1,k}\right) \mid k \in K\}$ 是 $G$ 的子群，映射 $h \mapsto \left( {h,1}\right)$ 对于 $h \in H$ 和 $k \mapsto \left( {1,k}\right)$ 对于 $k \in K$ 是这些子群与群 $H$ 和 $K$ 分别的同构。

Identifying $H$ and $K$ with their isomorphic copies in $G$ described in (2) we have

将 $H$ 和 $K$ 与 (2) 中描述的 $G$ 中的同构副本识别为相同的，我们有

(3) $H \trianglelefteq G$

(3) $H \trianglelefteq G$

(4) $H \cap K = 1$

(4) $H \cap K = 1$

(5) for all $h \in H$ and $k \in K,{kh}{k}^{-1} = k \cdot h = \varphi \left( k\right) \left( h\right)$ .

(5) 对于所有的 $h \in H$ 和 $k \in K,{kh}{k}^{-1} = k \cdot h = \varphi \left( k\right) \left( h\right)$。

Proof: It is straightforward to check that $G$ is a group under this multiplication using the fact that $\cdot$ is an action of $K$ on $H$ . For example,the associative law is verified as follows:

证明：通过直接验证可知 $G$ 在此乘法下是一个群，利用 $\cdot$ 是 $K$ 对 $H$ 的作用这一事实。例如，结合律的验证如下：

for all $\left( {a,x}\right) ,\left( {b,y}\right) ,\left( {c,z}\right) \in G$ . We leave as an exercise the verification that(1,1)is the identity of $G$ and that

对于所有 $\left( {a,x}\right) ,\left( {b,y}\right) ,\left( {c,z}\right) \in G$ 。我们将验证 (1,1) 是 $G$ 的单位元以及

for each $\left( {h,k}\right) \in G$ . The order of the group $G$ is clearly the product of the orders of $H$ and $K$ ,which proves (1).

对于每个 $\left( {h,k}\right) \in G$ 。显然，群 $G$ 的阶是 $H$ 和 $K$ 阶的乘积，这证明了 (1)。

Let $\widetilde{H} = \{ \left( {h,1}\right) \mid h \in H\}$ and $\widetilde{K} = \{ \left( {1,k}\right) \mid k \in K\}$ . We have

设 $\widetilde{H} = \{ \left( {h,1}\right) \mid h \in H\}$ 和 $\widetilde{K} = \{ \left( {1,k}\right) \mid k \in K\}$ 。我们有

for all $a,b \in H$ and

对于所有 $a,b \in H$ 和

for all $x,y \in K$ ,which show that $\widetilde{H}$ and $\widetilde{K}$ are subgroups of $G$ and that the maps in (2) are isomorphisms.

对于所有 $x,y \in K$ ，这表明 $\widetilde{H}$ 和 $\widetilde{K}$ 是 $G$ 的子群，并且 (2) 中的映射是同构。

It is clear that $\widetilde{H} \cap \widetilde{K} = 1$ ,which is (4). Now,

显然 $\widetilde{H} \cap \widetilde{K} = 1$ ，这就是 (4)。现在，

so that identifying(h,1)with $h$ and(1,k)with $k$ by the isomorphisms in (2) we have ${kh}{k}^{-1} = k \cdot h$ ,which is (5).

使得通过 (2) 中的同构将 (h,1) 与 $h$ 和 (1,k) 与 $k$ 相识别，我们得到 ${kh}{k}^{-1} = k \cdot h$ ，这就是 (5)。

Finally,we have just seen that (under the identifications in (2)) $K \leq {N}_{G}\left( H\right)$ . Since $G = {HK}$ and certainly $H \leq {N}_{G}\left( H\right)$ ,we have ${N}_{G}\left( H\right) = G$ ,i.e., $H \trianglelefteq G$ ,which proves (3) and completes the proof.

最后，我们刚刚看到（在 (2) 中的识别下） $K \leq {N}_{G}\left( H\right)$ 。由于 $G = {HK}$ 并且显然 $H \leq {N}_{G}\left( H\right)$ ，我们有 ${N}_{G}\left( H\right) = G$ ，即 $H \trianglelefteq G$ ，这证明了 (3) 并完成了证明。

Definition. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into $\operatorname{Aut}\left( H\right)$ . The group described in Theorem 10 is called the semidirect product of $H$ and $K$ with respect to $\varphi$ and will be denoted by $H{ \rtimes }_{\varphi }K$ (when there is no danger of confusion we shall simply write $H \rtimes K$ ).

定义。设 $H$ 和 $K$ 是群，$\varphi$ 是从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的同态。定理 10 中描述的群称为 $H$ 和 $K$ 关于 $\varphi$ 的半直积，记为 $H{ \rtimes }_{\varphi }K$（当不存在混淆的危险时，我们将简单地写作 $H \rtimes K$ ）。

The notation is chosen to remind us that the copy of $H$ in $H \rtimes K$ is the normal "factor" and that the construction of a semidirect product is not symmetric in $H$ and $K$ (unlike that of a direct product). Before giving some examples we clarify exactly when the semidirect product of $H$ and $K$ is their direct product (in particular,we see that direct products are a special case of semidirect products). See also Exercise 1.

该表示法的选择是为了提醒我们，$H$ 在 $H \rtimes K$ 中的副本是正常的“因子”，并且半直积的构造在 $H$ 和 $K$ 之间不是对称的（与直积不同）。在给出一些例子之前，我们明确说明在什么情况下 $H$ 和 $K$ 的半直积是它们的直积（特别是，我们发现直积是半直积的特殊情况）。参见练习1。

Proposition 11. Let $H$ and $K$ be groups and let $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ be a homomorphism. Then the following are equivalent:

命题11。设 $H$ 和 $K$ 是群，$\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ 是同态。那么以下条件是等价的：

(1) the identity (set) map between $H \rtimes K$ and $H \times K$ is a group homomorphism (hence an isomorphism)

（1）$H \rtimes K$ 和 $H \times K$ 之间的恒等（集合）映射是群同态（因此是同构）

(2) $\varphi$ is the trivial homomorphism from $K$ into $\operatorname{Aut}\left( H\right)$

（2）$\varphi$ 是从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的平凡同态

(3) $K \trianglelefteq H \rtimes K$ .

（3）$K \trianglelefteq H \rtimes K$ 。

Proof: (1) $\Rightarrow$ (2) By definition of the group operation in $H \rtimes K$

证明：（1）$\Rightarrow$ （2）根据 $H \rtimes K$ 中群运算的定义

for all ${h}_{1},{h}_{2} \in H$ and ${k}_{1},{k}_{2} \in K$ . By assumption (1), $\left( {{h}_{1},{k}_{1}}\right) \left( {{h}_{2},{k}_{2}}\right) = \left( {{h}_{1}{h}_{2},{k}_{1}{k}_{2}}\right)$ . Equating the first factors of these ordered pairs gives ${k}_{1} \cdot {h}_{2} = {h}_{2}$ for all ${h}_{2} \in H$ and all ${k}_{1} \in K$ ,i.e., $K$ acts trivially on $H$ . This is (2).

对于所有的 ${h}_{1},{h}_{2} \in H$ 和 ${k}_{1},{k}_{2} \in K$ 。根据假设（1），$\left( {{h}_{1},{k}_{1}}\right) \left( {{h}_{2},{k}_{2}}\right) = \left( {{h}_{1}{h}_{2},{k}_{1}{k}_{2}}\right)$ 。将这些有序对的第一因子相等，得到 ${k}_{1} \cdot {h}_{2} = {h}_{2}$ 对于所有的 ${h}_{2} \in H$ 和所有的 ${k}_{1} \in K$ ，即 $K$ 在 $H$ 上作用是平凡的。这是（2）。

$\left( 2\right) \Rightarrow \left( 3\right)$ If $\varphi$ is trivial,then the action of $K$ on $H$ is trivial,so that the elements of $H$ commute with those of $K$ by Theorem 10(5). In particular, $H$ normalizes $K$ . Since $K$ normalizes itself, $G = {HK}$ normalizes $K$ ,which is (3).

如果 $\varphi$ 是平凡的，那么 $K$ 在 $H$ 上的作用也是平凡的，因此根据定理10（5），$H$ 的元素与 $K$ 的元素交换。特别是，$H$ 正规化 $K$ 。由于 $K$ 正规化它自己，$G = {HK}$ 也正规化 $K$ ，这是（3）。

(3) $\Rightarrow$ (1) If $K$ is normal in $H \rtimes K$ then (as in the proof of Theorem 9) for all $h \in H$ and $k \in K,\left\lbrack {h,k}\right\rbrack \in H \cap K = 1$ . Thus ${hk} = {kh}$ and the action of $K$ on $H$ is trivial. The multiplication in the semidirect product is then the same as that in the direct product:

(3) $\Rightarrow$ (1) 如果 $K$ 在 $H \rtimes K$ 中是正规的（如同定理9的证明中那样），那么对于所有的 $h \in H$ 和 $k \in K,\left\lbrack {h,k}\right\rbrack \in H \cap K = 1$ 。因此 ${hk} = {kh}$ ，且 $K$ 对 $H$ 的作用是平凡的。半直积中的乘法与直积中的乘法相同：

for all ${h}_{1},{h}_{2} \in H$ and ${k}_{1},{k}_{2} \in K$ . This gives (1) and completes the proof.

对于所有的 ${h}_{1},{h}_{2} \in H$ 和 ${k}_{1},{k}_{2} \in K$ 。这给出了（1）并完成了证明。

Examples

示例

In all examples $H$ and $K$ are groups and $\varphi$ is a homomorphism from $K$ into Aut(H)with associated action of $K$ on $H$ denoted by a dot. Let $G = H \rtimes K$ and as in Theorem 10 we identify $H$ and $K$ as subgroups of $G$ . We shall use Propositions 4.16 and 4.17 to determine homomorphisms $\varphi$ for some specific groups $H$ . In each of the following examples the proof that $\varphi$ is a homomorphism is easy (since $K$ will often be cyclic) so the details are omitted.

在所有示例中 $H$ 和 $K$ 是群，$\varphi$ 是从 $K$ 到 Aut(H) 的同态，其相关的 $K$ 对 $H$ 的作用用点表示。设 $G = H \rtimes K$ ，如同定理10中，我们将 $H$ 和 $K$ 识别为 $G$ 的子群。我们将使用命题4.16和4.17来确定某些特定群 $H$ 的同态 $\varphi$ 。在以下每个示例中，证明 $\varphi$ 是同态是容易的（因为 $K$ 通常将是循环的），因此省略了细节。

(1) Let $H$ be any abelian group (even of infinite order) and let $K = \langle x\rangle \cong {Z}_{2}$ be the group of order 2. Define $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ by mapping $x$ to the automorphism of inversion on $H$ so that the associated action is $x \cdot h = {h}^{-1}$ ,for all $h \in H$ . Then $G$ contains the subgroup $H$ of index 2 and

(1) 设 $H$ 是任意的阿贝尔群（甚至是无限阶的），设 $K = \langle x\rangle \cong {Z}_{2}$ 是阶为2的群。定义 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ 通过将 $x$ 映射到 $H$ 上的自同构求逆，使得相关的作用是 $x \cdot h = {h}^{-1}$ ，对于所有的 $h \in H$ 。那么 $G$ 包含了指标为2的子群 $H$ 和

Of particular interest is the case when $H$ is cyclic: if $H = {Z}_{n}$ ,one recognizes $G$ as ${D}_{2n}$ and if $H = \mathbb{Z}$ we denote $G$ by ${D}_{\infty }$ .

特别值得注意的是当 $H$ 是循环的情况：如果 $H = {Z}_{n}$ ，则识别 $G$ 为 ${D}_{2n}$ ，如果 $H = \mathbb{Z}$ 我们将 $G$ 记作 ${D}_{\infty }$ 。

(2) We can generalize the preceding example in a number of ways. One way is to let $H$ be any abelian group and to let $K = \langle x\rangle \cong {Z}_{2n}$ be cyclic of order ${2n}$ . Define $\varphi$ again by mapping $x$ to inversion,so that ${x}^{2}$ acts as the identity on $H$ . In $G,{xh}{x}^{-1} = {h}^{-1}$ and ${x}^{2}h{x}^{-2} = h$ for all $h \in H$ . Thus ${x}^{2} \in Z\left( G\right)$ . In particular,if $H = {Z}_{3}$ and $K = {Z}_{4}$ , $G$ is a non-abelian group of order 12 which is not isomorphic to ${A}_{4}$ or ${D}_{12}$ (since its Sylow 2-subgroup, $K$ ,is cyclic of order 4).

（2）我们可以通过多种方式推广前面的例子。一种方式是让 $H$ 成为任意的阿贝尔群，并让 $K = \langle x\rangle \cong {Z}_{2n}$ 成为阶为 ${2n}$ 的循环群。再次通过映射 $x$ 到逆元来定义 $\varphi$ ，使得 ${x}^{2}$ 在 $H$ 上作为恒等映射。在 $G,{xh}{x}^{-1} = {h}^{-1}$ 和 ${x}^{2}h{x}^{-2} = h$ 中对于所有 $h \in H$ 成立。因此 ${x}^{2} \in Z\left( G\right)$ 。特别地，如果 $H = {Z}_{3}$ 且 $K = {Z}_{4}$ ，则 $G$ 是一个阶为12的非阿贝尔群，且不与 ${A}_{4}$ 或 ${D}_{12}$ 同构（因为它的Sylow 2-子群 $K$ 是阶为4的循环群）。

(3) Following up on the preceding example let $H = \langle h\rangle \cong {Z}_{{2}^{n}}$ and let $K = \langle x\rangle \cong {Z}_{4}$ with ${xh}{x}^{-1} = {h}^{-1}$ in $G$ . As noted above, ${x}^{2} \in Z\left( G\right)$ . Since $x$ inverts $h$ (i.e.,inverts $H),x$ inverts the unique subgroup $\langle z\rangle$ of order 2 in $H$ ,where $z = {h}^{{2}^{n - 1}}$ . Thus ${xz}{x}^{-1} = {z}^{-1} = z$ ,so $x$ centralizes $z$ . It follows that $z \in Z\left( G\right)$ . Thus ${x}^{2}z \in Z\left( G\right)$ hence $\left\langle {{x}^{2}z}\right\rangle \trianglelefteq G$ . Let $\bar{G} = G/\left\langle {{x}^{2}z}\right\rangle$ . Since ${x}^{2}$ and $z$ are distinct commuting elements of order 2,the order of ${x}^{2}z$ is 2,so $\left| \bar{G}\right| = \frac{1}{2}\left| G\right| = {2}^{n + 1}$ . By factoring out the product ${x}^{2}z$ to form $\bar{G}$ we identify ${x}^{2}$ and ${h}^{{2}^{n - 1}}$ in the quotient. In particular,when $n = 2$ ,both $\bar{x}$ and $\bar{h}$ have order 4, $\bar{x}$ inverts $\bar{h}$ and ${\bar{h}}^{2} = {\bar{x}}^{2}$ . It follows that $\bar{G} \cong {Q}_{8}$ in this case. In general,one can check that $\bar{G}$ has a unique subgroup of order 2 (namely ( ${\bar{x}}^{2}$ )) which equals the center of $\overline{G}$ . The group $\overline{G}$ is called the generalized quaternion group of order ${2}^{n + 1}$ and is denoted by ${Q}_{{2}^{n + 1}}$ :

(3) 继续前一个例子，设 $H = \langle h\rangle \cong {Z}_{{2}^{n}}$ 并且设 $K = \langle x\rangle \cong {Z}_{4}$ ，其中 ${xh}{x}^{-1} = {h}^{-1}$ 在 $G$ 中。如上所述，${x}^{2} \in Z\left( G\right)$ 。由于 $x$ 逆 $h$（即逆 $H),x$ 逆 $H$ 中阶为2的唯一子群 $\langle z\rangle$ ，其中 $z = {h}^{{2}^{n - 1}}$ ）。因此 ${xz}{x}^{-1} = {z}^{-1} = z$ ，所以 $x$ 使 $z$ 居中。由此得出 $z \in Z\left( G\right)$ 。因此 ${x}^{2}z \in Z\left( G\right)$ ，因此 $\left\langle {{x}^{2}z}\right\rangle \trianglelefteq G$ 。设 $\bar{G} = G/\left\langle {{x}^{2}z}\right\rangle$ 。由于 ${x}^{2}$ 和 $z$ 是两个不同的阶为2的交换元素，${x}^{2}z$ 的阶是2，所以 $\left| \bar{G}\right| = \frac{1}{2}\left| G\right| = {2}^{n + 1}$ 。通过将 ${x}^{2}z$ 的积因子分解成 $\bar{G}$ ，我们在商中识别 ${x}^{2}$ 和 ${h}^{{2}^{n - 1}}$ 。特别是当 $n = 2$ 时，$\bar{x}$ 和 $\bar{h}$ 的阶都是4，$\bar{x}$ 逆 $\bar{h}$ 并且 ${\bar{h}}^{2} = {\bar{x}}^{2}$ 。因此在这种情况下 $\bar{G} \cong {Q}_{8}$ 。一般来说，可以验证 $\bar{G}$ 有一个唯一的阶为2的子群（即 ( ${\bar{x}}^{2}$ )），它等于 $\overline{G}$ 的中心。这个群 $\overline{G}$ 被称为阶为 ${2}^{n + 1}$ 的广义四元数群，表示为 ${Q}_{{2}^{n + 1}}$ ：

(4) Let $H = \mathbb{Q}$ (under addition) and let $K = \langle x\rangle \cong \mathbb{Z}$ . Define $\varphi$ by mapping $x$ to the map "multiplication by 2 " on $H$ ,so that $x$ acts on $h \in H$ by $x \cdot h = {2h}$ . Note that multiplication by 2 is an automorphism of $H$ because it has a 2-sided inverse,namely multiplication by $\frac{1}{2}$ . In the group $G,\mathbb{Z} \leq \mathbb{Q}$ and the conjugate $x\mathbb{Z}{x}^{-1}$ of $\mathbb{Z}$ is a proper subgroup of $\mathbb{Z}$ (namely $2\mathbb{Z}$ ). Thus $x \notin {N}_{G}\left( \mathbb{Z}\right)$ even though $x\mathbb{Z}{x}^{-1} \leq \mathbb{Z}$ (note that ${x}^{-1}\mathbb{Z}x$ is not contained in $\mathbb{Z}$ ). This shows that in order to prove an element $g$ normalizes a subgroup $A$ in an infinite group it is not sufficient in general to show that the conjugate of $A$ by $g$ is just contained in $A$ (which is sufficient for finite groups).

(4) 设 $H = \mathbb{Q}$（在加法下）和 $K = \langle x\rangle \cong \mathbb{Z}$ 。定义 $\varphi$ 通过将 $x$ 映射到 $H$ 上的“乘以2”的映射 $h \in H$，使得 $x$ 通过 $x \cdot h = {2h}$ 作用于 $h \in H$ 。注意乘以2是 $H$ 的自同构，因为它有一个2边的逆元，即乘以 $\frac{1}{2}$ 。在群 $G,\mathbb{Z} \leq \mathbb{Q}$ 中，$x\mathbb{Z}{x}^{-1}$ 的共轭 $\mathbb{Z}$ 是 $\mathbb{Z}$ 的一个真子群（即 $2\mathbb{Z}$ ）。因此 $x \notin {N}_{G}\left( \mathbb{Z}\right)$，尽管 $x\mathbb{Z}{x}^{-1} \leq \mathbb{Z}$（注意 ${x}^{-1}\mathbb{Z}x$ 不包含在 $\mathbb{Z}$ 中）。这表明，在一般情况下，为了证明一个元素 $g$ 在一个无限群中正规化一个子群 $A$，仅仅证明 $A$ 通过 $g$ 的共轭包含在 $A$ 中是不够的（这对于有限群是充分的）。

(5) For $H$ any group let $K = \operatorname{Aut}\left( H\right)$ with $\varphi$ the identity map from $K$ to $\operatorname{Aut}\left( H\right)$ . The semidirect product $H \rtimes \operatorname{Aut}\left( H\right)$ is called the holomorph of $H$ and will be denoted by $\mathrm{{Hol}}\left( H\right)$ . Some holomorphs are described below; verifications of these isomorphisms are given as exercises at the end of this chapter.

(5) 对于 $H$ 任何群，设 $K = \operatorname{Aut}\left( H\right)$ 与 $\varphi$ 从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的恒等映射。半直积 $H \rtimes \operatorname{Aut}\left( H\right)$ 被称为 $H$ 的全形，并用 $\mathrm{{Hol}}\left( H\right)$ 表示。下面描述了一些全形；这些同构的验证将在本章末尾作为练习给出。

(a) $\operatorname{Hol}\left( {{Z}_{2} \times {Z}_{2}}\right) \cong {S}_{4}$ .

(a) $\operatorname{Hol}\left( {{Z}_{2} \times {Z}_{2}}\right) \cong {S}_{4}$ 。

(b) If $\left| G\right| = n$ and $\pi : G \rightarrow {S}_{n}$ is the left regular representation (Section 4.2),then ${N}_{{S}_{n}}\left( {\pi \left( G\right) }\right) \cong \operatorname{Hol}\left( G\right)$ . In particular,since the left regular representation of a generator of ${Z}_{n}$ is an $n$ -cycle in ${S}_{n}$ we obtain that for any $n$ -cycle $\left( {{12}\ldots n}\right)$ :

(b) 如果 $\left| G\right| = n$ 并且 $\pi : G \rightarrow {S}_{n}$ 是左正则表示（第4.2节），那么 ${N}_{{S}_{n}}\left( {\pi \left( G\right) }\right) \cong \operatorname{Hol}\left( G\right)$ 。特别地，由于一个生成元的左正则表示在 ${Z}_{n}$ 中是一个 $n$ -循环，我们得到对于任何 $n$ -循环 $\left( {{12}\ldots n}\right)$ ：

Note that the latter group has order ${n\varphi }\left( n\right)$ .

注意，后一个群有阶 ${n\varphi }\left( n\right)$。

(6) Let $p$ and $q$ be primes with $p < q$ ,let $H = {Z}_{q}$ and let $K = {Z}_{p}$ . We have already seen that if $p$ does not divide $q - 1$ then every group of order ${pq}$ is cyclic (see the example following Proposition 4.16). This is consistent with the fact that if $p$ does not divide $q - 1$ ,there is no nontrivial homomorphism from ${Z}_{p}$ into $\operatorname{Aut}\left( {Z}_{q}\right)$ (the latter group is cyclic of order $q - 1$ by Proposition 4.17). Assume now that $p \mid q - 1$ . By Cauchy’s Theorem, $\operatorname{Aut}\left( {Z}_{q}\right)$ contains a subgroup of order $p$ (which is unique because $\operatorname{Aut}\left( {Z}_{q}\right)$ is cyclic). Thus there is a nontrivial homomorphism, $\varphi$ ,from $K$ into $\operatorname{Aut}\left( H\right)$ . The associated group $G = H \rtimes K$ has order ${pq}$ and $K$ is not normal in $G$ (Proposition 11). In particular, $G$ is non-abelian. We shall prove shortly that $G$ is (up to isomorphism) the unique non-abelian group of order ${pq}$ . If $p = 2,G$ must be isomorphic to ${D}_{2q}$ .

(6) 设 $p$ 和 $q$ 是满足 $p < q$ 的素数，设 $H = {Z}_{q}$ 且设 $K = {Z}_{p}$。我们已经看到，如果 $p$ 不整除 $q - 1$，那么每个阶为 ${pq}$ 的群都是循环群（参见命题 4.16 后的示例）。这与如果 $p$ 不整除 $q - 1$，则不存在从 ${Z}_{p}$ 到 $\operatorname{Aut}\left( {Z}_{q}\right)$ 的非平凡同态的事实一致（根据命题 4.17，后者是阶为 $q - 1$ 的循环群）。现在假设 $p \mid q - 1$。根据柯西定理，$\operatorname{Aut}\left( {Z}_{q}\right)$ 包含一个阶为 $p$ 的子群（因为 $\operatorname{Aut}\left( {Z}_{q}\right)$ 是循环群，所以这个子群是唯一的）。因此存在一个非平凡同态 $\varphi$，从 $K$ 到 $\operatorname{Aut}\left( H\right)$。相关的群 $G = H \rtimes K$ 有阶 ${pq}$，且 $K$ 在 $G$ 中不是正规子群（命题 11）。特别地，$G$ 是非阿贝尔群。我们很快将证明 $G$ 是（同构意义下的）唯一的阶为 ${pq}$ 的非阿贝尔群。如果 $p = 2,G$，则必须同构于 ${D}_{2q}$。

(7) Let $p$ be an odd prime. We construct two nonisomorphic non-abelian groups of order ${p}^{3}$ (we shall later prove that any non-abelian group of order ${p}^{3}$ is isomorphic to one of these two).

(7) 设 $p$ 是一个奇素数。我们构造两个不同构的非阿贝尔群，它们的阶为 ${p}^{3}$（我们稍后将证明任何阶为 ${p}^{3}$ 的非阿贝尔群都同构于这两个群之一）。

Let $H = {Z}_{p} \times {Z}_{p}$ and let $K = {Z}_{p}$ . By Proposition 4.17,Aut $\left( H\right) \cong G{L}_{2}\left( {\mathbb{F}}_{p}\right)$ and $\left| {G{L}_{2}\left( {\mathbb{F}}_{p}\right) }\right| = \left( {{p}^{2} - 1}\right) \left( {{p}^{2} - p}\right)$ . Since $p \mid \left| {\operatorname{Aut}\left( H\right) }\right|$ ,by Cauchy’s Theorem $H$ has an automorphism of order $p$ . Thus there is a nontrivial homomorphism, $\varphi$ ,from $K$ into $\operatorname{Aut}\left( H\right)$ and so the associated group $H \rtimes K$ is a non-abelian group of order ${p}^{3}$ . More explicitly,if $H = \langle a\rangle \times \langle b\rangle$ ,and $x$ is a generator for $K$ then $x$ acts on $a$ and $b$ by

令 $H = {Z}_{p} \times {Z}_{p}$ 且令 $K = {Z}_{p}$ 。根据命题 4.17，Aut $\left( H\right) \cong G{L}_{2}\left( {\mathbb{F}}_{p}\right)$ 和 $\left| {G{L}_{2}\left( {\mathbb{F}}_{p}\right) }\right| = \left( {{p}^{2} - 1}\right) \left( {{p}^{2} - p}\right)$ 。由于 $p \mid \left| {\operatorname{Aut}\left( H\right) }\right|$ ，根据柯西定理 $H$ 存在一个阶为 $p$ 的自同构。因此存在一个非平凡同态，$\varphi$ ，从 $K$ 到 $\operatorname{Aut}\left( H\right)$ ，因此相关的群 $H \rtimes K$ 是一个阶为 ${p}^{3}$ 的非阿贝尔群。更具体地说，如果 $H = \langle a\rangle \times \langle b\rangle$ ，且 $x$ 是 $K$ 的生成元，那么 $x$ 通过

which defines the action of $x$ on all of $H$ . With respect to the ${\mathbb{F}}_{p}$ -basis $a,b$ of the 2-dimensional vector space $H$ the action of $x$ (which can be considered in additive notation as a nonsingular linear transformation) has matrix

定义了 $x$ 对所有 $H$ 的作用。关于 ${\mathbb{F}}_{p}$ -基 $a,b$ 的 2 维向量空间 $H$ 的作用（可以认为是加法表示下的非奇异线性变换）具有矩阵

The resulting semidirect product has the presentation

生成的半直积具有如下表示

(in fact,this group is generated by $\{ x,a\}$ ,and is called the Heisenberg group over $\mathbb{Z}/p\mathbb{Z}$ ,cf. Exercise 25).

（实际上，这个群是由 $\{ x,a\}$ 生成的，并被称为 $\mathbb{Z}/p\mathbb{Z}$ 上的海森堡群，参见练习 25）。

Next let $H = {Z}_{{p}^{2}}$ and $K = {Z}_{p}$ . Again by Proposition 4.17,Aut $\left( H\right) \cong {Z}_{p\left( {p - 1}\right) }$ , so $H$ admits an automorphism of order $p$ . Thus there is a nontrivial homomorphism, $\varphi$ ,from $K\operatorname{into}\operatorname{Aut}\left( H\right)$ and so the group $H \rtimes K$ is non-abelian and of order ${p}^{3}$ . More explicitly,if $H = \langle y\rangle$ ,and $x$ is a generator for $K$ then $x$ acts on $y$ by

接下来令 $H = {Z}_{{p}^{2}}$ 和 $K = {Z}_{p}$ 。再次根据命题 4.17，Aut $\left( H\right) \cong {Z}_{p\left( {p - 1}\right) }$ ，所以 $H$ 具有一个阶为 $p$ 的自同构。因此存在一个非平凡同态，$\varphi$ ，从 $K\operatorname{into}\operatorname{Aut}\left( H\right)$ ，因此群 $H \rtimes K$ 是非阿贝尔的，且阶为 ${p}^{3}$ 。更具体地说，如果 $H = \langle y\rangle$ ，且 $x$ 是 $K$ 的生成元，那么 $x$ 通过

The resulting semidirect product has the presentation

生成的半直积具有如下表示

These two groups are not isomorphic (the former contains no element of order ${p}^{2}$ ,cf. Exercise 25,and the latter clearly does,namely $y$ ).

这两个群不是同构的（前者不包含阶为 ${p}^{2}$ 的元素，参见练习25，而后者显然包含，即 $y$ ）。

( ) Let $H = {Q}_{8} \times \left( {{Z}_{2} \times {Z}_{2}}\right) = \langle i,j\rangle \times \left( {\langle a\rangle \times \langle b\rangle }\right)$ and let $K = \langle y\rangle \cong {Z}_{3}$ . The map defined by

（ ）设 $H = {Q}_{8} \times \left( {{Z}_{2} \times {Z}_{2}}\right) = \langle i,j\rangle \times \left( {\langle a\rangle \times \langle b\rangle }\right)$ 且设 $K = \langle y\rangle \cong {Z}_{3}$ 。由以下映射定义的映射

is easily seen to give an automorphism of $H$ of order 3 . Let $\varphi$ be the homomorphism from $K$ to $\operatorname{Aut}\left( H\right)$ defined by mapping $y$ to this automorphism,and let $G$ be the associated semidirect product,so that $y \in G$ acts by

可以轻易看出它给出了一个 $H$ 的阶为3的自同构。设 $\varphi$ 是从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的同态，定义为将 $y$ 映射到这个自同构，设 $G$ 为相关的半直积，使得 $y \in G$ 通过

The group $G = H \rtimes K$ is a non-abelian group of order 96 with the property that the element ${i}^{2}a \in {G}^{\prime }$ but ${i}^{2}a$ cannot be expressed as a single commutator $\left\lbrack {x,y}\right\rbrack$ ,for any $x,y \in G$ (checking the latter assertion is an elementary calculation).

群 $G = H \rtimes K$ 是一个阶为96的非阿贝尔群，具有如下性质：元素 ${i}^{2}a \in {G}^{\prime }$ 但 ${i}^{2}a$ 不能表示为单个换位子 $\left\lbrack {x,y}\right\rbrack$ ，对于任何 $x,y \in G$（验证后一断言是一个基本计算）。

As in the case of direct products we now prove a recognition theorem for semidirect products. This theorem will enable us to “break down” or “factor” all groups of certain orders and, as a result, classify groups of those orders. The strategy is discussed in greater detail following this theorem.

与直积的情况类似，我们现在为半直积证明一个识别定理。这个定理将使我们能够“分解”或“分解”所有某些阶的群，并因此对这些阶的群进行分类。在定理之后的策略会有更详细的讨论。

Theorem 12. Suppose $G$ is a group with subgroups $H$ and $K$ such that

定理12。假设 $G$ 是一个群，其子群为 $H$ 和 $K$ ，使得

(1) $H \trianglelefteq G$ ,and

（1）$H \trianglelefteq G$ ，并且

(2) $H \cap K = 1$ .

（2）$H \cap K = 1$ 。

Let $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ be the homomorphism defined by mapping $k \in K$ to the automorphism of left conjugation by $k$ on $H$ . Then ${HK} \cong H \rtimes K$ . In particular,if $G = {HK}$ with $H$ and $K$ satisfying (1) and (2),then $G$ is the semidirect product of $H$ and $K$ .

设 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ 为一个同态，定义为将 $k \in K$ 映射到由 $k$ 对 $H$ 进行左共轭的自同构。那么 ${HK} \cong H \rtimes K$ 。特别地，如果 $G = {HK}$ 且 $H$ 和 $K$ 满足（1）和（2），那么 $G$ 是 $H$ 和 $K$ 的半直积。

Proof: Note that since $H \trianglelefteq G,{HK}$ is a subgroup of $G$ . By Proposition 8 every element of ${HK}$ can be written uniquely in the form ${hk}$ ,for some $h \in H$ and $k \in K$ . Thus the map ${hk} \mapsto \left( {h,k}\right)$ is a set bijection from ${HK}$ onto $H \rtimes K$ . The fact that this map is a homomorphism is the computation at the beginning of this section which led us to the formulation of the definition of the semidirect product.

证明：注意由于 $H \trianglelefteq G,{HK}$ 是 $G$ 的子群。根据命题8，${HK}$ 的每个元素都可以唯一地写成 ${hk}$ 的形式，对于某些 $h \in H$ 和 $k \in K$ 。因此映射 ${hk} \mapsto \left( {h,k}\right)$ 是从 ${HK}$ 到 $H \rtimes K$ 的集合双射。这个映射是一个同态的事实是本节开头进行的计算，它引导我们形成了半直积定义的表述。

Definition. Let $H$ be a subgroup of the group $G$ . A subgroup $K$ of $G$ is called a complement for $H$ in $G$ if $G = {HK}$ and $H \cap K = 1$ .

定义。设 $H$ 是群 $G$ 的子群。如果 $K$ 是 $G$ 的一个子群，并且满足 $G = {HK}$ 和 $H \cap K = 1$ ，则称 $K$ 为 $H$ 在 $G$ 中的补。

With this terminology, the criterion for recognizing a semidirect product is simply that there must exist a complement for some proper normal subgroup of $G$ . Not every group is the semidirect product of two of its proper subgroups (for example, if the group is simple), but as we have seen, the notion of a semidirect product greatly increases our list of known groups.

使用这个术语，识别半直积的标准仅仅是必须存在某个 $G$ 的适当正规子群的补。并非每个群都是其两个适当子群的半直积（例如，如果群是简单的），但正如我们所见，半直积的概念大大增加了我们已知的群列表。

Some Classifications

一些分类

We now apply Theorem 12 to classify groups of order $n$ for certain values of $n$ . The basic idea in each of the following arguments is to

我们现在将定理12应用于分类阶为 $n$ 的群，对于某些 $n$ 的值。以下每个论证的基本思想是

(a) show every group of order $n$ has proper subgroups $H$ and $K$ satisfying the hypothesis of Theorem 12 with $G = {HK}$

(a) 证明每个阶为 $n$ 的群都有满足定理12假设的适当子群 $H$ 和 $K$ ，且 $G = {HK}$

(b) find all possible isomorphism types for $H$ and $K$

(b) 找出 $H$ 和 $K$ 的所有可能的同构类型

(c) for each pair $H,K$ found in (b) find all possible homomorphisms $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$

(c) 对于在（b）中找到的每对 $H,K$ ，找出所有可能的同态 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$

(d) for each triple $H,K,\varphi$ found in (c) form the semidirect product $H \rtimes K$ (so any group $G$ of order $n$ is isomorphic to one of these explicitly constructed groups) and among all these semidirect products determine which pairs are isomorphic. This results in a list of the distinct isomorphism types of groups of order $n$ .

(d) 对于在 (c) 中找到的每个三元组 $H,K,\varphi$，形成半直积 $H \rtimes K$（因此任何阶数为 $n$ 的群 $G$ 都与这些显式构造的群同构）并在所有这些半直积中确定哪些对是同构的。这产生了一个阶数为 $n$ 的不同同构类型群的列表。

In order to start this process we must first find subgroups $H$ and $K$ (of an arbitrary group $G$ of order $n$ ) satisfying the above conditions. In the case of "small" values of $n$ we can often do this by Sylow’s Theorem. To show normality of $H$ we use the conjugacy part of Sylow’s Theorem or other normality criteria established in Chapter 4 (e.g., Corollary 4.5). Some of this work has already been done in the examples in Section 4.5. In many of the examples that follow, $\left| H\right|$ and $\left| K\right|$ are relatively prime,so $H \cap K = 1$ holds by Lagrange’s Theorem.

为了开始这个过程，我们首先必须找到子群 $H$ 和 $K$（阶数为 $n$ 的任意群 $G$ 的子群）满足上述条件。在 $n$ 的“小”值的情况下，我们通常可以通过 Sylow 定理来完成这一点。为了显示 $H$ 的正规性，我们使用 Sylow 定理的共轭部分或第 4 章中建立的其它正规性准则（例如，推论 4.5）。在 4.5 节的例子中已经完成了一些这样的工作。在接下来的许多例子中，$\left| H\right|$ 和 $\left| K\right|$ 是互质的，因此根据 Lagrange 定理，$H \cap K = 1$ 成立。

Since $H$ and $K$ are proper subgroups of $G$ one should think of the determination of $H$ and $K$ as being achieved inductively. In the examples we discuss, $H$ and $K$ will have sufficiently small order that we shall know all possible isomorphism types from previous results. For example,in most instances $H$ and $K$ will be of prime or prime squared order.

由于 $H$ 和 $K$ 是 $G$ 的真子群，人们应该认为确定 $H$ 和 $K$ 是通过归纳实现的。在我们讨论的例子中，$H$ 和 $K$ 将有足够小的阶数，以至于我们可以从之前的结果中知道所有可能的同构类型。例如，在大多数情况下，$H$ 和 $K$ 将是素数或素数平方的阶数。

There will be relatively few possible homomorphisms $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ in our examples, particularly after we take into account certain symmetries (such as replacing one generator of $K$ by another when $K$ is cyclic).

在我们的例子中，可能的同态 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ 将相对较少，特别是我们在考虑某些对称性之后（例如，当 $K$ 是循环群时，用另一个生成元替换 $K$ 的一个生成元）。

Finally, the semidirect products which emerge from this process will, in our examples, be small in number and we shall find that, for the most part, they are (pairwise) not isomorphic. In general, this can be a more delicate problem, as Exercise 4 indicates.

最后，在这个过程中出现的半直积在我们的例子中数量较少，我们将发现，在大多数情况下，它们（两两之间）不是同构的。一般来说，这可以是一个更微妙的问题，正如练习4所示。

We emphasize that this approach to “factoring” every group of some given order $n$ as a semidirect product does not work for arbitrary $n$ . For example, ${Q}_{8}$ is not a semidirect product since no proper subgroup has a complement (although we saw that it is a quotient of a semidirect product). Empirically, this process generally works well when the group order $n$ is not divisible by a large power of any prime. At the other extreme,only a small percentage of the groups of order ${p}^{\alpha }$ for large $\alpha$ ( $p$ a prime) are nontrivial semidirect products.

我们强调，这种将某些给定阶的每个群 $n$ 作为半直积进行“分解”的方法并不适用于任意的 $n$ 。例如，${Q}_{8}$ 不是一个半直积，因为没有适当的子群具有补集（尽管我们看到它是半直积的商）。从经验上讲，当群的阶 $n$ 不能被任何素数的较大幂整除时，这个过程通常工作得很好。在另一个极端，只有一小部分的阶为 ${p}^{\alpha }$ 的群对于大的 $\alpha$ （$p$ 是一个素数）是非平凡半直积。

Example: (Groups of Order ${pq},p$ and $q$ primes with $p < q$ )

示例：（阶为 ${pq},p$ 且 $q$ 为素数的群）

Let $G$ be any group of order ${pq}$ ,let $P \in {\operatorname{Syl}}_{p}\left( G\right)$ and let $Q \in {\operatorname{Syl}}_{q}\left( G\right)$ . In Example 1 of the applications of Sylow’s Theorems we proved that $G \cong Q \rtimes P$ ,for some $\varphi : P \rightarrow \operatorname{Aut}\left( Q\right)$ . Since $P$ and $Q$ are of prime order,they are cyclic. The group $\operatorname{Aut}\left( Q\right)$ is cyclic of order $q - 1$ . If $p$ does not divide $q - 1$ ,the only homomorphism from $P$ to $\operatorname{Aut}\left( Q\right)$ is the trivial homomorphism, hence the only semidirect product in this case is the direct product, i.e., $G$ is cyclic.

设 $G$ 为任意阶为 ${pq}$ 的群，令 $P \in {\operatorname{Syl}}_{p}\left( G\right)$ 并且令 $Q \in {\operatorname{Syl}}_{q}\left( G\right)$ 。在应用 Sylow 定理的例1中，我们证明了 $G \cong Q \rtimes P$ ，对于某个 $\varphi : P \rightarrow \operatorname{Aut}\left( Q\right)$ 。由于 $P$ 和 $Q$ 的阶为素数，它们是循环的。群 $\operatorname{Aut}\left( Q\right)$ 是阶为 $q - 1$ 的循环群。如果 $p$ 不整除 $q - 1$ ，那么从 $P$ 到 $\operatorname{Aut}\left( Q\right)$ 的唯一同态是平凡同态，因此在这种情况下唯一的半直积是直积，即 $G$ 是循环的。

Consider now the case when $p \mid q - 1$ and let $P = \langle y\rangle$ . Since $\operatorname{Aut}\left( Q\right)$ is cyclic it contains a unique subgroup of order $p$ ,say $\langle \gamma \rangle$ ,and any homomorphism $\varphi : P \rightarrow \operatorname{Aut}\left( Q\right)$ must map $y$ to a power of $\gamma$ . There are therefore $p$ homomorphisms ${\varphi }_{i} : P \rightarrow \operatorname{Aut}\left( Q\right)$ given by ${\varphi }_{i}\left( y\right) = {\gamma }^{i},0 \leq i \leq p - 1$ . Since ${\varphi }_{0}$ is the trivial homomorphism, $Q{ \rtimes }_{{\varphi }_{0}}P \cong Q \times P$ as before. Each ${\varphi }_{i}$ for $i \neq 0$ gives rise to a non-abelian group, ${G}_{i}$ ,of order ${pq}$ . It is straightforward to check that these groups are all isomorphic because for each ${\varphi }_{i},i > 0$ , there is some generator ${y}_{i}$ of $P$ such that ${\varphi }_{i}\left( {y}_{i}\right) = \gamma$ . Thus,up to a choice for the (arbitrary) generator of $P$ ,these semidirect products are all the same (see Exercise 6. See also Exercise 28 of Section 4.3).

现在考虑当 $p \mid q - 1$ 且让 $P = \langle y\rangle$ 的情况。由于 $\operatorname{Aut}\left( Q\right)$ 是循环的，它包含一个阶为 $p$ 的唯一子群，称为 $\langle \gamma \rangle$，并且任何同态 $\varphi : P \rightarrow \operatorname{Aut}\left( Q\right)$ 必须将 $y$ 映射到 $\gamma$ 的幂。因此有 $p$ 个同态 ${\varphi }_{i} : P \rightarrow \operatorname{Aut}\left( Q\right)$，由 ${\varphi }_{i}\left( y\right) = {\gamma }^{i},0 \leq i \leq p - 1$ 给出。由于 ${\varphi }_{0}$ 是平凡同态，$Q{ \rtimes }_{{\varphi }_{0}}P \cong Q \times P$ 如前所述。每个 ${\varphi }_{i}$ 对于 $i \neq 0$ 产生一个非阿贝尔群，${G}_{i}$，其阶为 ${pq}$。可以简单地验证这些群都是同构的，因为对于每个 ${\varphi }_{i},i > 0$，存在 $P$ 的某个生成元 ${y}_{i}$，使得 ${\varphi }_{i}\left( {y}_{i}\right) = \gamma$。因此，除了对 $P$ 的（任意）生成元的选择之外，这些半直积都是相同的（参见练习6。也见第4.3节的练习28）。

Example: (Groups of Order 30)

示例：（阶为30的群）

By the examples following Sylow’s Theorem every group $G$ of order 30 contains a subgroup $H$ of order 15 . By the preceding example $H$ is cyclic and $H$ is normal in $G$ (index 2). By Sylow’s Theorem there is a subgroup $K$ of $G$ of order 2. Thus $G = {HK}$ and $H \cap K = 1$ so $G \cong H \rtimes K$ ,for some $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ . By Proposition 4.16,

根据 Sylow 定理后的例子，每个阶为30的群 $G$ 包含一个阶为15的子群 $H$。根据前一个例子，$H$ 是循环的，并且 $H$ 在 $G$ 中是正规子群（指数为2）。根据 Sylow 定理，存在 $G$ 的一个阶为2的子群 $K$。因此 $G = {HK}$ 并且 $H \cap K = 1$，所以 $G \cong H \rtimes K$，对于某个 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$。根据命题4.16，

The latter isomorphism can be computed directly, or one can use Exercise 11 of the preceding section: writing $H$ as $\langle a\rangle \times \langle b\rangle \cong {Z}_{5} \times {Z}_{3}$ ,we have (since these two subgroups are characteristic in $H$ )

后一个同构可以直接计算，或者可以使用上一节的练习11：将 $H$ 写作 $\langle a\rangle \times \langle b\rangle \cong {Z}_{5} \times {Z}_{3}$，我们得到（因为这两个子群在 $H$ 中是特征子群）。

In particular, $\operatorname{Aut}\left( H\right)$ contains precisely three elements of order 2,whose actions on the group $H = \langle a\rangle \times \langle b\rangle$ are the following:

特别是，$\operatorname{Aut}\left( H\right)$ 精确地包含三个阶为2的元素，它们在群 $H = \langle a\rangle \times \langle b\rangle$ 上的作用如下：

Thus there are three nontrivial homomorphisms from $K$ into $\operatorname{Aut}\left( H\right)$ given by sending the generator of $K$ into one of these three elements of order 2 (as usual,the trivial homomorphism gives the direct product: $\left. {H \times K \cong {Z}_{30}}\right)$ .

因此存在三个从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的非平凡同态，它们通过将 $K$ 的生成元映射到这三个阶为2的元素之一来实现（如通常情况，平凡同态给出直积：$\left. {H \times K \cong {Z}_{30}}\right)$）。

Let $K = \langle k\rangle$ . If the homomorphism ${\varphi }_{1} : K \rightarrow \operatorname{Aut}\left( H\right)$ is defined by mapping $k$ to the first automorphism above (so that $k \cdot a = a$ and $k \cdot b = {b}^{-1}$ gives the action of $k$ on $H$ ) then ${G}_{1} = H{ \rtimes }_{{\varphi }_{1}}K$ is easily seen to be isomorphic to ${Z}_{5} \times {D}_{6}$ (note that in this semidirect product $k$ centralizes the element $a$ of $H$ of order 5,so the factorization as a direct product is $\langle a\rangle \times \langle b,k\rangle )$ .

设 $K = \langle k\rangle$ 。如果同态 ${\varphi }_{1} : K \rightarrow \operatorname{Aut}\left( H\right)$ 通过将 $k$ 映射到上述第一个自同构来定义（这样 $k \cdot a = a$ 和 $k \cdot b = {b}^{-1}$ 给出了 $k$ 在 $H$ 上的作用），那么 ${G}_{1} = H{ \rtimes }_{{\varphi }_{1}}K$ 很容易看出与 ${Z}_{5} \times {D}_{6}$ 同构（注意在这个半直积中 $k$ centralized $a$ 的元素 $H$ 阶为5，因此分解为直积是 $\langle a\rangle \times \langle b,k\rangle )$）。

If ${\varphi }_{2}$ is defined by mapping $k$ to the second automorphism above,then ${G}_{2} = H{ \rtimes }_{{\varphi }_{2}}K$ is easily seen to be isomorphic to ${Z}_{3} \times {D}_{10}$ (note that in this semidirect product $k$ centralizes the element $b$ of $H$ of order 3,so the factorization as a direct product is $\langle b\rangle \times \langle a,k\rangle$ ).

如果 ${\varphi }_{2}$ 通过将 $k$ 映射到上述第二个自同构来定义，那么 ${G}_{2} = H{ \rtimes }_{{\varphi }_{2}}K$ 很容易看出与 ${Z}_{3} \times {D}_{10}$ 同构（注意在这个半直积中 $k$ centralized $b$ 的元素 $H$ 阶为3，因此分解为直积是 $\langle b\rangle \times \langle a,k\rangle$）。

If ${\varphi }_{3}$ is defined by mapping $k$ to the third automorphism above then ${G}_{3} = H{ \rtimes }_{{\varphi }_{3}}K$ is easily seen to be isomorphic to ${D}_{30}$ .

如果 ${\varphi }_{3}$ 通过将 $k$ 映射到上述第三个自同构来定义，那么 ${G}_{3} = H{ \rtimes }_{{\varphi }_{3}}K$ 很容易看出与 ${D}_{30}$ 同构。

Note that these groups are all nonisomorphic since their centers have orders 30 (in the abelian case),5(for ${G}_{1}$ ),3(for ${G}_{2}$ ),and 1 (for ${G}_{3}$ ).

注意这些群都是非同构的，因为它们的中心阶分别为30（在阿贝尔情况下）、5（对于 ${G}_{1}$）、3（对于 ${G}_{2}$）和1（对于 ${G}_{3}$）。

We emphasize that although (in hindsight) this procedure does not give rise to any groups we could not already have constructed using only direct products, the argument proves that this is the complete list of isomorphism types of groups of order 30.

我们强调，尽管（事后看来）这个程序并没有产生我们无法仅使用直积构造的任何群，但该论证证明了这是阶数为30的群的同构类型的完整列表。

Example: (Groups of Order 12)

示例：（阶数为12的群）

Let $G$ be a group of order 12,let $V \in {\operatorname{Syl}}_{2}\left( G\right)$ and let $T \in {\operatorname{Syl}}_{3}\left( G\right)$ . By the discussion of groups of order 12 in Section 4.5 we know that either $V$ or $T$ is normal in $G$ (for purposes of illustration we shall not invoke the full force of our results from Chapter 4, namely that either $T \trianglelefteq G$ or $G \cong {A}_{4}$ ). By Lagrange’s Theorem $V \cap T = 1$ . Thus $G$ is a semidirect product. Note that $V \cong {Z}_{4}$ or ${Z}_{2} \times {Z}_{2}$ and $T \cong {Z}_{3}$ . Case 1: $V \trianglelefteq G$

设 $G$ 是一个阶数为12的群，设 $V \in {\operatorname{Syl}}_{2}\left( G\right)$ 并且设 $T \in {\operatorname{Syl}}_{3}\left( G\right)$ 。根据第4.5节中关于阶数为12的群的讨论，我们知道 $V$ 或 $T$ 在 $G$ 中是正规群（为了说明目的，我们不会动用第4章的全部结果，即 $T \trianglelefteq G$ 或 $G \cong {A}_{4}$）。根据拉格朗日定理 $V \cap T = 1$ 。因此 $G$ 是一个半直积。注意 $V \cong {Z}_{4}$ 或 ${Z}_{2} \times {Z}_{2}$ 并且 $T \cong {Z}_{3}$ 。情形1：$V \trianglelefteq G$

We must determine all possible homomorphisms from $T$ into $\operatorname{Aut}\left( V\right)$ . If $V \cong {Z}_{4}$ , then $\operatorname{Aut}\left( V\right) \cong {Z}_{2}$ and there are no nontrivial homomorphisms from $T$ into $\operatorname{Aut}\left( V\right)$ . Thus the only group of order 12 with a normal cyclic Sylow 2-subgroup is ${Z}_{12}$ .

我们必须确定从 $T$ 到 $\operatorname{Aut}\left( V\right)$ 的所有可能同态。如果 $V \cong {Z}_{4}$ ，那么 $\operatorname{Aut}\left( V\right) \cong {Z}_{2}$ 并且不存在从 $T$ 到 $\operatorname{Aut}\left( V\right)$ 的非平凡同态。因此，唯一具有正规循环Sylow 2-子群的阶数为12的群是 ${Z}_{12}$。

Assume therefore that $V \cong {Z}_{2} \times {Z}_{2}$ . In this case $\operatorname{Aut}\left( V\right) \cong {S}_{3}$ and there is a unique subgroup of $\operatorname{Aut}\left( V\right)$ of order3,say $\langle \gamma \rangle$ . Thus if $T = \langle \gamma \rangle$ ,there are three possible homomorphisms from $T$ into $\operatorname{Aut}\left( V\right)$ :

因此假设 $V \cong {Z}_{2} \times {Z}_{2}$ 。在这种情况下 $\operatorname{Aut}\left( V\right) \cong {S}_{3}$ 并且在 $\operatorname{Aut}\left( V\right)$ 中存在一个唯一的阶数为3的子群，设为 $\langle \gamma \rangle$ 。因此，如果 $T = \langle \gamma \rangle$ ，那么从 $T$ 到 $\operatorname{Aut}\left( V\right)$ 有三种可能的同态：

As usual, ${\varphi }_{0}$ is the trivial homomorphism,which gives rise to the direct product ${Z}_{2} \times {Z}_{2} \times {Z}_{3}$ . Homomorphisms ${\varphi }_{1}$ and ${\varphi }_{2}$ give rise to isomorphic semidirect products because they differ only in the choice of a generator for $T$ (i.e., ${\varphi }_{1}\left( y\right) = \gamma$ and ${\varphi }_{2}\left( {y}^{\prime }\right) = \gamma$ , where ${y}^{\prime } = {y}^{2}$ and ${y}^{\prime }$ is another choice of generator for $T$ -see also Exercise 6). The unique non-abelian group in this case is ${A}_{4}$ .

如往常一样，${\varphi }_{0}$ 是平凡同态，它产生了直积 ${Z}_{2} \times {Z}_{2} \times {Z}_{3}$。同态 ${\varphi }_{1}$ 和 ${\varphi }_{2}$ 产生了同构的半直积，因为它们只在 $T$ 的生成元选择上有所不同（即 ${\varphi }_{1}\left( y\right) = \gamma$ 和 ${\varphi }_{2}\left( {y}^{\prime }\right) = \gamma$，其中 ${y}^{\prime } = {y}^{2}$ 和 ${y}^{\prime }$ 是 $T$ 的另一生成元选择 - 参见练习6）。在此情况下唯一的非阿贝尔群是 ${A}_{4}$。

Case 2: $T \trianglelefteq G$

情况2：$T \trianglelefteq G$

We must determine all possible homomorphisms from $V$ into $\operatorname{Aut}\left( T\right)$ . Note that $\operatorname{Aut}\left( T\right) = \langle \lambda \rangle \cong {Z}_{2}$ ,where $\lambda$ inverts $T$ . If $V = \langle x\rangle \cong {Z}_{4}$ ,there are precisely two homomorphisms from $V$ into $\operatorname{Aut}\left( T\right)$ : the trivial homomorphism and the homomorphism which sends $x$ to $\lambda$ . As usual,the trivial homomorphism gives rise to the direct product: ${Z}_{3} \times {Z}_{4} \cong {Z}_{12}$ . The nontrivial homomorphism gives the semidirect product which was discussed in Example 2 following Proposition 11 of this section.

我们必须确定所有从 $V$ 到 $\operatorname{Aut}\left( T\right)$ 的可能同态。注意 $\operatorname{Aut}\left( T\right) = \langle \lambda \rangle \cong {Z}_{2}$，其中 $\lambda$ 逆转 $T$。如果 $V = \langle x\rangle \cong {Z}_{4}$，则存在恰好两个从 $V$ 到 $\operatorname{Aut}\left( T\right)$ 的同态：平凡同态和将 $x$ 映射到 $\lambda$ 的同态。如往常一样，平凡同态产生了直积：${Z}_{3} \times {Z}_{4} \cong {Z}_{12}$。非平凡同态给出了在本文第11命题后例2中讨论的半直积。

Finally,assume $V = \langle a\rangle \times \langle b\rangle \cong {Z}_{2} \times {Z}_{2}$ . There are precisely three nontrivial homomorphisms from $V$ into $\operatorname{Aut}\left( T\right)$ determined by specifying their kernels as one of the three subgroups of order2in $V.$ For example, ${\varphi }_{1}\left( a\right) = \lambda$ and ${\varphi }_{1}\left( b\right) = \lambda$ has kernel $\langle \;{ab}\;\rangle ,$ that is,in this semidirect product both $a$ and $b$ act by inverting $T$ and ${ab}$ centralizes $T$ . If ${\varphi }_{2}$ and ${\varphi }_{3}$ have kernels $\langle a\rangle$ and $\langle b\rangle$ ,respectively,then one easily checks that the resulting three semidirect products are all isomorphic to ${S}_{3} \times {Z}_{2}$ ,where the ${Z}_{2}$ direct factor is the kernel of ${\varphi }_{i}$ . For example,

最后，假设 $V = \langle a\rangle \times \langle b\rangle \cong {Z}_{2} \times {Z}_{2}$ 。存在恰好三个非平凡同态从 $V$ 到 $\operatorname{Aut}\left( T\right)$，通过指定它们的核为 $V.$ 中阶为2的三个子群之一来确定。例如，${\varphi }_{1}\left( a\right) = \lambda$ 和 ${\varphi }_{1}\left( b\right) = \lambda$ 的核是 $\langle \;{ab}\;\rangle ,$，也就是说，在这个半直积中，$a$ 和 $b$ 通过求逆 $T$ 和 ${ab}$ 来作用，并且 ${ab}$ 使 $T$居中。如果 ${\varphi }_{2}$ 和 ${\varphi }_{3}$ 的核分别是 $\langle a\rangle$ 和 $\langle b\rangle$，那么很容易验证，得到的三个半直积都与 ${S}_{3} \times {Z}_{2}$ 同构，其中 ${Z}_{2}$ 直因子的核是 ${\varphi }_{i}$ 。例如，

In summary, there are precisely 5 groups of order 12, three of which are non-abelian.

总结起来，存在恰好5个12阶群，其中三个是非阿贝尔群。

Example: (Groups of Order ${p}^{3},p$ an odd prime)

示例：(奇素数阶的群)

Let $G$ be a group of order ${p}^{3},p$ an odd prime,and assume $G$ is not cyclic. By Exercise 9 of the previous section the map $x \mapsto {x}^{p}$ is a homomorphism from $G$ into $Z\left( G\right)$ and the kernel of this homomorphism has order ${p}^{2}$ or ${p}^{3}$ . In the former case $G$ must contain an element of order ${p}^{2}$ and in the latter case every nonidentity element of $G$ has order $p$ .

设 $G$ 是一个奇素数阶 ${p}^{3},p$ 的群，并假设 $G$ 不是循环群。根据上一节的练习9，映射 $x \mapsto {x}^{p}$ 是从 $G$ 到 $Z\left( G\right)$ 的同态，这个同态的核的阶是 ${p}^{2}$ 或 ${p}^{3}$。在前一种情况下 $G$ 必须包含一个阶为 ${p}^{2}$ 的元素，而在后一种情况下 $G$ 的每个非单位元素都有阶 $p$。

Case 1: $G$ has an element of order ${p}^{2}$

情况1：$G$ 有一个阶为 ${p}^{2}$ 的元素

Let $x$ be an element of order ${p}^{2}$ and let $H = \langle x\rangle$ . Note that since $H$ has index $p,H$ is normal in $G$ by Corollary 4.5. If $E$ is the kernel of the ${p}^{\text{th }}$ power map,then in this case $E \cong {Z}_{p} \times {Z}_{p}$ and $E \cap H = \left\langle {x}^{p}\right\rangle$ . Let $y$ be any element of $E - H$ and let $K = \langle y\rangle$ . By construction, $H \cap K = 1$ and so $G$ is isomorphic to ${Z}_{{p}^{2}} \rtimes {Z}_{p}$ ,for some $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ . If $\varphi$ is the trivial homomorphism, $G \cong {Z}_{{p}^{2}} \times {Z}_{p}$ ,so we need only consider the nontrivial homomorphisms. By Proposition 4.17 Aut $\left( H\right) \cong {\mathbb{Z}}_{p\left( {p - 1}\right) }$ is cyclic and so contains a unique subgroup of order $p$ ,explicitly given by $\langle \gamma \rangle$ where

设 $x$ 是阶为 ${p}^{2}$ 的元素，设 $H = \langle x\rangle$ 。注意由于 $H$ 的指数为 $p,H$ ，根据推论 4.5，$G$ 在 $E$ 中是正规子群。如果 ${p}^{\text{th }}$ 是 $E \cong {Z}_{p} \times {Z}_{p}$ 幂映射的核，那么在这种情况下 $E \cap H = \left\langle {x}^{p}\right\rangle$ 和 $y$ 。设 $E - H$ 的任意元素为 $K = \langle y\rangle$ ，设 $H \cap K = 1$ 。根据构造，${Z}_{{p}^{2}} \rtimes {Z}_{p}$ ，因此 $G$ 同构于 $\varphi : K \rightarrow \operatorname{Aut}\left( H\right)$ ，对于某个 $\varphi$ 。如果 $G \cong {Z}_{{p}^{2}} \times {Z}_{p}$ 是平凡同态，那么 $\left( H\right) \cong {\mathbb{Z}}_{p\left( {p - 1}\right) }$ ，所以我们只需要考虑非平凡同态。根据命题 4.17，Aut $p$ 是循环群，因此包含一个唯一的阶为 $\langle \gamma \rangle$ 的子群，具体由 给出，其中

As usual,up to choice of a generator for the cyclic group $K$ ,there is only one nontrivial homomorphism, $\varphi$ ,from $K$ into $\operatorname{Aut}\left( H\right)$ ,given by $\varphi \left( y\right) = \gamma$ ; hence up to isomorphism there is a unique non-abelian group $H \rtimes K$ in this case. This group is described in Example 7 above.

如往常一样，直到选择循环群 $K$ 的生成元，只有一个非平凡同态 $\varphi$ ，从 $K$ 到 $\operatorname{Aut}\left( H\right)$ ，由 $\varphi \left( y\right) = \gamma$ 给出；因此，在同构的意义下，这种情况下只有一个非阿贝尔群 $H \rtimes K$ 。这个群在上述示例 7 中有所描述。

Case 2: every nonidentity element of $G$ has order $p$

情况 2：$G$ 的每个非单位元素都有阶 $p$

In this case let $H$ be any subgroup of $G$ of order ${p}^{2}$ (see Exercise 29,Section 4.3). Necessarily $H \cong {Z}_{p} \times {Z}_{p}$ . Let $K = \langle y\rangle$ for any element $y$ of $G - H$ . Since $H$ has index $p,H \trianglelefteq G$ and since $K$ has order $p$ but is not contained in $H,H \cap K = 1$ . Then $G$ is isomorphic to $\left( {{\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}}\right) \rtimes {\mathbb{Z}}_{p}$ ,for some $\varphi : K \rightarrow \mathrm{{Aut}}\left( H\right)$ . If $\varphi$ is trivial, $G \cong {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}$ (the elementary abelian group),so we may assume $\varphi$ is nontrivial. By Proposition 4.17,

在此情况下，设 $H$ 为 $G$ 的任意子群，其阶为 ${p}^{2}$（参见练习29，第4.3节）。必然 $H \cong {Z}_{p} \times {Z}_{p}$ 。设 $K = \langle y\rangle$ 对于 $G - H$ 中的任意元素 $y$ 。由于 $H$ 的指数为 $p,H \trianglelefteq G$ ，且由于 $K$ 的阶为 $p$ 但不包含在 $H,H \cap K = 1$ 中。那么 $G$ 与 $\left( {{\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}}\right) \rtimes {\mathbb{Z}}_{p}$ 同构，对于某个 $\varphi : K \rightarrow \mathrm{{Aut}}\left( H\right)$ 。如果 $\varphi$ 是平凡的，$G \cong {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p}$（初等阿贝尔群），因此我们可以假设 $\varphi$ 是非平凡的。根据命题4.17，

so $\left| {\operatorname{Aut}\left( H\right) }\right| = \left( {{p}^{2} - 1}\right) \left( {{p}^{2} - p}\right)$ . Note that a Sylow $p$ -subgroup of $\operatorname{Aut}\left( H\right)$ has order $p$ so all subgroups of order $p$ in $\operatorname{Aut}\left( H\right)$ are conjugate in $\operatorname{Aut}\left( H\right)$ by Sylow’s Theorem. Explicitly,(as discussed in Example 7 above) every subgroup of order $p$ in $\operatorname{Aut}\left( H\right)$ is conjugate to $\langle \gamma \rangle$ ,where if $H = \langle a\rangle \times \langle b\rangle$ ,the automorphism $\gamma$ is defined by

所以 $\left| {\operatorname{Aut}\left( H\right) }\right| = \left( {{p}^{2} - 1}\right) \left( {{p}^{2} - p}\right)$ 。注意，$\operatorname{Aut}\left( H\right)$ 的 Sylow $p$ -子群的阶为 $p$ ，因此 $\operatorname{Aut}\left( H\right)$ 中所有阶为 $p$ 的子群都由 Sylow定理在 $\operatorname{Aut}\left( H\right)$ 中共轭。具体来说（如上例7所述），$\operatorname{Aut}\left( H\right)$ 中每个阶为 $p$ 的子群都与 $\langle \gamma \rangle$ 共轭，其中如果 $H = \langle a\rangle \times \langle b\rangle$ ，则自同构 $\gamma$ 定义为

With respect to the ${\mathbb{F}}_{p}$ -basis $a,b$ of the 2-dimensional vector space $H$ the automorphism

关于 ${\mathbb{F}}_{p}$ -基 $a,b$ 的2维向量空间 $H$ 的自同构

has matrix

其矩阵为

Thus (again quoting Exercise 6) there is a unique isomorphism type of semidirect product in this case.

因此（再次引用练习6），在这种情况下，半直积的同构类型是唯一的。

Finally, since the two non-abelian groups have different orders for the kernels of the ${p}^{\text{th }}$ power maps,they are not isomorphic. A presentation for this group is also given in Example 7 above.

最后，由于这两个非阿贝尔群对于 ${p}^{\text{th }}$ 次幂映射的核有不同的阶，它们不是同构的。这个群的表示也在上面的例7中给出。

EXERCISES

练习

Let $H$ and $K$ be groups,let $\varphi$ be a homomorphism from $K$ into $\operatorname{Aut}\left( H\right)$ and,as usual,identify $H$ and $K$ as subgroups of $G = H{ \rtimes }_{\varphi }K$ .

设 $H$ 和 $K$ 为群，$\varphi$ 为从 $K$ 到 $\operatorname{Aut}\left( H\right)$ 的同态，并且如往常一样，将 $H$ 和 $K$ 视为 $G = H{ \rtimes }_{\varphi }K$ 的子群。

1. Prove that ${C}_{K}\left( H\right) = \ker \varphi$ (recall that ${C}_{K}\left( H\right) = {C}_{G}\left( H\right) \cap K$ ).

2. 证明 ${C}_{K}\left( H\right) = \ker \varphi$（回忆 ${C}_{K}\left( H\right) = {C}_{G}\left( H\right) \cap K$）。

3. Prove that ${C}_{H}\left( K\right) = {N}_{H}\left( K\right)$ .

4. 证明 ${C}_{H}\left( K\right) = {N}_{H}\left( K\right)$。

5. In Example 1 following the proof of Proposition 11 prove that every element of $G - H$ has order 2. Prove that $G$ is abelian if and only if ${h}^{2} = 1$ for all $h \in H$ .

6. 在命题 11 的证明后的例 1 中，证明 $G - H$ 的每个元素都有阶 2。证明当且仅当对于所有 $h \in H$ ，$G$ 是阿贝尔群时 ${h}^{2} = 1$。

7. Let $p = 2$ and check that the construction of the two non-abelian groups of order ${p}^{3}$ is valid in this case. Prove that both resulting groups are isomorphic to ${D}_{8}$ .

8. 设 $p = 2$ 并验证在这种情况下，构造的两个非阿贝尔群阶数为 ${p}^{3}$ 的有效性。证明这两个结果群同构于 ${D}_{8}$。

9. Let $G = \operatorname{Hol}\left( {{Z}_{2} \times {Z}_{2}}\right)$ .

10. 设 $G = \operatorname{Hol}\left( {{Z}_{2} \times {Z}_{2}}\right)$。

(a) Prove that $G = H \rtimes K$ where $H = {Z}_{2} \times {Z}_{2}$ and $K \cong {S}_{3}$ . Deduce that $\left| G\right| = {24}$ .

(a) 证明 $G = H \rtimes K$ 其中 $H = {Z}_{2} \times {Z}_{2}$ 和 $K \cong {S}_{3}$。推导出 $\left| G\right| = {24}$。

(b) Prove that $G$ is isomorphic to ${S}_{4}$ . [Obtain a homomorphism from $G$ into ${S}_{4}$ by letting $G$ act on the left cosets of $K$ . Use Exercise 1 to show this representation is faithful.]

(b) 证明 $G$ 同构于 ${S}_{4}$。[通过让 $G$ 作用于 $K$ 的左陪集来获得从 $G$ 到 ${S}_{4}$ 的同态。使用练习 1 来证明这个表示是忠实的。]

6. Assume that $K$ is a cyclic group, $H$ is an arbitrary group and ${\varphi }_{1}$ and ${\varphi }_{2}$ are homomorphisms from $K$ into $\mathrm{{Aut}}\left( H\right)$ such that ${\varphi }_{1}\left( K\right)$ and ${\varphi }_{2}\left( K\right)$ are conjugate subgroups of $\mathrm{{Aut}}\left( H\right)$ . If $K$ is infinite assume ${\varphi }_{1}$ and ${\varphi }_{2}$ are injective. Prove by constructing an explicit isomorphism that $\begin{matrix} H{ \rtimes }_{{\varphi }_{1}}K \cong H{ \rtimes }_{{\varphi }_{2}}K\text{ (in particular,if the subgroups }{\varphi }_{1}\left( K\right) \text{ and }{\varphi }_{2}\left( K\right) \text{ are equal in }\mathrm{{Aut}}\left( H\right) , \end{matrix}$ then the resulting semidirect products are isomorphic). [Suppose $\sigma {\varphi }_{1}\left( K\right) {\sigma }^{-1} = {\varphi }_{2}\left( K\right)$ so that for some $a \in \mathbb{Z}$ we have $\sigma {\varphi }_{1}\left( k\right) {\sigma }^{-1} = {\varphi }_{2}{\left( k\right) }^{a}$ for all $k \in K$ . Show that the map $\psi : H{ \rtimes }_{{\varphi }_{1}}K \rightarrow H{ \rtimes }_{{\varphi }_{2}}K$ defined by $\psi \left( \left( {h,k}\right) \right) = \left( {\sigma \left( h\right) ,{k}^{a}}\right)$ is a homomorphism. Show $\psi$ is bijective by constructing a 2-sided inverse.]

7. 假设 $K$ 是一个循环群，$H$ 是一个任意群，${\varphi }_{1}$ 和 ${\varphi }_{2}$ 是从 $K$ 到 $\mathrm{{Aut}}\left( H\right)$ 的同态，使得 ${\varphi }_{1}\left( K\right)$ 和 ${\varphi }_{2}\left( K\right)$ 是 $\mathrm{{Aut}}\left( H\right)$ 的共轭子群。如果 $K$ 是无限的，假设 ${\varphi }_{1}$ 和 ${\varphi }_{2}$ 是单射。通过构造一个明确的同构来证明 $\begin{matrix} H{ \rtimes }_{{\varphi }_{1}}K \cong H{ \rtimes }_{{\varphi }_{2}}K\text{ (in particular,if the subgroups }{\varphi }_{1}\left( K\right) \text{ and }{\varphi }_{2}\left( K\right) \text{ are equal in }\mathrm{{Aut}}\left( H\right) , \end{matrix}$，那么得到的半直积是同构的）。[假设 $\sigma {\varphi }_{1}\left( K\right) {\sigma }^{-1} = {\varphi }_{2}\left( K\right)$，因此对于某些 $a \in \mathbb{Z}$ 我们有 $\sigma {\varphi }_{1}\left( k\right) {\sigma }^{-1} = {\varphi }_{2}{\left( k\right) }^{a}$ 对于所有 $k \in K$。证明由 $\psi : H{ \rtimes }_{{\varphi }_{1}}K \rightarrow H{ \rtimes }_{{\varphi }_{2}}K$ 定义的映射 $\psi \left( \left( {h,k}\right) \right) = \left( {\sigma \left( h\right) ,{k}^{a}}\right)$ 是一个同态。通过构造一个双边逆元来证明 $\psi$ 是双射。]

8. This exercise describes thirteen isomorphism types of groups of order 56 . (It is not too difficult to show that every group of order 56 is isomorphic to one of these.)

9. 这个练习描述了13种阶为56的群的同构类型。（证明每个阶为56的群同构于这些群之一并不太困难。）

(a) Prove that there are three abelian groups of order 56 .

(a) 证明存在三个阶为56的阿贝尔群。

(b) Prove that every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.

(b) 证明每个阶为56的群要么有一个正规 Sylow 2-子群，要么有一个正规 Sylow 7-子群。

(c) Construct the following non-abelian groups of order 56 which have a normal Sylow 7-subgroup and whose Sylow 2-subgroup $S$ is as specified:

(c) 构造以下阶为56的非阿贝尔群，这些群有一个正规的 Sylow 7-子群，且其 Sylow 2-子群 $S$ 如下所述：

one group when $S \cong {Z}_{2} \times {Z}_{2} \times {Z}_{2}$

当 $S \cong {Z}_{2} \times {Z}_{2} \times {Z}_{2}$ 时一个群。

two nonisomorphic groups when $S \cong {Z}_{4} \times {Z}_{2}$

当 $S \cong {Z}_{4} \times {Z}_{2}$ 时两个非同构的群。

one group when $S \cong {Z}_{8}$

当 $S \cong {Z}_{8}$ 时一个群。

two nonisomorphic groups when $S \cong {Q}_{8}$

当 $S \cong {Q}_{8}$ 时两个非同构的群。

three nonisomorphic groups when $S \cong {D}_{8}$ .

当 $S \cong {D}_{8}$ 时三个非同构的群。

[For a particular $S$ ,two groups are not isomorphic if the kernels of the maps from $S$ into $\operatorname{Aut}\left( {Z}_{7}\right)$ are not isomorphic.]

[对于特定的 $S$，如果从 $S$ 到 $\operatorname{Aut}\left( {Z}_{7}\right)$ 的映射的核不同构，则两个群不是同构的。]

(d) Let $G$ be a group of order 56 with a nonnormal Sylow 7-subgroup. Prove that if $S$ is the Sylow 2-subgroup of $G$ then $S \cong {Z}_{2} \times {Z}_{2} \times {Z}_{2}$ . [Let an element of order 7 act by conjugation on the seven nonidentity elements of $S$ and deduce that they all have the same order.]

(d) 设 $G$ 是一个阶数为56的群，且具有一个非正规的Sylow 7-子群。证明如果 $S$ 是 $G$ 的Sylow 2-子群，那么 $S \cong {Z}_{2} \times {Z}_{2} \times {Z}_{2}$ 。[让一个阶数为7的元素通过共轭作用于 $S$ 的七个非单位元素，并推导出它们都有相同的阶数。]

(e) Prove that there is a unique group of order 56 with a nonnormal Sylow 7-subgroup. [For existence use the fact that $\left| {G{L}_{3}\left( {\mathbb{F}}_{2}\right) }\right| = {168}$ ; for uniqueness use Exercise 6.]

(e) 证明存在一个唯一的阶数为56且具有非正规Sylow 7-子群的群。[存在性使用 $\left| {G{L}_{3}\left( {\mathbb{F}}_{2}\right) }\right| = {168}$ 的事实；唯一性使用练习6。]

8. Construct a non-abelian group of order 75 . Classify all groups of order 75 (there are three of them). [Use Exercise 6 to show that the non-abelian group is unique.] (The classification of groups of order $p{q}^{2}$ ,where $p$ and $q$ are primes with $p < q$ and $p$ not dividing $q - 1$ , is quite similar.)

9. 构造一个阶数为75的非阿贝尔群。分类所有阶数为75的群（有三个）。[使用练习6来证明非阿贝尔群是唯一的。]（阶数为 $p{q}^{2}$ 的群的分类，其中 $p$ 和 $q$ 是素数且 $p < q$ 和 $p$ 不整除 $q - 1$ ，是非常相似的。）

10. Show that the matrix $\left( \begin{array}{rr} 0 & - 1 \\ 1 & 4 \end{array}\right)$ is an element of order 5 in $G{L}_{2}\left( {\mathbb{F}}_{19}\right)$ . Use this matrix to construct a non-abelian group of order 1805 and give a presentation of this group. Classify groups of order 1805 (there are three isomorphism types). [Use Exercise 6 to prove uniqueness of the non-abelian group.] (A general method for finding elements of prime order in $G{L}_{n}\left( {\mathbb{F}}_{p}\right)$ is described in the exercises in Section 12.2; this particular matrix of order 5 in $G{L}_{2}\left( {\mathbb{F}}_{19}\right)$ appears in Exercise 16 of that section as an illustration of the method.)

11. 证明矩阵 $\left( \begin{array}{rr} 0 & - 1 \\ 1 & 4 \end{array}\right)$ 是 $G{L}_{2}\left( {\mathbb{F}}_{19}\right)$ 中阶数为5的元素。使用这个矩阵构造一个阶数为1805的非阿贝尔群，并给出这个群的表示。分类阶数为1805的群（有三个同构类型）。[使用练习6来证明非阿贝尔群的唯一性。]（在12.2节的练习中描述了在 $G{L}_{n}\left( {\mathbb{F}}_{p}\right)$ 中寻找素数阶元素的一般方法；这个特定的阶数为5的矩阵在12.2节的练习16中出现，作为方法的例证。）

12. This exercise classifies the groups of order 147 (there are six isomorphism types).

13. 这个练习分类了阶数为147的群（有六个同构类型）。

(a) Prove that there are two abelian groups of order 147.

(a) 证明存在两个阶数为147的阿贝尔群。

(b) Prove that every group of order 147 has a normal Sylow 7-subgroup.

(b) 证明每个阶数为147的群都有一个正规Sylow 7-子群。

(c) Prove that there is a unique non-abelian group whose Sylow 7-subgroup is cyclic.

(c) 证明存在一个唯一的非阿贝尔群，其Sylow 7-子群是循环的。

(d) $\operatorname{Let}{t}_{1} = \left( \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right)$ and ${t}_{2} = \left( \begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)$ be elements of $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ . Prove $P = \left\langle {{t}_{1},{t}_{2}}\right\rangle$ is a Sylow 3-subgroup of $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ and that $P \cong {Z}_{3} \times {Z}_{3}$ . Deduce that every subgroup of $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ of order 3 is conjugate in $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ to a subgroup of $P$ .

(d) 设 $\operatorname{Let}{t}_{1} = \left( \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right)$ 和 ${t}_{2} = \left( \begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)$ 是 $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ 的元素。证明 $P = \left\langle {{t}_{1},{t}_{2}}\right\rangle$ 是 $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ 的一个Sylow 3-子群，并且 $P \cong {Z}_{3} \times {Z}_{3}$ 。推导出 $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ 的每个阶数为3的子群在 $G{L}_{2}\left( {\mathbb{F}}_{7}\right)$ 中与 $P$ 的一个子群共轭。

(e) By Example 3 in Section 1 the group $P$ has four subgroups of order 3 and these are: ${P}_{1} = \left\langle {t}_{1}\right\rangle ,{P}_{2} = \left\langle {t}_{2}\right\rangle ,{P}_{3} = \left\langle {{t}_{1}{t}_{2}}\right\rangle$ ,and ${P}_{4} = \left\langle {{t}_{1}{t}_{2}^{2}}\right\rangle$ . For $i = 1,2,3,4$ let ${G}_{i} = \left( {{Z}_{7} \times {Z}_{7}}\right) { \rtimes }_{{\varphi }_{i}}{Z}_{3}$ ,where ${\varphi }_{i}$ is an isomorphism of ${Z}_{3}$ with the subgroup ${P}_{i}$ of $\operatorname{Aut}\left( {{Z}_{7} \times {Z}_{7}}\right)$ . For each $i$ describe ${G}_{i}$ in terms of generators and relations. Deduce that ${G}_{1} \cong {G}_{2}$ .

(e) 根据第1节中的示例3，群 $P$ 有四个阶数为3的子群，分别是：${P}_{1} = \left\langle {t}_{1}\right\rangle ,{P}_{2} = \left\langle {t}_{2}\right\rangle ,{P}_{3} = \left\langle {{t}_{1}{t}_{2}}\right\rangle$、${P}_{4} = \left\langle {{t}_{1}{t}_{2}^{2}}\right\rangle$。对于 $i = 1,2,3,4$，设 ${G}_{i} = \left( {{Z}_{7} \times {Z}_{7}}\right) { \rtimes }_{{\varphi }_{i}}{Z}_{3}$，其中 ${\varphi }_{i}$ 是 ${Z}_{3}$ 与 $\operatorname{Aut}\left( {{Z}_{7} \times {Z}_{7}}\right)$ 的子群 ${P}_{i}$ 的同构。对于每个 $i$，用生成元和关系描述 ${G}_{i}$。推导出 ${G}_{1} \cong {G}_{2}$。

(f) Prove that ${G}_{1}$ is not isomorphic to either ${G}_{3}$ or ${G}_{4}$ . [Show that the center of ${G}_{1}$ has

(f) 证明 ${G}_{1}$ 与 ${G}_{3}$ 或 ${G}_{4}$ 不同构。[证明 ${G}_{1}$ 的中心有

order 7 whereas the centers of ${G}_{3}$ and ${G}_{4}$ are trivial.]

(g) Prove that ${G}_{3}$ is not isomorphic to ${G}_{4}$ . [Show that every subgroup of order 7 in ${G}_{3}$ is normal in ${G}_{3}$ but that ${G}_{4}$ has subgroups of order 7 that are not normal.]

(h) Classify the groups of order 147 by showing that the six nonisomorphic groups described above (two from part (a),one from part (c) and ${G}_{1},{G}_{3}$ ,and ${G}_{4}$ ) are all the groups of order 147. [Use Exercise 6 and part (d).] (The classification of groups of order $p{q}^{2}$ ,where $p$ and $q$ are primes with $p < q$ and $p \mid q - 1$ ,is quite similar.)

11. Classify groups of order 28 (there are four isomorphism types).

12. Classify the groups of order 20 (there are five isomorphism types).

13. Classify groups of order ${4p}$ ,where $p$ is a prime greater than 3. [There are four isomorphism types when $p \equiv 3\left( {\;\operatorname{mod}\;4}\right)$ and five isomorphism types when $p \equiv 1\left( {\;\operatorname{mod}\;4}\right)$ .]

14. This exercise classifies the groups of order 60 (there are thirteen isomorphism types). Let $G$ be a group of order 60,let $P$ be a Sylow 5-subgroup of $G$ and let $Q$ be a Sylow 3-subgroup of $G$ .

(a) Prove that if $P$ is not normal in $G$ then $G \cong {A}_{5}$ . [See Section 4.5.]

(b) Prove that if $P \trianglelefteq G$ but $Q$ is not normal in $G$ then $G \cong {A}_{4} \times {Z}_{5}$ . [Show in this case that $\left. {P \leq Z\left( G\right) ,G/P \cong {A}_{4}\text{,a Sylow 2-subgroup}T\text{ of }G\text{ is normal and }T\;Q \cong {A}_{4}.}\right\rbrack$

(c) Prove that if both $P$ and $Q$ are normal in $G$ then $G \cong {Z}_{15} \rtimes T$ where $T \cong {Z}_{4}$ or ${Z}_{2} \times {Z}_{2}$ . Show in this case that there are six isomorphism types when $T$ is cyclic (one abelian) and there are five isomorphism types when $T$ is the Klein 4-group (one abelian). [Use the same ideas as in the classifications of groups of orders 30 and 20.]

15. Let $p$ be an odd prime. Prove that every element of order 2 in $G{L}_{2}\left( {\mathbb{F}}_{p}\right)$ is conjugate to a diagonal matrix with $\pm 1$ ’s on the diagonal. Classify the groups of order $2{p}^{2}$ . [If $A$ is a $2 \times 2$ matrix with ${A}^{2} = I$ and ${v}_{1},{v}_{2}$ is a basis for the underlying vector space,look at $A$ acting on the vectors ${w}_{1} = {v}_{1} + {v}_{2}$ and ${w}_{2} = {v}_{1} - {v}_{2}$ .]

16. Show that there are exactly 4 distinct homomorphisms from ${Z}_{2}$ into $\operatorname{Aut}\left( {Z}_{8}\right)$ . Prove that the resulting semidirect products are the groups: ${Z}_{8} \times {Z}_{2},{D}_{16}$ ,the quasidihedral group $Q{D}_{16}$ and the modular group $M$ (cf. the exercises in Section 2.5).

17. Show that for any $n \geq 3$ there are exactly 4 distinct homomorphisms from ${Z}_{2}$ into $\operatorname{Aut}\left( {Z}_{{2}^{n}}\right)$ . Prove that the resulting semidirect products give 4 nonisomorphic groups of order ${2}^{n + 1}.$ [Recall Exercises 21 to 23 in Section 2.3.] (These four groups together with the cyclic group and the generalized quaternion group, ${Q}_{{2}^{n + 1}}$ ,are all the groups of order ${2}^{n + 1}$ which possess a cyclic subgroup of index 2.)

18. Show that if $H$ is any group then there is a group $G$ that contains $H$ as a normal subgroup with the property that for every automorphism $\sigma$ of $H$ there is an element $g \in G$ such that conjugation by $g$ when restricted to $H$ is the given automorphism $\sigma$ ,i.e.,every automorphism of $H$ is obtained as an inner automorphism of $G$ restricted to $H$ .

19. Let $H$ be a group of order $n$ ,let $K = \operatorname{Aut}\left( H\right)$ and form $G = \operatorname{Hol}\left( H\right) = H \rtimes K$ (where $\varphi$ is the identity homomorphism). Let $G$ act by left multiplication on the left cosets of $K$ in $G$ and let $\pi$ be the associated permutation representation $\pi : G \rightarrow {S}_{n}$ .

(a) Prove the elements of $H$ are coset representatives for the left cosets of $K$ in $G$ and with this choice of coset representatives $\pi$ restricted to $H$ is the regular representation of $H$ .

(b) Prove $\pi \left( G\right)$ is the normalizer in ${S}_{n}$ of $\pi \left( H\right)$ . Deduce that under the regular representation of any finite group $H$ of order $n$ ,the normalizer in ${S}_{n}$ of the image of $H$ is isomorphic to $\operatorname{Hol}\left( H\right)$ . [Show $\left| G\right| = \left| {{N}_{{S}_{n}}\left( {\pi \left( H\right) }\right) }\right|$ using Exercises 1 and 2 above.]

(c) Deduce that the normalizer of the group generated by an $n$ -cycle in ${S}_{n}$ is isomorphic to $\operatorname{Hol}\left( {Z}_{n}\right)$ and has order ${n\varphi }\left( n\right)$ .

20. Let $p$ be an odd prime. Prove that if $P$ is a non-cyclic $p$ -group then $P$ contains a normal subgroup $U$ with $U \cong {Z}_{p} \times {Z}_{p}$ . Deduce that for odd primes $p$ a $p$ -group that contains a unique subgroup of order $p$ is cyclic. (For $p = 2$ it is a theorem that the generalized quaternion groups ${Q}_{{2}^{n}}$ are the only non-cyclic 2 -groups which contain a unique subgroup of order 2). [Proceed by induction on $\left| P\right|$ . Let $Z$ be a subgroup of order $p$ in $Z\left( P\right)$ and let $\bar{P} = P/Z$ . If $\bar{P}$ is cyclic then $P$ is abelian by Exercise 36 in Section 3.1 - show the result is true for abelian groups. When $\bar{P}$ is not cyclic use induction to produce a normal subgroup $\bar{H}$ of $\bar{P}$ with $\bar{H} \cong {Z}_{p} \times {Z}_{p}$ . Let $H$ be the complete preimage of $\bar{H}$ in $P$ ,so $\left| H\right| = {p}^{3}$ . Let ${H}_{0} = \left\{ {x \in H \mid {x}^{p} = 1}\right\}$ so that ${H}_{0}$ is a characteristic subgroup of $H$ of order ${p}^{2}$ or ${p}^{3}$ by Exercise 9 in Section 4. Show that a suitable subgroup of ${H}_{0}$ gives the desired normal subgroup $U$ .]

21. 设 $p$ 为一个奇素数。证明如果 $P$ 是一个非循环的 $p$ -群，那么 $P$ 包含一个正规子群 $U$ 满足 $U \cong {Z}_{p} \times {Z}_{p}$ 。推导出对于奇素数 $p$ ，一个包含唯一阶为 $p$ 的子群的 $p$ -群是循环的。（对于 $p = 2$ ，有一个定理指出广义四元数群 ${Q}_{{2}^{n}}$ 是唯一包含唯一阶为2的非循环2-群）。[通过对 $\left| P\right|$ 进行归纳。设 $Z$ 是 $Z\left( P\right)$ 中阶为 $p$ 的子群，令 $\bar{P} = P/Z$ 。如果 $\bar{P}$ 是循环的，那么根据3.1节练习36，$P$ 是阿贝尔群 - 证明该结果对于阿贝尔群成立。当 $\bar{P}$ 不是循环群时，使用归纳法构造 $\bar{P}$ 的一个正规子群 $\bar{H}$ 满足 $\bar{H} \cong {Z}_{p} \times {Z}_{p}$ 。设 $H$ 是 $\bar{H}$ 在 $P$ 中的完全原像，因此 $\left| H\right| = {p}^{3}$ 。设 ${H}_{0} = \left\{ {x \in H \mid {x}^{p} = 1}\right\}$ ，使得 ${H}_{0}$ 是 $H$ 的一个特征子群，其阶为 ${p}^{2}$ 或 ${p}^{3}$ ，根据4.1节练习9。证明 ${H}_{0}$ 的一个适当子群给出了所需的正规子群 $U$ 。]

22. Let $p$ be an odd prime and let $P$ be a $p$ -group. Prove that if every subgroup of $P$ is normal then $P$ is abelian. (Note that ${Q}_{8}$ is a non-abelian 2-group with this property,so the result is false for $p = 2$ .) [Use the preceding exercises and Exercise 15 of Section 4.]

23. 设 $p$ 为一个奇素数，设 $P$ 为一个 $p$ -群。证明如果 $P$ 的每个子群都是正规群，那么 $P$ 是阿贝尔群。（注意 ${Q}_{8}$ 是一个具有这种性质的非常数 2-群，因此该结果对于 $p = 2$ 是错误的。）[使用前面的练习以及第4节的练习15。]

24. Let $F$ be a field let $n$ be a positive integer and let $G$ be the group of upper triangular matrices in $G{L}_{n}\left( F\right)$ (cf. Exercise 16,Section 2.1)

25. 设 $F$ 为一个域，设 $n$ 为一个正整数，设 $G$ 为 $G{L}_{n}\left( F\right)$ 中的上三角矩阵群（参见第2.1节的练习16）。

(a) Prove that $G$ is the semidirect product $U \rtimes D$ where $U$ is the set of upper triangular matrices with 1’s down the diagonal (cf. Exercise 17,Section 2.1) and $D$ is the set of diagonal matrices in $G{L}_{n}\left( F\right)$ .

(a) 证明 $G$ 是半直积 $U \rtimes D$，其中 $U$ 是对角线上为1的上三角矩阵集合（参见第2.1节的练习17），而 $D$ 是 $G{L}_{n}\left( F\right)$ 中的对角矩阵集合。

(b) Let $n = 2$ . Recall that $U \cong F$ and $D \cong {F}^{ \times } \times {F}^{ \times }$ (cf. Exercise 11 in Section 3.1). Describe the homomorphism from $D$ into $\operatorname{Aut}\left( U\right)$ explicitly in terms of these isomorphisms (i.e.,show how each element of ${F}^{ \times } \times {F}^{ \times }$ acts as an automorphism on $F$ ).

(b) 设 $n = 2$。回忆 $U \cong F$ 和 $D \cong {F}^{ \times } \times {F}^{ \times }$（参见第3.1节的练习11）。用这些同构明确描述从 $D$ 到 $\operatorname{Aut}\left( U\right)$ 的同态（即，展示 ${F}^{ \times } \times {F}^{ \times }$ 的每个元素如何在 $F$ 上作为自同构作用）。

23. Let $K$ and $L$ be groups,let $n$ be a positive integer,let $\rho : K \rightarrow {S}_{n}$ be a homomorphism and let $H$ be the direct product of $n$ copies of $L$ . In Exercise 8 of Section 1 an injective homomorphism $\psi$ from ${S}_{n}$ into $\operatorname{Aut}\left( H\right)$ was constructed by letting the elements of ${S}_{n}$ permute the $n$ factors of $H$ . The composition $\psi \circ \rho$ is a homomorphism from $G$ into Aut(H). The wreath product of $L$ by $K$ is the semidirect product $H \rtimes K$ with respect to this homomorphism and is denoted by $L\wr K$ (this wreath product depends on the choice of permutation representation $\rho$ of $K$ - if none is given explicitly, $\rho$ is assumed to be the left regular representation of $K$ ).

令 $K$ 和 $L$ 为群，$n$ 为正整数，$\rho : K \rightarrow {S}_{n}$ 为同态，$H$ 为 $n$ 个 $L$ 的直积。在第 1 部分的练习 8 中，通过让 ${S}_{n}$ 的元素对 $H$ 的 $n$ 个因子进行排列，构造了一个从 ${S}_{n}$ 到 $\operatorname{Aut}\left( H\right)$ 的单射同态 $\psi$。合成 $\psi \circ \rho$ 是从 $G$ 到 Aut(H) 的同态。由 $L$ 关于 $K$ 的冠积是相对于此同态的半直积 $H \rtimes K$，并记作 $L\wr K$（这个冠积依赖于 $K$ 的置换表示 $\rho$ 的选择 - 如果没有明确给出，$\rho$ 被假定为 $K$ 的左正则表示）。

(a) Assume $K$ and $L$ are finite groups and $\rho$ is the left regular representation of $K$ . Find $\left| {L\wr K}\right|$ in terms of $\left| K\right|$ and $\left| L\right|$ .

(a) 假设 $K$ 和 $L$ 是有限群，$\rho$ 是 $K$ 的左正则表示。找到 $\left| {L\wr K}\right|$ 关于 $\left| K\right|$ 和 $\left| L\right|$ 的表达式。

(b) Let $p$ be a prime,let $K = L = {Z}_{p}$ and let $\rho$ be the left regular representation of $K$ . Prove that ${Z}_{p}\wr {Z}_{p}$ is a non-abelian group of order ${p}^{p + 1}$ and is isomorphic to a Sylow $p$ -subgroup of ${S}_{{p}^{2}}$ . [The $p$ copies of ${Z}_{p}$ whose direct product makes up $H$ may be represented by $p$ disjoint $p$ -cycles; these are cyclically permuted by $K$ .]

(b) 令 $p$ 为素数，$K = L = {Z}_{p}$ 和 $\rho$ 为 $K$ 的左正则表示。证明 ${Z}_{p}\wr {Z}_{p}$ 是一个阶为 ${p}^{p + 1}$ 的非阿贝尔群，并且同构于 ${S}_{{p}^{2}}$ 的一个 Sylow $p$ -子群。[由 $p$ 个 ${Z}_{p}$ 的副本组成的直积构成 $H$，这些副本可以表示为 $p$ 个不相交的 $p$ -循环；这些循环被 $K$ 循环排列。]

24. Let $n$ be an integer $> 1$ . Prove the following classification: every group of order $n$ is abelian if and only if $n = {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\ldots {p}_{r}^{{\alpha }_{r}}$ ,where ${p}_{1},\ldots ,{p}_{r}$ are distinct primes, ${\alpha }_{i} = 1$ or 2 for all $i \in \{ 1,\ldots ,r\}$ and ${p}_{i}$ does not divide ${p}_{j}^{{\alpha }_{j}} - 1$ for all $i$ and $j$ . [See Exercise 56 in Section 4.5.]

25. 设 $n$ 为一个整数 $> 1$ 。证明以下分类：每个阶为 $n$ 的群是阿贝尔群当且仅当 $n = {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\ldots {p}_{r}^{{\alpha }_{r}}$，其中 ${p}_{1},\ldots ,{p}_{r}$ 是不同的素数，${\alpha }_{i} = 1$ 或对所有 $i \in \{ 1,\ldots ,r\}$ 和 ${p}_{i}$ 不整除 ${p}_{j}^{{\alpha }_{j}} - 1$ 对所有 $i$ 和 $j$ 成立。 [参见第4.5节练习56。]

26. Let $H\left( {\mathbb{F}}_{p}\right)$ be the Heisenberg group over the finite field ${\mathbb{F}}_{p} = \mathbb{Z}/p\mathbb{Z}$ (cf. Exercise 20 in Section 4). Prove that $H\left( {\mathbb{F}}_{2}\right) \cong {D}_{8}$ ,and that $H\left( {\mathbb{F}}_{p}\right)$ has exponent $p$ and is isomorphic to the first non-abelian group in Example 7.

27. 设 $H\left( {\mathbb{F}}_{p}\right)$ 为有限域 ${\mathbb{F}}_{p} = \mathbb{Z}/p\mathbb{Z}$ 上的海森堡群（参见第4节的练习20）。证明 $H\left( {\mathbb{F}}_{2}\right) \cong {D}_{8}$，并且 $H\left( {\mathbb{F}}_{p}\right)$ 的指数为 $p$ 且同构于示例7中的第一个非阿贝尔群。