中等阶群中的应用

6.2 APPLICATIONS IN GROUPS OF MEDIUM ORDER

6.2 中等阶群中的应用

The purpose of this section is to work through a number of examples which illustrate many of the techniques we have developed. These examples use Sylow’s Theorems extensively and demonstrate how they are applied in the study of finite groups. Motivated by the Hölder Program we address primarily the problem of showing that for certain $n$ every group of order $n$ has a proper,nontrivial normal subgroup (i.e.,there are no simple groups of order $n$ ). In most cases we shall stop once this has been accomplished. However readers should be aware that in the process of achieving this result we shall $\frac{\text{already have determined}}{\text{a great deal of information about arbitrary groups of given}}$ order $n$ for the $n$ that we consider. This information could be built upon to classify groups of these orders (but in general this requires techniques beyond the simple use of semidirect products to construct groups).

本节的目的是通过大量示例来说明我们已经开发的许多技术。这些示例广泛使用 Sylow 定理，并展示了它们在有限群研究中的应用。受到 Hölder 计划的启发，我们主要解决这样一个问题：对于某些 $n$，每个 $n$ 阶的群都有一个适当、非平凡的正规子群（即不存在 $n$ 阶的简单群）。在大多数情况下，我们将在此完成后停止。然而，读者应当注意，在实现这一结果的过程中，我们将对考虑的 $n$ 进行 $\frac{\text{already have determined}}{\text{a great deal of information about arbitrary groups of given}}$ 阶的排序。这些信息可以用来对这类阶的群进行分类（但在一般情况下，这需要超出仅使用半直积构造群的技术）。

Since for $p$ a prime we have already proved that there are no simple $p$ -groups (other than the cyclic group of order $p,{Z}_{p}$ ) and since the structure of $p$ -groups can be very complicated (recall the table in Section 5.3), we shall not study the structure of $p$ -groups explicitly. Rather,the theory of $p$ -groups developed in the preceding section will be applied to subgroups of groups of non-prime-power order.

由于我们已经证明了对 $p$ 的素数，不存在简单的 $p$ -群（除了阶为 $p,{Z}_{p}$ 的循环群），并且 $p$ -群的结构可能非常复杂（回顾第5.3节的表格），我们不会明确研究 $p$ -群的结构。相反，上一节中发展的 $p$ -群理论将被应用于非素数次幂阶群子群的研究。

Finally,for certain $n$ (e.g.,60,168,360,504,...) there do exist simple groups of order $n$ so,of course,we cannot force every group of these orders to be nonsimple. As in Section 4.5 we can, in certain cases, prove there is a unique simple group of order $n$ and unravel some of its internal structure (Sylow numbers,etc.). We shall study simple groups of order 168 as an additional test case. Thus the Sylow Theorems will be applied in a number of different contexts to show how groups of a given order may be manipulated.

最后，对于某些 $n$（例如，60, 168, 360, 504,...），确实存在阶为 $n$ 的简单群，所以，显然，我们不能强制这些阶数的每个群都是非简单的。正如在第4.5节中，我们可以在某些情况下证明存在唯一的阶为 $n$ 的简单群，并揭示其一些内部结构（ Sylow 数等）。我们将研究阶数为168的简单群作为一个额外的测试案例。因此，Sylow 定理将在多个不同的背景下应用，以展示如何操作给定阶数的群。

We shall end this section with some comments on the existence problem for groups, particularly for finite simple groups.

我们将以一些关于群存在问题的评论结束本节，特别是对于有限简单群的存在问题。

For $n < {10000}$ there are 60 odd,non-prime-power numbers for which the congruence conditions of Sylow’s Theorems do not force at least one of the Sylow subgroups to be normal i.e., ${n}_{p}$ can be $> 1$ for all primes $p \mid n$ (recall that ${n}_{p}$ denotes the number of Sylow $p$ -subgroups). For example,no numbers of the form ${pq}$ ,where $p$ and $q$ are distinct primes occur in our list by results of Section 4.5. In contrast, for even numbers $< {500}$ there are already 46 candidates for orders of simple groups (the congruence conditions allow many more possibilities). Many of our numerical examples arise from these lists of numbers and we often use odd numbers because the Sylow congruence conditions allow fewer values for ${n}_{p}$ . The purpose of these examples is to illustrate the use of the results we have proved. Many of these examples can be dealt with by more advanced techniques (for example, the Feit-Thompson Theorem proves that there are no simple groups of odd composite order).

对于 $n < {10000}$ 而言，存在60个奇数，非素数幂的数，它们不满足 Sylow 定理的同余条件，从而不能保证至少有一个 Sylow 子群是正规子群，即 ${n}_{p}$ 可以是 $> 1$ 对于所有素数 $p \mid n$（请注意 ${n}_{p}$ 表示 Sylow $p$ -子群的数量）。例如，根据第4.5节的结果，没有任何形式的数 ${pq}$，其中 $p$ 和 $q$ 是不同的素数，出现在我们的列表中。相比之下，对于偶数 $< {500}$，已经有46个简单群阶数的候选者（同余条件允许更多的可能性）。我们的大部分数值例子都来源于这些数字列表，我们经常使用奇数，因为 Sylow 同余条件允许 ${n}_{p}$ 的值更少。这些例子的目的是说明我们所证明结果的使用。这些例子中的许多可以通过更高级的技术来处理（例如，Feit-Thompson 定理证明了不存在奇数合数阶的简单群）。

As we saw in the case $n = {30}$ in Section 4.5,even though Sylow’s Theorem permitted ${n}_{5} = 6$ and ${n}_{3} = {10},$ further examination showed that any group of order30 must have both ${n}_{5} = 1$ and ${n}_{3} = 1$ . Thus the congruence part of Sylow’s Theorem is a sufficient but by no means necessary condition for normality of a Sylow subgroup. For many $n$ (e.g., $n = {120}$ ) we can prove that there are no simple groups of order $n$ ,so there is a nontrivial normal subgroup but this subgroup may not be a Sylow subgroup. For example, ${S}_{5}$ and $S{L}_{2}\left( {\mathbb{F}}_{5}\right)$ both have order 120 . The group ${S}_{5}$ has a unique nontrivial proper normal subgroup of order 60 $\left( {A}_{5}\right)$ and $S{L}_{2}\left( {\mathbb{F}}_{5}\right)$ has a unique nontrivial proper normal subgroup of order 2 $\left( {Z\left( {S{L}_{2}\left( {\mathbb{F}}_{5}\right) }\right) \cong {Z}_{2}}\right) ,$ neither of which is a Sylow subgroup. Our techniques for producing normal subgroups must be flexible enough to cover such diverse possibilities. In this section we shall examine Sylow subgroups for different primes dividing $n$ ,intersections of Sylow subgroups,normalizers of $p$ -subgroups and many other less obvious subgroups. The elementary methods we outline are by no means exhaustive, even for groups of "medium" order.

如我们在第4.5节中的案例 $n = {30}$ 所见，尽管 Sylow 定理允许 ${n}_{5} = 6$ 和 ${n}_{3} = {10},$ 进一步的检查表明，任何阶为30的群都必须有 ${n}_{5} = 1$ 和 ${n}_{3} = 1$ 。因此，Sylow 定理的同余部分是对 Sylow 子群正规性的一个充分条件，但绝不是必要条件。对于许多 $n$（例如，$n = {120}$），我们可以证明不存在阶为 $n$ 的简单群，因此存在一个非平凡的正规子群，但这个子群可能不是 Sylow 子群。例如，${S}_{5}$ 和 $S{L}_{2}\left( {\mathbb{F}}_{5}\right)$ 都有阶120。群 ${S}_{5}$ 有一个唯一的非平凡真正规子群，其阶为60 $\left( {A}_{5}\right)$，而 $S{L}_{2}\left( {\mathbb{F}}_{5}\right)$ 有一个唯一的非平凡真正规子群，其阶为2 $\left( {Z\left( {S{L}_{2}\left( {\mathbb{F}}_{5}\right) }\right) \cong {Z}_{2}}\right) ,$，这两个都不是 Sylow 子群。我们生成正规子群的技术必须足够灵活，以涵盖如此多样的可能性。在本节中，我们将考察不同素数除 $n$ 的 Sylow 子群、Sylow 子群的交集、$p$ 子群的正规化子群以及许多其他不那么明显的子群。我们概述的初等方法对于即使是“中等”阶的群来说也远非详尽。

Some Techniques

一些技术

Before listing some techniques for producing normal subgroups in groups of a given (“medium”) order we note that in all the problems where one deals with groups of order $n$ ,for some specific $n$ ,it is first necessary to factor $n$ into prime powers and then to compute the permissible values of ${n}_{p}$ ,for all primes $p$ dividing $n$ . We emphasize the need to be comfortable computing mod $p$ when carrying out the last step. The techniques we describe may be listed as follows:

在列举一些在给定（“中等”）阶数的群中产生正规子群的技巧之前，我们注意到，在所有处理阶数为 $n$ 的群的问题中，对于某些特定的 $n$ ，首先需要将 $n$ 分解为素数的幂次，然后计算对于所有除 $n$ 的素数 $p$ 的 ${n}_{p}$ 的允许值。我们强调在进行最后一步时需要熟练计算模 $p$ 。我们描述的技巧可以列举如下：

(1) Counting elements.

（1）计算元素数量。

(2) Exploiting subgroups of small index.

（2）利用小指数的子群。

(3) Permutation representations.

（3）排列表示。

(4) Playing $p$ -subgroups off against each other for different primes $p$ .

（4）对于不同的素数 $p$ ，将 $p$ -子群相互对抗。

(5) Studying normalizers of intersections of Sylow $p$ -subgroups.

（5）研究 Sylow $p$ -子群交集的正规化子。

Counting Elements

计算元素数量

Let $G$ be a group of order $n$ ,let $p$ be a prime dividing $n$ and let $P \in {\operatorname{Syl}}_{p}\left( G\right)$ . If $\left| P\right| = p$ ,then every nonidentity element of $P$ has order $p$ and every element of $G$ of order $p$ lies in some conjugate of $P$ . By Lagrange’s Theorem distinct conjugates of $P$ intersect in the identity,hence in this case the number of elements of $G$ of order $p$ is ${n}_{p}\left( {p - 1}\right)$ .

设 $G$ 是一个阶数为 $n$ 的群，设 $p$ 是一个除 $n$ 的素数，设 $P \in {\operatorname{Syl}}_{p}\left( G\right)$ 。如果 $\left| P\right| = p$ ，那么 $P$ 的每个非单位元素都有阶数 $p$ ，并且 $G$ 中每个阶数为 $p$ 的元素都位于 $P$ 的某个共轭子群中。根据拉格朗日定理，不同的 $P$ 的共轭子群在单位元相交，因此在这种情况下，$G$ 中阶数为 $p$ 的元素的数量是 ${n}_{p}\left( {p - 1}\right)$ 。

If Sylow $p$ -subgroups for different primes $p$ have prime order and we assume none of these is normal, we can sometimes show that the number of elements of prime order is $> \left| G\right|$ . This contradiction would show that at least one of the ${n}_{p}$ ’s must be 1 (i.e., some Sylow subgroup is normal in $G$ ).

如果不同素数 $p$ 的 Sylow $p$ -子群都有素数阶数，并且我们假设这些子群中没有一个是正规子群，我们有时可以证明素数阶元素的数量是 $> \left| G\right|$ 。这个矛盾将表明 ${n}_{p}$ 中至少有一个必须是 1（即某些 Sylow 子群在 $G$ 中是正规子群）。

This is the argument we used (in Section 4.5) to prove that there are no simple groups of order 30. For another example,suppose $\left| G\right| = {105} = 3 \cdot 5 \cdot 7$ . If $G$ were simple,we must have ${n}_{3} = 7,{n}_{5} = {21}$ and ${n}_{7} = {15}$ . Thus the number of elements of order 3 is $7 \cdot 2 = {14}$ the number of elements of order 5 is ${21} \cdot 4\; = \;{84}$ the number of elements of order 7 is ${15} \cdot 6 = {90}$ the number of elements of prime order is ${188} > \left| G\right|$ .

这是我们在第4.5节中使用的论据，用以证明不存在阶数为30的简单群。举另一个例子，假设 $\left| G\right| = {105} = 3 \cdot 5 \cdot 7$ 。如果 $G$ 是简单的，那么我们必须有 ${n}_{3} = 7,{n}_{5} = {21}$ 和 ${n}_{7} = {15}$ 。因此，阶数为3的元素的数量是 $7 \cdot 2 = {14}$ ，阶数为5的元素的数量是 ${21} \cdot 4\; = \;{84}$ ，阶数为7的元素的数量是 ${15} \cdot 6 = {90}$ ，素数阶的元素的数量是 ${188} > \left| G\right|$ 。

Sometimes counting elements of prime order does not lead to too many elements. However, there may be so few elements remaining that there must be a normal subgroup involving these elements. This was (in essence) the technique used in Section 4.5 to show that in a group of order 12 either ${n}_{2} = 1$ or ${n}_{3} = 1$ . This technique works particularly well when $G$ has a Sylow $p$ -subgroup $P$ of order $p$ such that ${N}_{G}\left( P\right) = P$ . For example,let $\left| G\right| = {56}$ . If $G$ were simple,the only possibility for the number of Sylow 7-subgroups is 8 , so

有时候计算素数阶的元素并不会导致元素数量过多。然而，剩余的元素可能如此之少，以至于必然存在一个包含这些元素的正常子群。这（本质上）是第4.5节中用来证明在阶数为12的群中要么 ${n}_{2} = 1$ 要么 ${n}_{3} = 1$ 的技术。当 $G$ 有一个Sylow $p$ -子群 $P$ ，其阶数为 $p$ 且满足 ${N}_{G}\left( P\right) = P$ 时，这种技术特别有效。例如，假设 $\left| G\right| = {56}$ 。如果 $G$ 是简单的，那么Sylow 7-子群的数量只能是8，所以

the number of elements of order 7 is $8 \cdot 6 = {48}$ .

阶数为7的元素的数量是 $8 \cdot 6 = {48}$ 。

Thus there are ${56} - {48} = 8$ elements remaining in $G$ . Since a Sylow 2-subgroup contains 8 elements (none of which have order 7), there can be at most one Sylow 2-subgroup,hence $G$ has a normal Sylow 2-subgroup.

因此在 $G$ 中还剩下 ${56} - {48} = 8$ 个元素。由于一个Sylow 2-子群包含8个元素（其中没有阶数为7的），所以最多只能有一个Sylow 2-子群，因此 $G$ 有一个正规Sylow 2-子群。

Exploiting Subgroups of Small Index

利用小指数子群

Recall that the results of Section 4.2 show that if $G$ has a subgroup $H$ of index $k$ , then there is a homomorphism from $G$ into the symmetric group ${S}_{k}$ whose kernel is contained in $H$ . If $k > 1$ ,this kernel is a proper normal subgroup of $G$ and if we are trying to prove that $G$ is not simple,we may,by way of contradiction,assume that this kernel is the identity. Then,by the First Isomorphism Theorem, $G$ is isomorphic to a subgroup of ${S}_{k}$ . In particular,the order of $G$ divides $k$ !. This argument shows that if $k$ is the smallest integer with $\left| G\right|$ dividing $k$ ! for a finite simple group $G$ then $G$ contains no proper subgroups of index less than $k$ . This smallest permissible index $k$ should be calculated at the outset of the study of groups of a given order $n$ . In the examples we consider this is usually quite easy: $n$ will often factor as

回顾第4.2节的结果，如果 $G$ 有一个指标为 $k$ 的子群 $H$ ，那么存在一个从 $G$ 到对称群 ${S}_{k}$ 的同态，其核包含在 $H$ 中。如果 $k > 1$ ，那么这个核是 $G$ 的一个真正规子群，如果我们试图证明 $G$ 不是单群，我们可以通过反证法假设这个核是单位元。那么，根据第一同构定理， $G$ 与 ${S}_{k}$ 的一个子群同构。特别是，$G$ 的阶数整除 $k$ !。这个论证表明，如果 $k$ 是使得 $\left| G\right|$ 整除 $k$ ! 的最小整数，对于一个有限单群 $G$ ，那么 $G$ 不包含指标小于 $k$ 的真子群。这个允许的最小指标 $k$ 应在研究给定阶数的群之前计算出来。在我们考虑的例子中这通常很容易：$n$ 通常可以分解为

and ${\alpha }_{s}$ is usually equal to 1 or 2 in our examples. In this case the minimal index of a proper subgroup will have to be at least ${p}_{s}$ (respectively $2{p}_{s}$ ) and this is often its exact value.

并且在我们的例子中 ${\alpha }_{s}$ 通常等于1或2。在这种情况下，一个真子群的最小指标必须至少为 ${p}_{s}$ （分别地 $2{p}_{s}$），这通常是它的确切值。

For example,there is no simple group of order 3393,because if $n = {3393} =$ ${3}^{2} \cdot {13} \cdot {29}$ ,then the minimal index of a proper subgroup is29( $n$ does not divide 28 ! because 29 does not divide 28!). However any simple group of order 3393 must have ${n}_{3} = {13}$ ,so for $P \in {\operatorname{Syl}}_{3}\left( G\right) ,{N}_{G}\left( P\right)$ has index 13,a contradiction.

例如，不存在阶数为3393的单群，因为如果 $n = {3393} =$ ${3}^{2} \cdot {13} \cdot {29}$ ，那么一个真子群的最小指标是29（$n$ 不整除28!，因为29不整除28!）。然而任何阶数为3393的单群必须有 ${n}_{3} = {13}$ ，所以对于 $P \in {\operatorname{Syl}}_{3}\left( G\right) ,{N}_{G}\left( P\right)$ 有指标13，这是一个矛盾。

Permutation Representations

排列表示

This method is a refinement of the preceding one. As above,if $G$ is a simple group of order $n$ with a proper subgroup of index $k$ ,then $G$ is isomorphic to a subgroup of ${S}_{k}$ . We may identify $G$ with this subgroup and so assume $G \leq {S}_{k}$ . Rather than relying only on Lagrange's Theorem for our contradiction (this was what we did for the preceding technique) we can sometimes show by calculating within ${S}_{k}$ that ${S}_{k}$ contains no simple subgroup of order $n$ . Two restrictions which may enable one to show such a result are

此方法是对前述方法的一种改进。如上所述，如果 $G$ 是一个阶数为 $n$ 的简单群，且具有一个指标为 $k$ 的适当子群，那么 $G$ 与 ${S}_{k}$ 的一个子群同构。我们可以将 $G$ 与这个子群等同起来，从而假设 $G \leq {S}_{k}$ 。与仅依赖拉格朗日定理来得到矛盾（这是我们前一个技术中使用的）不同，我们有时可以通过在 ${S}_{k}$ 内计算来证明 ${S}_{k}$ 不包含阶数为 $n$ 的简单子群。两个可能使人们能够证明此类结果的条件是

(1) if $G$ contains an element or subgroup of a particular order,so must ${S}_{k}$ ,and

（1）如果 $G$ 包含一个特定阶数的元素或子群，那么 ${S}_{k}$ 也必须包含；

(2) if $P \in {\operatorname{Syl}}_{p}\left( G\right)$ and if $P$ is also a Sylow $p$ -subgroup of ${S}_{k}$ ,then $\left| {{N}_{G}\left( P\right) }\right|$ must divide $\left| {{N}_{{S}_{k}}\left( P\right) }\right|$ .

（2）如果 $P \in {\operatorname{Syl}}_{p}\left( G\right)$ ，并且 $P$ 也是 ${S}_{k}$ 的一个 Sylow $p$ -子群，那么 $\left| {{N}_{G}\left( P\right) }\right|$ 必须整除 $\left| {{N}_{{S}_{k}}\left( P\right) }\right|$ 。

Condition (2) arises frequently when $p$ is a prime, $k = p$ or $p + 1$ and $G$ has a subgroup of index $k$ . In this case ${p}^{2}$ does not divide $\mathrm{k}$ !,so Sylow $p$ -subgroups of $G$ are also Sylow $p$ -subgroups of ${S}_{k}$ . Since now Sylow $p$ -subgroups of ${S}_{k}$ are precisely the groups generated by a $p$ -cycle,and distinct Sylow $p$ -subgroups intersect in the identity,

当 $p$ 是一个素数 $k = p$ 或 $p + 1$ ，并且 $G$ 有一个指标为 $k$ 的子群时，条件（2）经常出现。在这种情况下 ${p}^{2}$ 不整除 $\mathrm{k}$ ！因此 $G$ 的 Sylow $p$ -子群也是 ${S}_{k}$ 的 Sylow $p$ -子群。既然 ${S}_{k}$ 的 Sylow $p$ -子群正好是由一个 $p$ -循环生成的群，且不同的 Sylow $p$ -子群在单位元处相交，

the no. of Sylow $p$ -subgroups of ${S}_{k} = \frac{\text{ the no. of }p\text{-cycles in a Sylow }p\text{-subgroup }}{\text{ the no. of }p\text{-cycles in a Sylow }p}$

${S}_{k} = \frac{\text{ the no. of }p\text{-cycles in a Sylow }p\text{-subgroup }}{\text{ the no. of }p\text{-cycles in a Sylow }p}$ 的 Sylow $p$ -子群的数量

This number gives the index in ${S}_{k}$ of the normalizer of a Sylow $p$ -subgroup of ${S}_{k}$ . Thus for $k = p$ or $p + 1$

这个数字给出了 ${S}_{k}$ 中一个 Sylow $p$ -子群的正规化子的索引。因此对于 $k = p$ 或 $p + 1$

(cf. also the corresponding discussion for centralizers of elements in symmetric groups in Section 4.3 and the last exercises in Section 4.3). This proves, under the above hypotheses,that $\left| {{N}_{G}\left( P\right) }\right|$ must divide $p\left( {p - 1}\right)$ .

（参见第4.3节中关于对称群中元素的中心化子的相应讨论以及第4.3节最后的练习）。这证明了，在上述假设下，$\left| {{N}_{G}\left( P\right) }\right|$ 必须整除 $p\left( {p - 1}\right)$。

For example,if $G$ were a simple group of order ${396} = {2}^{2} \cdot {3}^{2} \cdot {11}$ ,we must have ${n}_{11} = {12}$ ,so if $P \in {Sy}{l}_{11}\left( G\right) ,\left| {G : {N}_{G}\left( P\right) }\right| = {12}$ and $\left| {{N}_{G}\left( P\right) }\right| = {33}$ . Since $G$ has a subgroup of index 12, $G$ is isomorphic to a subgroup of ${S}_{12}$ . But then (considering $G$ as actually contained in $\left. {S}_{12}\right) P \in {\operatorname{Syl}}_{11}\left( {S}_{12}\right)$ and $\left| {{N}_{{S}_{12}}\left( P\right) }\right| = {110}$ . Since ${N}_{G}\left( P\right) \leq$ ${N}_{{S}_{12}}\left( P\right)$ ,this would imply 33 | 110,clearly impossible,so we cannot have a simple group of order 396.

例如，如果 $G$ 是一个阶为 ${396} = {2}^{2} \cdot {3}^{2} \cdot {11}$ 的简单群，那么我们必须有 ${n}_{11} = {12}$，所以如果 $P \in {Sy}{l}_{11}\left( G\right) ,\left| {G : {N}_{G}\left( P\right) }\right| = {12}$ 和 $\left| {{N}_{G}\left( P\right) }\right| = {33}$。由于 $G$ 有一个指数为12的子群，$G$ 与 ${S}_{12}$ 的一个子群同构。但是，如果将 $G$ 实际视为包含在 $\left. {S}_{12}\right) P \in {\operatorname{Syl}}_{11}\left( {S}_{12}\right)$ 中，并且 $\left| {{N}_{{S}_{12}}\left( P\right) }\right| = {110}$。由于 ${N}_{G}\left( P\right) \leq$ ${N}_{{S}_{12}}\left( P\right)$，这将意味着 33 能整除 110，显然这是不可能的，所以我们不能有一个阶为396的简单群。

We can sometimes squeeze a little bit more out of this method by working in ${A}_{k}$ rather than ${S}_{k}$ . This slight improvement helps only occasionally and only for groups of even order. It is based on the following observations (the first of which we have made earlier in the text).

我们有时可以通过在 ${A}_{k}$ 中而不是在 ${S}_{k}$ 中工作，从这个方法中挤出更多东西。这种轻微的改进只在偶尔情况下有用，且仅适用于偶数阶的群。它基于以下观察（其中第一个我们在正文前面已经提到）。

Proposition 12.

命题12。

(1) If $G$ has no subgroup of index 2 and $G \leq {S}_{k}$ ,then $G \leq {A}_{k}$ .

(1) 如果 $G$ 没有指数为2的子群且 $G \leq {S}_{k}$，那么 $G \leq {A}_{k}$。

(2) If $P \in {Sy}{l}_{p}\left( {S}_{k}\right)$ for some odd prime $p$ ,then $P \in {Sy}{l}_{p}\left( {A}_{k}\right)$ and $\left| {{N}_{{A}_{k}}\left( P\right) }\right| =$ $\frac{1}{2}\left| {{N}_{{S}_{k}}\left( P\right) }\right|$

(2) 如果 $P \in {Sy}{l}_{p}\left( {S}_{k}\right)$ 对于某个奇素数 $p$，那么 $P \in {Sy}{l}_{p}\left( {A}_{k}\right)$ 且 $\left| {{N}_{{A}_{k}}\left( P\right) }\right| =$ $\frac{1}{2}\left| {{N}_{{S}_{k}}\left( P\right) }\right|$

Proof: The first assertion follows from the Second Isomorphism Theorem: if $G$ is not contained in ${A}_{k}$ ,then ${A}_{k} < G{A}_{k}$ so we must have $G{A}_{k} = {S}_{k}$ . But now

证明：第一个断言遵循第二同构定理：如果 $G$ 不包含在 ${A}_{k}$ 中，那么 ${A}_{k} < G{A}_{k}$，所以我们必须有 $G{A}_{k} = {S}_{k}$。但是，现在

so $G$ has a subgroup, $G \cap {A}_{k}$ ,of index 2 .

so $G$ has a subgroup, $G \cap {A}_{k}$ ,of index 2 .

To prove (2) note that if $P \in {\operatorname{Syl}}_{p}\left( {S}_{k}\right)$ ,for some odd prime $p$ ,by (1) (or order considerations) $P \leq {A}_{k}$ ,hence $P \in {\operatorname{Syl}}_{p}\left( {A}_{k}\right)$ as well. By Frattini’s Argument (Proposition 6)

To prove (2) note that if $P \in {\operatorname{Syl}}_{p}\left( {S}_{k}\right)$ ,for some odd prime $p$ ,by (1) (or order considerations) $P \leq {A}_{k}$ ,hence $P \in {\operatorname{Syl}}_{p}\left( {A}_{k}\right)$ as well. By Frattini’s Argument (Proposition 6)

so,in particular, ${N}_{{S}_{k}}\left( P\right)$ is not contained in ${A}_{k}$ . This forces ${N}_{{S}_{k}}\left( P\right) \cap {A}_{k}\left( { = {N}_{{A}_{k}}\left( P\right) }\right)$ to be a subgroup of index 2 in ${N}_{{S}_{k}}\left( P\right)$ .

so,in particular, ${N}_{{S}_{k}}\left( P\right)$ is not contained in ${A}_{k}$ . This forces ${N}_{{S}_{k}}\left( P\right) \cap {A}_{k}\left( { = {N}_{{A}_{k}}\left( P\right) }\right)$ to be a subgroup of index 2 in ${N}_{{S}_{k}}\left( P\right)$ .

For example,there is no simple group of order 264. Suppose $G$ were a simple group of order 264 $= {2}^{3} \cdot 3 \cdot {11}$ . We must have ${n}_{11} = {12}$ . As usual, $G$ would be isomorphic to a subgroup of ${S}_{12}$ . Since $G$ is simple (hence contains no subgroup of index 2), $G \leq {A}_{12}$ . Let $P \in {\operatorname{Syl}}_{11}\left( G\right)$ . Since ${n}_{11} = {12} = \left| {G : {N}_{G}\left( P\right) }\right|$ ,we have $\left| {{N}_{G}\left( P\right) }\right| = {22}$ . As above,

For example,there is no simple group of order 264. Suppose $G$ were a simple group of order 264 $= {2}^{3} \cdot 3 \cdot {11}$ . We must have ${n}_{11} = {12}$ . As usual, $G$ would be isomorphic to a subgroup of ${S}_{12}$ . Since $G$ is simple (hence contains no subgroup of index 2), $G \leq {A}_{12}$ . Let $P \in {\operatorname{Syl}}_{11}\left( G\right)$ . Since ${n}_{11} = {12} = \left| {G : {N}_{G}\left( P\right) }\right|$ ,we have $\left| {{N}_{G}\left( P\right) }\right| = {22}$ . As above,

however,22 does not divide 55,a contradiction to ${N}_{G}\left( P\right) \leq {N}_{{A}_{12}}\left( P\right)$ .

however,22 does not divide 55,a contradiction to ${N}_{G}\left( P\right) \leq {N}_{{A}_{12}}\left( P\right)$ .

Finally, we emphasize that we have only barely touched upon the combinatorial information available from certain permutation representations. Whenever possible in the remaining examples we shall illustrate other applications of this technique.

Finally, we emphasize that we have only barely touched upon the combinatorial information available from certain permutation representations. Whenever possible in the remaining examples we shall illustrate other applications of this technique.

Playing $p$ -Subgroups Off Against Each Other for Different Primes $p$

Playing $p$ -Subgroups Off Against Each Other for Different Primes $p$

Suppose $p$ and $q$ are distinct primes such that every group of order ${pq}$ is cyclic. This is equivalent to $p \nmid q - 1$ ,where $p < q$ . If $G$ has a Sylow $q$ -subgroup $Q$ of order $q$ and $p\left| {{N}_{G}\left( Q\right) }\right|$ ,applying Cauchy’s Theorem in ${N}_{G}\left( Q\right)$ gives a group $P$ of order $p$ normalizing $Q$ (note that $P$ need not be a Sylow $p$ -subgroup of $G$ ). Thus ${PQ}$ is a group and if ${PQ}$ is abelian,we obtain

假设 $p$ 和 $q$ 是不同的素数，使得每个阶为 ${pq}$ 的群都是循环群。这等价于 $p \nmid q - 1$ ，其中 $p < q$ 。如果 $G$ 有一个阶为 $q$ 的 Sylow $q$ -子群 $Q$ 且 $p\left| {{N}_{G}\left( Q\right) }\right|$ ，在 ${N}_{G}\left( Q\right)$ 中应用 Cauchy 定理得到一个阶为 $p$ 的群 $P$ 规范化 $Q$（注意 $P$ 不一定是 $G$ 的 Sylow $p$ -子群）。因此 ${PQ}$ 是一个群，如果 ${PQ}$ 是阿贝尔群，我们得到

(A symmetric argument applies if Sylow $p$ -subgroups of $G$ have order $p$ and $q$ divides the order of a Sylow $p$ -normalizer). This numerical information alone may be sufficient to force ${N}_{G}\left( P\right) = G$ (i.e., $P \trianglelefteq G$ ),or at least to force ${N}_{G}\left( P\right)$ to have index smaller than the minimal index permitted by permutation representations, giving a contradiction by a preceding technique.

（如果 $G$ 的 Sylow $p$ -子群的阶为 $p$ 且 $q$ 能整除一个 Sylow $p$ -正规化子的阶，则对称的论证也适用）。仅凭这些数值信息可能足以迫使 ${N}_{G}\left( P\right) = G$（即 $P \trianglelefteq G$ ），或者至少迫使 ${N}_{G}\left( P\right)$ 的指数小于由排列表示法允许的最小指数，通过先前技术给出矛盾。

For example,there are no simple groups of order 1785. If there were,let $G$ be a simple group of order 1785 $= 3 \cdot 5 \cdot 7 \cdot {17}$ . The only possible value for ${n}_{17}$ is 35,so if $\;Q$ is a Sylow 17-subgroup, $\left| {G\; : \;{N}_{G}\left( Q\right) }\right| = {35}.$ Thus $\left| {{N}_{G}\left( Q\right) }\right| = 3 \cdot {17}.$ Let $P$ be a Sylow 3-subgroup of ${N}_{G}\left( Q\right)$ . The group ${PQ}$ is abelian since 3 does not divide ${17} - 1$ ,so $Q \leq {N}_{G}\left( P\right)$ and ${17}\left| {{N}_{G}\left( P\right) }\right|$ . In this case $P \in {\operatorname{Syl}}_{3}\left( G\right)$ . The permissible values of ${n}_{3}$ are 7,85 and 595; however,since 17 $\left| {{N}_{G}\left( P\right) }\right|$ ,we cannot have 17 $\left| {G : {N}_{G}\left( P\right) }\right| = {n}_{3}$ . Thus ${n}_{3} = 7$ . But $G$ has no proper subgroup of index $<$ 17 (the minimal index of a proper subgroup is 17 for this order),a contradiction. Alternatively,if ${n}_{3} = 7$ ,then $\left| {{N}_{G}\left( P\right) }\right| = 3 \cdot 5 \cdot {17}$ ,and by Sylow’s Theorem applied in ${N}_{G}\left( P\right)$ we have $Q \trianglelefteq {N}_{G}\left( P\right)$ . This contradicts the fact that $\left| {{N}_{G}\left( Q\right) }\right| = 3 \cdot {17}$ .

例如，不存在阶数为1785的简单群。如果存在，设 $G$ 为阶数为1785的简单群 $= 3 \cdot 5 \cdot 7 \cdot {17}$ 。${n}_{17}$ 的唯一可能值是35，所以如果 $\;Q$ 是一个Sylow 17-子群， $\left| {G\; : \;{N}_{G}\left( Q\right) }\right| = {35}.$ 因此 $\left| {{N}_{G}\left( Q\right) }\right| = 3 \cdot {17}.$ 设 $P$ 为 ${N}_{G}\left( Q\right)$ 的一个Sylow 3-子群。由于3不整除 ${17} - 1$ ，所以群 ${PQ}$ 是阿贝尔群，因此 $Q \leq {N}_{G}\left( P\right)$ 和 ${17}\left| {{N}_{G}\left( P\right) }\right|$ 。在这种情况下 $P \in {\operatorname{Syl}}_{3}\left( G\right)$ 。${n}_{3}$ 的可允许值是7、85和595；然而，由于17 $\left| {{N}_{G}\left( P\right) }\right|$ ，我们不能有17 $\left| {G : {N}_{G}\left( P\right) }\right| = {n}_{3}$ 。因此 ${n}_{3} = 7$ 。但是 $G$ 没有指数为 $<$ 17的适当子群（对于此阶数，适当子群的最小指数是17），这是一个矛盾。另外，如果 ${n}_{3} = 7$ ，那么 $\left| {{N}_{G}\left( P\right) }\right| = 3 \cdot 5 \cdot {17}$ ，并且通过在 ${N}_{G}\left( P\right)$ 中应用Sylow定理，我们得到 $Q \trianglelefteq {N}_{G}\left( P\right)$ 。这与 $\left| {{N}_{G}\left( Q\right) }\right| = 3 \cdot {17}$ 的事实相矛盾。

We can refine this method by not requiring $P$ and $Q$ to be of prime order. Namely, if $p$ and $q$ are distinct primes dividing $\left| G\right|$ such that $Q \in {Sy}{l}_{q}\left( G\right)$ and $p \mid \left| {{N}_{G}\left( Q\right) }\right|$ , let $P \in {Sy}{l}_{p}\left( {{N}_{G}\left( Q\right) }\right)$ . We can then apply Sylow’s Theorems in ${N}_{G}\left( Q\right)$ to see whether $P \trianglelefteq {N}_{G}\left( Q\right)$ ,and if so,force ${N}_{G}\left( P\right)$ to be of small index. If $P$ is a Sylow $p$ -subgroup of the whole group $G$ ,we can use the congruence part of Sylow’s Theorem to put further restrictions on $\left| {{N}_{G}\left( P\right) }\right|$ (as we did in the preceding example). If $P$ is not a Sylow $p$ -subgroup of $G$ ,then by the second part of Sylow’s Theorem $P \leq {P}^{ * } \in {\operatorname{Syl}}_{p}\left( G\right)$ . In this case since $P < {P}^{ * }$ ,Theorem 1(4) shows that $P < {N}_{{P}^{ * }}\left( P\right)$ . Thus ${N}_{G}\left( P\right)$ (which contains ${N}_{{P}^{ * }}\left( P\right) )$ has order divisible by a larger power of $p$ than divides $\left| P\right|$ (as well as being divisible by $\left| Q\right|$ ).

我们可以通过不要求 $P$ 和 $Q$ 是素数阶的来改进这个方法。具体来说，如果 $p$ 和 $q$ 是不同的素数，它们整除 $\left| G\right|$ 且满足 $Q \in {Sy}{l}_{q}\left( G\right)$ 和 $p \mid \left| {{N}_{G}\left( Q\right) }\right|$ ，那么设 $P \in {Sy}{l}_{p}\left( {{N}_{G}\left( Q\right) }\right)$ 。然后我们可以应用 ${N}_{G}\left( Q\right)$ 中的 Sylow 定理来观察 $P \trianglelefteq {N}_{G}\left( Q\right)$ 是否成立，如果是的话，强制 ${N}_{G}\left( P\right)$ 为小指标。如果 $P$ 是整个群 $G$ 的一个 Sylow $p$ 子群，我们可以使用 Sylow 定理的同余部分对 $\left| {{N}_{G}\left( P\right) }\right|$ 进行进一步限制（就像我们在前一个例子中做的那样）。如果 $P$ 不是 $G$ 的一个 Sylow $p$ 子群，那么根据 Sylow 定理的第二部分 $P \leq {P}^{ * } \in {\operatorname{Syl}}_{p}\left( G\right)$ 。在这种情况下，由于 $P < {P}^{ * }$ ，定理 1(4) 显示 $P < {N}_{{P}^{ * }}\left( P\right)$ 。因此 ${N}_{G}\left( P\right)$（包含 ${N}_{{P}^{ * }}\left( P\right) )$）的阶被 $p$ 的更高次幂整除，而不是被 $\left| P\right|$ 整除（同时也被 $\left| Q\right|$ 整除）。

For example,there are no simple groups of order 3675. If there were,let $G$ be a simple group of order ${3675} = 3 \cdot {5}^{2} \cdot {7}^{2}$ . The only possibility for ${n}_{7}$ is 15,so for $Q \in {\operatorname{Syl}}_{7}\left( G\right) ,\left| {G : {N}_{G}\left( Q\right) }\right| = {15}$ and $\left| {{N}_{G}\left( Q\right) }\right| = {245} = 5 \cdot {7}^{2}$ . Let $N = {N}_{G}\left( Q\right)$ and let $P \in {\operatorname{Syl}}_{5}\left( N\right)$ . By the congruence conditions of Sylow’s Theorem applied in $N$ we get $P \trianglelefteq N$ . Since $\left| P\right| = 5,P$ is not itself a Sylow 5-subgroup of $G$ so $P$ is contained in some Sylow 5-subgroup ${P}^{ * }$ of $G$ . Since $P$ is of index 5 in the 5-group ${P}^{ * },P \trianglelefteq {P}^{ * }$ by Theorem 1,that is ${P}^{ * } \leq {N}_{G}\left( P\right)$ . This proves

例如，不存在阶数为3675的简单群。如果存在，设 $G$ 是阶数为 ${3675} = 3 \cdot {5}^{2} \cdot {7}^{2}$ 的简单群。 ${n}_{7}$ 的唯一可能是15，因此对于 $Q \in {\operatorname{Syl}}_{7}\left( G\right) ,\left| {G : {N}_{G}\left( Q\right) }\right| = {15}$ 和 $\left| {{N}_{G}\left( Q\right) }\right| = {245} = 5 \cdot {7}^{2}$ 。设 $N = {N}_{G}\left( Q\right)$ 并且设 $P \in {\operatorname{Syl}}_{5}\left( N\right)$ 。根据应用于 $N$ 的 Sylow 定理的同余条件，我们得到 $P \trianglelefteq N$ 。由于 $\left| P\right| = 5,P$ 本身不是 $G$ 的 Sylow 5-子群，所以 $P$ 包含在 $G$ 的某个 Sylow 5-子群 ${P}^{ * }$ 中。由于 $P$ 在5-群 ${P}^{ * },P \trianglelefteq {P}^{ * }$ 中的指数为5，根据定理1，即 ${P}^{ * } \leq {N}_{G}\left( P\right)$ 。这证明了

Thus $\left| {G : {N}_{G}\left( P\right) }\right|$ 3,which is impossible since $P$ is not normal and $G$ has no subgroup of index 3 .

因此 $\left| {G : {N}_{G}\left( P\right) }\right|$ 3，这是不可能的，因为 $P$ 不是正规群，并且 $G$ 没有指数为3的子群。

Studying Normalizers of Intersections of Sylow $p$ -Subgroups

研究 Sylow $p$ -子群交集的正规化子

One of the reasons the counting arguments in the first method above do not immediately generalize to Sylow subgroups which are not of prime order is because if $P \in {Sy}{l}_{p}\left( G\right)$ for some prime $p$ and $\left| P\right| = {p}^{a},a \geq 2$ ,then it need not be the case that distinct conjugates of $P$ intersect in the identity subgroup. If distinct conjugates of $P$ do intersect in the identity, we can again count to find that the number of elements of $p$ -power order is ${n}_{p}\left( {\left| P\right| - 1}\right)$ .

上述第一种方法中的计数论证不能立即推广到非素数阶的 Sylow 子群的一个原因是，如果 $P \in {Sy}{l}_{p}\left( G\right)$ 对于某个素数 $p$ 和 $\left| P\right| = {p}^{a},a \geq 2$ 成立，那么不同共轭的 $P$ 不一定在单位子群中相交。如果不同共轭的 $P$ 确实在单位子群中相交，我们可以再次计数以找到 $p$ -次幂阶的元素数量是 ${n}_{p}\left( {\left| P\right| - 1}\right)$ 。

Suppose,however,there exists $R \in {\operatorname{Syl}}_{p}\left( G\right)$ with $R \neq P$ and $P \cap R \neq 1$ . Let ${P}_{0} = P \cap R$ . Then ${P}_{0} < P$ and ${P}_{0} < R$ ,hence by Theorem 1

假设然而存在 $R \in {\operatorname{Syl}}_{p}\left( G\right)$ 使得 $R \neq P$ 和 $P \cap R \neq 1$ 成立。设 ${P}_{0} = P \cap R$ 。那么 ${P}_{0} < P$ 和 ${P}_{0} < R$ ，因此根据定理1

One can try to use this to prove that the normalizer in $G$ of ${P}_{0}$ is sufficiently large (i.e., of sufficiently small index) to obtain a contradiction by previous methods (note that this normalizer is a proper subgroup since ${P}_{0} \neq 1$ ).

可以尝试使用这一点来证明 $G$ 中 ${P}_{0}$ 的正规化子足够大（即，指数足够小），通过之前的方法得到矛盾（注意，由于 ${P}_{0} \neq 1$ ，这个正规化子是一个真子群）。

One special case where this works particularly well is when $\left| {P}_{0}\right| = {p}^{a - 1}$ i.e.,the two Sylow $p$ -subgroups $R$ and $P$ have large intersection. In this case set $N = {N}_{G}\left( {P}_{0}\right)$ . Then by the above reasoning (i.e.,since ${P}_{0}$ is a maximal subgroup of the $p$ -groups $P$ and $R$ ), ${P}_{0} \trianglelefteq P$ and ${P}_{0} \trianglelefteq R$ ,that is,

这种方法特别有效的一个特例是当 $\left| {P}_{0}\right| = {p}^{a - 1}$ ，即两个 Sylow $p$ 子群 $R$ 和 $P$ 有较大交集时。在这种情况下，设 $N = {N}_{G}\left( {P}_{0}\right)$ 。那么根据上述推理（即，由于 ${P}_{0}$ 是 $p$ -群 $P$ 和 $R$ 的极大子群）， ${P}_{0} \trianglelefteq P$ 和 ${P}_{0} \trianglelefteq R$ ，也就是说，

$N$ has 2 distinct Sylow $p$ -subgroups: $P$ and $R$ .

$N$ 有两个不同的 Sylow $p$ 子群：$P$ 和 $R$。

In particular, $\left| N\right| = {p}^{a}k$ ,where (by Sylow’s Theorem) $k \geq p + 1$ .

特别是，$\left| N\right| = {p}^{a}k$ ，根据 Sylow 定理 $k \geq p + 1$ 。

Recapitulating,if Sylow $p$ -subgroups pairwise intersect in the identity,then counting elements of $p$ -power order is possible; otherwise there is some intersection of Sylow $\begin{matrix} p\text{-subgroups whose normalizer is “large.” Since for an arbitrary group order one cannot} \end{matrix}$ necessarily tell which of these two phenomena occurs, it may be necessary to split the nonsimplicity argument into two (mutually exclusive) cases and derive a contradiction in each. This process is especially amenable when the order of a Sylow $p$ -subgroup is ${p}^{2}$ (for example,this line of reasoning was used to count elements of 2-power order in the proof that a simple group of order 60 is isomorphic to ${A}_{5}$ -Proposition 23,Section 4.5).

总结一下，如果 Sylow $p$ 子群两两在单位元处相交，那么计算 $p$ -次幂阶的元素是可能的；否则，Sylow $\begin{matrix} p\text{-subgroups whose normalizer is “large.” Since for an arbitrary group order one cannot} \end{matrix}$ 必然存在一些交集。有时可能无法确定这两种现象中的哪一种会发生，此时可能需要将非单性论证分为两个（相互排斥的）情况，并在每种情况下导出矛盾。当 Sylow $p$ 子群的阶为 ${p}^{2}$ 时，这种方法尤其适用（例如，在证明阶数为60的简单群同构于 ${A}_{5}$ -命题23，第4.5节时，就使用了这一推理来计算2次幂阶的元素）。

Before proceeding with an example we state a lemma which gives a sufficient condition to force a nontrivial Sylow intersection.

在举例子之前，我们陈述一个引理，它给出了迫使 Sylow 交集非平凡的一个充分条件。

Lemma 13. In a finite group $G$ if ${n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right)$ ,then there are distinct Sylow $p$ -subgroups $P$ and $R$ of $G$ such that $P \cap R$ is of index $p$ in both $P$ and $R$ (hence is normal in each).

定理13。在一个有限群 $G$ 中，如果 ${n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right)$，那么存在不同的 Sylow $p$ 子群 $P$ 和 $R$ 在 $G$ 中，使得 $P \cap R$ 在 $P$ 和 $R$ 中的指数为 $p$（因此在每个群中都是正规群）。

Proof: The argument is an easy refinement of the proof of the congruence part of Sylow’s Theorem (cf. the exercises at the end of Section 4.5). Let $P$ act by conjugation on the set ${\operatorname{Syl}}_{p}\left( G\right)$ . Let ${\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{s}$ be the orbits under this action with ${\mathcal{O}}_{1} = \{ P\}$ . If ${p}^{2}$ divides $\left| {P : P \cap R}\right|$ for all Sylow $p$ -subgroups $R$ of $G$ different from $P$ ,then each ${\mathcal{O}}_{i}$ has size divisible by ${p}^{2},i = 2,3,\ldots ,s$ . In this case,since ${n}_{p}$ is the sum of the lengths of the orbits we would have ${n}_{p} = 1 + k{p}^{2}$ ,contrary to assumption. Thus for some $R \in {\operatorname{Syl}}_{p}\left( G\right) ,\left| {P : P \cap R}\right| = p$ .

证明：这个论证是对 Sylow 定理同余部分的证明的一个简单改进（参见第4.5节末尾的练习）。设 $P$ 通过共轭作用在集合 ${\operatorname{Syl}}_{p}\left( G\right)$ 上。设 ${\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{s}$ 是在此作用下的轨道，且 ${\mathcal{O}}_{1} = \{ P\}$。如果 ${p}^{2}$ 整除 $\left| {P : P \cap R}\right|$ 对于所有不同于 $P$ 的 $p$ -子群 $R$ 的 Sylow 子群，那么每个 ${\mathcal{O}}_{i}$ 的大小都能被 ${p}^{2},i = 2,3,\ldots ,s$ 整除。在这种情况下，由于 ${n}_{p}$ 是轨道长度的和，我们将有 ${n}_{p} = 1 + k{p}^{2}$，这与假设相反。因此对于某些 $R \in {\operatorname{Syl}}_{p}\left( G\right) ,\left| {P : P \cap R}\right| = p$。

For example,there are no simple groups of order 1053. If there were,let $G$ be a simple group of order ${1053} = {3}^{4} \cdot {13}$ and let $P \in {\operatorname{Syl}}_{3}\left( G\right)$ . We must have ${n}_{3} = {13}$ . But ${13} ≢ 1\left( {\;\operatorname{mod}\;{3}^{2}}\right)$ so there exist $P,R \in {\operatorname{Syl}}_{3}\left( G\right)$ such that $\left| {P \cap R}\right| = {3}^{3}$ . Let $N = {N}_{G}\left( {P \cap R}\right)$ ,so by the above arguments $P,R \leq N$ . Thus ${3}^{4}\left| N\right|$ and $\left| N\right| > {3}^{4}$ . The only possibility is $N = G$ ,i.e., $P \cap R \trianglelefteq G$ ,a contradiction.

例如，不存在阶为1053的简单群。如果存在，设 $G$ 是一个阶为 ${1053} = {3}^{4} \cdot {13}$ 的简单群，并且设 $P \in {\operatorname{Syl}}_{3}\left( G\right)$。我们必须有 ${n}_{3} = {13}$。但是 ${13} ≢ 1\left( {\;\operatorname{mod}\;{3}^{2}}\right)$，所以存在 $P,R \in {\operatorname{Syl}}_{3}\left( G\right)$ 使得 $\left| {P \cap R}\right| = {3}^{3}$。设 $N = {N}_{G}\left( {P \cap R}\right)$，所以根据上述论证 $P,R \leq N$。因此 ${3}^{4}\left| N\right|$ 并且 $\left| N\right| > {3}^{4}$。唯一可能的情况是 $N = G$，即 $P \cap R \trianglelefteq G$，这是一个矛盾。

Simple Groups of Order 168

阶为168的简单群

We now show how many of our techniques can be used to unravel the structure of and then classify certain simple groups by classifying the simple groups of order 168. Because there are no nontrivial normal subgroups in simple groups,this process departs from the methods in Section 5.5, but the overall approach typifies methods used in the study of finite simple groups.

我们现在展示如何将我们的许多技术用于解析某些简单群的结构，并通过分类阶数为168的简单群来对这些简单群进行分类。因为简单群中不存在非平凡正规子群，这一过程与第5.5节中的方法不同，但整体方法典型地代表了在有限简单群研究中使用的方法。

We begin by assuming there is a simple group $G$ of order 168 $= {2}^{3} \cdot 3 \cdot 7$ . We first work out many of its properties: the number and structure of its Sylow subgroups, the conjugacy classes, etc. All of these calculations are based only on the order and simplicity of $G$ . We use these results to first prove the uniqueness of $G$ ; and ultimately we prove the existence of the simple group of order 168.

我们从假设存在一个阶数为168的简单群 $G$ 开始。我们首先研究它的许多性质：它的 Sylow 子群的数量和结构、共轭类等。所有这些计算仅基于 $G$ 的阶数和简单性。我们利用这些结果首先证明了 $G$ 的唯一性；最终我们证明了阶数为168的简单群的存在。

Because $\left| G\right|$ does not divide 6 ! we have

因为 $\left| G\right|$ 不能整除 6 !，所以我们有

$G$ has no proper subgroup of index less than 7,

$G$ 没有指数小于7的正规子群，

since otherwise the action of $G$ on the cosets of the subgroup would give a (necessarily injective since $G$ is simple) homomorphism from $G$ into some ${S}_{n}$ with $n \leq 6$ .

否则 $G$ 对子群的陪集的作用将给出一个从 $G$ 到某个 ${S}_{n}$ 的（必然是单射的，因为 $G$ 是简单群）同态，其中 $n \leq 6$ 。

The simplicity of $G$ and Sylow’s Theorem also immediately imply that

$G$ 的简单性和 Sylow 定理也立即意味着

${n}_{7} = 8$ ,so the normalizer of a Sylow 7-subgroup has order 21. In particular,no element of order 2 normalizes a Sylow 7-subgroup and G has no elements of order 14. If $G$ had an element of order 21 then the normalizer of a Sylow 3-subgroup of $G$ would have order divisible by 7 . Thus ${n}_{3}$ would be relatively prime to 7 . Since then ${n}_{3} \mid 8$ we would have ${n}_{3} = 4$ contrary to (1). This proves:

${n}_{7} = 8$，所以一个 Sylow 7-子群的正规化子群有阶数21。特别地，没有任何阶数为2的元素正规化一个 Sylow 7-子群，且 G 没有阶数为14的元素。如果 $G$ 有一个阶数为21的元素，那么 $G$ 的一个 Sylow 3-子群的正规化子群的阶数将能被7整除。因此 ${n}_{3}$ 将与7互质。既然如此 ${n}_{3} \mid 8$，我们将有 ${n}_{3} = 4$ ，这与（1）矛盾。这证明了：

(3) G has no elements of order 21 .

(3) G 没有阶为 21 的元素。

By Sylow’s Theorem ${n}_{3} = 7$ or 28; we next rule out the former possibility. Assume ${n}_{3} = 7$ ,let $P \in {\operatorname{Syl}}_{3}\left( G\right)$ and let $T$ be a Sylow 2-subgroup of the group ${N}_{G}\left( P\right)$ of order 24. Each Sylow 3-subgroup normalizes some Sylow 7-subgroup of $G$ so $P$ normalizes a Sylow 7-subgroup $R$ of $G.$ For every $t \in T$ we also have that $P = t{\text{ Pt}}^{-1}$ normalizes ${tR}{t}^{-1}$ . The subgroup $T$ acts by conjugation on the set of eight Sylow 7-subgroups of $G$ and since no element of order 2 in $G$ normalizes a Sylow 7-subgroup by (2),it follows that $T$ acts transitively,i.e.,every Sylow 7-subgroup of $G$ is one of the ${tR}{t}^{-1}$ . Hence $P$ normalizes every Sylow 7-subgroup of $G$ ,i.e., $P$ is contained in the intersection of the normalizers of all Sylow 7-subgroups. But this intersection is a proper normal subgroup of $G$ ,so it must be trivial. This contradiction proves:

由 Sylow 定理 ${n}_{3} = 7$ 或 28；我们接下来排除前一种可能性。假设 ${n}_{3} = 7$ ，设 $P \in {\operatorname{Syl}}_{3}\left( G\right)$ 并且设 $T$ 是阶为 24 的群 ${N}_{G}\left( P\right)$ 的一个 Sylow 2-子群。每个 Sylow 3-子群正规化 $G$ 的某个 Sylow 7-子群，因此 $P$ 正规化一个 Sylow 7-子群 $R$ 的 $G.$ 对于每个 $t \in T$ ，我们也有 $P = t{\text{ Pt}}^{-1}$ 正规化 ${tR}{t}^{-1}$ 。子群 $T$ 通过共轭作用于 $G$ 的八个 Sylow 7-子群集合上，由于 $G$ 中没有阶为 2 的元素通过 (2) 正规化一个 Sylow 7-子群，因此可以得出 $T$ 作用 transitively，即 $G$ 的每个 Sylow 7-子群都是 ${tR}{t}^{-1}$ 之一。因此 $P$ 正规化 $G$ 的每个 Sylow 7-子群，即 $P$ 包含在所有 Sylow 7-子群正规化的交集中。但是这个交集是 $G$ 的一个真正规子群，所以它必须是平凡群。这个矛盾证明了：

Since ${n}_{2} = 7$ or 21,we have ${n}_{2} ≢ 1{\;\operatorname{mod}\;8}$ ,so by Exercise 21 there is a pair of distinct Sylow 2-subgroups that have nontrivial intersection; over all such pairs let ${T}_{1}$ and ${T}_{2}$ be chosen with $U = {T}_{1} \cap {T}_{2}$ of maximal order. We next prove

由于 ${n}_{2} = 7$ 或 21，我们有 ${n}_{2} ≢ 1{\;\operatorname{mod}\;8}$ ，因此由练习 21 可知存在一对不同的 Sylow 2-子群它们具有非平凡交集；在所有这样的对中，设 ${T}_{1}$ 和 ${T}_{2}$ 为选择的最大阶的 $U = {T}_{1} \cap {T}_{2}$ 。我们接下来证明

(5)

$U$ is a Klein 4-group and ${N}_{G}\left( U\right) \cong {S}_{4}$ .

$U$ 是一个 Klein 四元群且 ${N}_{G}\left( U\right) \cong {S}_{4}$ 。

Let $N = {N}_{G}\left( U\right)$ . Since $\left| U\right| = 2$ or 4 and $N$ permutes the nonidentity elements of $U$ by conjugation,a subgroup of order 7 in $N$ would commute with some element of order 2 in $U$ ,contradicting (2). It follows that the order of $N$ is not divisible by 7. By Exercise 13, $N$ has more than one Sylow 2-subgroup,hence $\left| N\right| = {2}^{a} \cdot 3$ ,where $a = 2$ or 3. Let $P \in {\operatorname{Syl}}_{3}\left( N\right)$ . Since $P$ is a Sylow 3-subgroup of $G$ ,by (4) the group ${N}_{N}\left( P\right)$ has order 3 or 6 (with $P$ as its unique subgroup of order 3). Thus by Sylow’s Theorem $N$ must have four Sylow 3-subgroups,and these are permuted transitively by $N$ under conjugation. Since any group of order 12 must have either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup (cf. Section 4.5), $\left| N\right| = {24}.$ Let $K$ be the kernel of $N$ acting by conjugation on its four Sylow 3-subgroups,so $K$ is the intersection of the normalizers of the Sylow 3-subgroups of $N$ . If $K = 1$ then $N \cong {S}_{4}$ as asserted; so consider when $K \neq 1$ . Since $K \leq {N}_{N}\left( P\right)$ ,the group $K$ has order dividing 6,and since $P$ does not normalize another Sylow 3-subgroup, $P$ is not contained in $K$ . It follows that $\left| K\right| = 2$ . But now $N/K$ is a group of order 12 which is seen to have more than one Sylow 2-subgroup and four Sylow 3-subgroups,contrary to the property of groups of order 12 cited earlier. This proves $N \cong {S}_{4}$ . Since ${S}_{4}$ has a unique nontrivial normal 2-subgroup, ${V}_{4}$ ,(5) holds. Since $N \cong {S}_{4}$ ,it follows that $N$ contains a Sylow 2-subgroup of $G$ and also that ${N}_{N}\left( P\right) \cong {S}_{3}$ (so also ${N}_{G}\left( P\right) \cong {S}_{3}$ by (4)). Hence we obtain (6) Sylow 2-subgroups of $G$ are isomorphic to ${D}_{8}$ ,and

令 $N = {N}_{G}\left( U\right)$ 。由于 $\left| U\right| = 2$ 或 4 并且 $N$ 通过共轭作用置换 $U$ 的非单位元素，$N$ 中的一个阶为 7 的子群将与 $U$ 中某个阶为 2 的元素交换，这与 (2) 矛盾。因此，$N$ 的阶不可被 7 整除。由练习 13 知，$N$ 有多个 Sylow 2-子群，因此 $\left| N\right| = {2}^{a} \cdot 3$ ，其中 $a = 2$ 或 3。令 $P \in {\operatorname{Syl}}_{3}\left( N\right)$ 。由于 $P$ 是 $G$ 的一个 Sylow 3-子群，根据 (4)，群 ${N}_{N}\left( P\right)$ 的阶为 3 或 6（以 $P$ 作为其唯一的阶为 3 的子群）。因此根据 Sylow 定理，$N$ 必须有四个 Sylow 3-子群，并且这些子群在 $N$ 的共轭作用下可传递置换。由于任何阶为 12 的群必须有一个正规的 Sylow 2-子群或一个正规的 Sylow 3-子群（参见第 4.5 节），$\left| N\right| = {24}.$ 令 $K$ 为 $N$ 通过共轭作用在其四个 Sylow 3-子群上的作用核，因此 $K$ 是 $N$ 的 Sylow 3-子群的正规化子的交集。如果 $K = 1$ ，那么 $N \cong {S}_{4}$ 如所断言；因此考虑当 $K \neq 1$ 时。由于 $K \leq {N}_{N}\left( P\right)$ ，群 $K$ 的阶可被 6 整除，并且由于 $P$ 不正规化另一个 Sylow 3-子群，$P$ 不包含在 $K$ 中。因此 $\left| K\right| = 2$ 。但现在 $N/K$ 是一个阶为 12 的群，它被发现有多个 Sylow 2-子群和四个 Sylow 3-子群，这与前面引用的阶为 12 的群的性质相矛盾。这证明了 $N \cong {S}_{4}$ 。由于 ${S}_{4}$ 有一个唯一的非平凡正规 2-子群，${V}_{4}$ ，(5) 成立。由于 $N \cong {S}_{4}$ ，因此 $N$ 包含 $G$ 的一个 Sylow 2-子群，并且还有 ${N}_{N}\left( P\right) \cong {S}_{3}$（因此也由 (4) 得到 ${N}_{G}\left( P\right) \cong {S}_{3}$）。因此我们得到 (6) $G$ 的 Sylow 2-子群同构于 ${D}_{8}$ ，并且

(7) the normalizer in $G$ of a Sylow 3-subgroup is isomorphic to ${S}_{3}$ and so $G$ has no elements of order6. By (2) and (7), no element of order 2 commutes with an element of odd prime order. If $T \in {Sy}{l}_{2}\left( G\right)$ ,then $T \cong {D}_{8}$ by (6),so $Z\left( T\right) = \langle z\rangle$ where $z$ is an element of order 2. Then $T \leq {C}_{G}\left( z\right)$ and $\left| {{C}_{G}\left( z\right) }\right|$ has no odd prime factors by what was just said, so ${C}_{G}\left( z\right) = T$ . Since any element normalizing $T$ would normalize its center,hence commute with $z$ ,it follows that Sylow 2-subgroups of $G$ are self-normalizing. This gives

(7) 一个 Sylow 3-子群的正规化子 $G$ 同构于 ${S}_{3}$，因此 $G$ 没有阶为6的元素。由 (2) 和 (7)，没有阶为2的元素与奇素数阶的元素交换。如果 $T \in {Sy}{l}_{2}\left( G\right)$ ，那么 $T \cong {D}_{8}$ 由 (6) 可得，所以 $Z\left( T\right) = \langle z\rangle$ ，其中 $z$ 是一个阶为2的元素。那么 $T \leq {C}_{G}\left( z\right)$ ，由于刚才所说的，$\left| {{C}_{G}\left( z\right) }\right|$ 没有奇素数因子，所以 ${C}_{G}\left( z\right) = T$ 。由于任何正规化 $T$ 的元素都会正规化其中心，因此与 $z$ 交换，因此 $G$ 的 Sylow 2-子群是自正规化的。这给出了

Since $\left| {{C}_{G}\left( z\right) }\right| = 8$ ,the element $z$ in (8) has 21 conjugates. By (6), $G$ has one conjugacy class of elements of order 4, which by (6) and (8) contains 42 elements. By (2) there are 48 elements of order 7,and by (4) there are 56 elements of order 3. These account for all 167 nonidentity elements of $G$ ,and so every element of order 2 must be conjugate to $z$ ,i.e.,

由于 $\left| {{C}_{G}\left( z\right) }\right| = 8$ ，(8) 中的元素 $z$ 有21个共轭。由 (6)，$G$ 有一个阶为4的元素的共轭类，由 (6) 和 (8) 包含42个元素。由 (2) 有48个阶为7的元素，由 (4) 有56个阶为3的元素。这些解释了 $G$ 的所有167个非单位元素，因此每个阶为2的元素必须与 $z$ 共轭，即，

$G$ has a unique conjugacy class of elements of order 2.

$G$ 有一个唯一的阶为2的元素的共轭类。

Continuing with the same notation,let $T \in {\operatorname{Syl}}_{2}\left( G\right)$ with $U \leq T$ and let $W$ be the other Klein 4-group in $T$ . It follows from Sylow’s Theorem that $U$ and $W$ are not conjugate in $G$ since they are not conjugate in ${N}_{G}\left( T\right) = T$ (cf. Exercise 50 in Section 4.5). We argue next that

继续使用相同的记号，设 $T \in {\operatorname{Syl}}_{2}\left( G\right)$ 有 $U \leq T$ ，并设 $W$ 是 $T$ 中的另一个克莱因四元群。根据 Sylow 定理，$U$ 和 $W$ 在 $G$ 中不是共轭的，因为它们在 ${N}_{G}\left( T\right) = T$ 中也不是共轭的（参见第4.5节练习50）。我们接下来论证

(10) ${N}_{G}\left( W\right) \cong {S}_{4}$ .

(10) ${N}_{G}\left( W\right) \cong {S}_{4}$ .

To see this let $W = \langle z,w\rangle$ where,as before, $\langle z\rangle = Z\left( T\right)$ . Since $w$ is conjugate in $G$ to $z,{C}_{G}\left( w\right) = {T}_{0}$ is another Sylow 2-subgroup of $G$ containing $W$ but different from $T$ . Thus $W = T \cap {T}_{0}$ . Since $U$ was an arbitrary maximal intersection of Sylow 2-subgroups of $G$ ,the argument giving (5) implies (10).

为了看到这一点，设 $W = \langle z,w\rangle$ ，如之前所述， $\langle z\rangle = Z\left( T\right)$ 。由于 $w$ 在 $G$ 中与 $z,{C}_{G}\left( w\right) = {T}_{0}$ 共轭，是 $G$ 中包含 $W$ 但不同于 $T$ 的另一个 Sylow 2-子群。因此 $W = T \cap {T}_{0}$ 。由于 $U$ 是 $G$ 中 Sylow 2-子群的任意极大交集，给出 (5) 的论证也隐含了 (10)。

We now record results which we have proved or which are easy consequences of (1) to (10).

我们现在记录我们已经证明的或者是由 (1) 到 (10) 的简单推论的结果。

Proposition 14. If $G$ is a simple group of order 168,then the following hold:

命题 14。如果 $G$ 是一个阶数为 168 的简单群，那么以下成立：

(1) ${n}_{2} = {21},{n}_{3} = 7$ and ${n}_{7} = 8$

(1) ${n}_{2} = {21},{n}_{3} = 7$ 和 ${n}_{7} = 8$

(2) Sylow 2-subgroups of $G$ are dihedral,Sylow 3- and 7-subgroups are cyclic

(2) $G$ 的 Sylow 2-子群是二面体群，Sylow 3-和 7-子群是循环群

(3) $G$ is isomorphic to a subgroup of ${A}_{7}$ and $G$ has no subgroup of index $\leq 6$

(3) $G$ 同构于 ${A}_{7}$ 的一个子群，并且 $G$ 没有指数为 $\leq 6$ 的子群

(4) the conjugacy classes of $G$ are the following: the identity; two classes of elements of order 7 each of which contains 24 elements (represented by any element of order 7 and its inverse); one class of elements of order 3 containing 56 elements; one class of elements of order 4 containing 42 elements; one class of elements of order 2 containing 21 elements

(4) $G$ 的共轭类如下：单位元；两个包含 7 阶元素的类，每个类包含 24 个元素（由任意一个 7 阶元素及其逆元表示）；一个包含 56 个 3 阶元素的类；一个包含 42 个 4 阶元素的类；一个包含 21 个 2 阶元素的类

(in particular,every element of $G$ has order a power of a prime)

（特别是，$G$ 的每个元素都有素数幂的阶）

(5) if $T \in {\operatorname{Syl}}_{2}\left( G\right)$ and $U,W$ are the two Klein 4-groups in $T$ ,then $U$ and $W$ are not conjugate in $G$ and ${N}_{G}\left( U\right) \cong {N}_{G}\left( W\right) \cong {S}_{4}$

(5) 如果 $T \in {\operatorname{Syl}}_{2}\left( G\right)$ 和 $U,W$ 是 $T$ 中的两个 Klein 4-群，那么 $U$ 和 $W$ 在 $G$ 和 ${N}_{G}\left( U\right) \cong {N}_{G}\left( W\right) \cong {S}_{4}$ 中不是共轭的。

(6) $G$ has precisely three conjugacy classes of maximal subgroups,two of which are isomorphic to ${S}_{4}$ and one of which is isomorphic to the non-abelian group of order 21.

(6) $G$ 有恰好三个共轭的最大子群的类，其中两个同构于 ${S}_{4}$，另一个同构于阶为21的非阿贝尔群。

All of the calculations above were predicated on the assumption that there exists a simple group of order 168. The fact that none of these arguments leads to a contradiction does not prove the existence of such a group, but rather just gives strong evidence that there may be a simple group of this order. We next illustrate how the internal subgroup structure of $G$ gives rise to a geometry on which $G$ acts,and so leads to a proof that a simple group of order 168 is unique, if it exists (which we shall also show).

以上所有的计算都是基于存在一个168阶的简单群的假设。这些论证中没有一个是导致矛盾的，这并不证明这样一个群的存在，而只是给出了可能存在这个阶的简单群的有力证据。接下来我们说明 $G$ 的内部子群结构如何在 $G$ 上产生一个几何结构，并因此导致一个证明，即如果存在168阶的简单群，那么它是唯一的（我们也将证明这一点）。

Continuing the above notation let ${U}_{1},\ldots ,{U}_{7}$ be the conjugates of $U$ and let ${W}_{1},\ldots ,{W}_{7}$ be the conjugates of $W$ . Call the ${U}_{i}$ points and the ${W}_{j}$ lines. Define an "incidence relation" by specifying that

继续上述记号，设 ${U}_{1},\ldots ,{U}_{7}$ 为 $U$ 的共轭，${W}_{1},\ldots ,{W}_{7}$ 为 $W$ 的共轭。称 ${U}_{i}$ 为点，${W}_{j}$ 为线。通过指定以下内容定义一个“ incidence relation”：

the point ${U}_{i}$ is on the line ${W}_{j}$ if and only if ${U}_{i}$ normalizes ${W}_{j}$ .

点 ${U}_{i}$ 在线 ${W}_{j}$ 上当且仅当 ${U}_{i}$ 正规化 ${W}_{j}$。

Note that ${U}_{i}$ normalizes ${W}_{j}$ if and only if ${U}_{i}{W}_{j} \cong {D}_{8}$ ,which in turn occurs if and only if ${W}_{j}$ normalizes ${U}_{i}$ . In each point or line stabilizer—which is isomorphic to ${S}_{4}$ — there is a unique normal 4-group, $V$ ,and precisely three other (nonnormal) 4-groups ${A}_{1},{A}_{2},{A}_{3}$ . The groups $V{A}_{i}$ are the three Sylow 2-subgroups of the ${S}_{4}$ . We therefore have:

注意 ${U}_{i}$ 正规化 ${W}_{j}$ 当且仅当 ${U}_{i}{W}_{j} \cong {D}_{8}$，而这又当且仅当 ${W}_{j}$ 正规化 ${U}_{i}$。在每个点或线的稳定子群中——它与 ${S}_{4}$ 同构——有一个唯一的正规4-群 $V$，以及恰好三个其他（非正规）的4-群 ${A}_{1},{A}_{2},{A}_{3}$。$V{A}_{i}$ 群是 ${S}_{4}$ 的三个 Sylow 2-子群。因此我们得到：

each line contains exactly 3 points and each point lies on exactly 3 lines.

每条线包含恰好3个点，每个点恰好位于3条线上。

Since any two nonnormal 4-groups in an ${S}_{4}$ generate the ${S}_{4}$ ,hence uniquely determine the other two Klein groups in that ${S}_{4}$ ,we obtain

由于任何两个非正规4-群在 ${S}_{4}$ 中生成 ${S}_{4}$ ，因此唯一确定了该 ${S}_{4}$ 中的其他两个克莱因群，我们得到

12. any 2 points on a line uniquely determine the line (and the third point on it).

13. 直线上任意两点唯一确定该直线（以及其上的第三点）。

Since there are 7 points and 7 lines, elementary counting now shows that

由于有7个点和7条线，简单的计数现在表明

(13) each pair of points lies on a unique line, and each pair of lines intersects in a unique point.

(13) 每对点都位于唯一的一条直线上，每对直线在唯一的点处相交。

(This configuration of points and lines thus satisfies axioms for what is termed a projective plane.) It is now straightforward to show that the incidence geometry is uniquely determined and may be represented by the graph in Figure 1, where points are vertices and lines are the six sides and medians of the triangle together with the inscribed circle—see Exercise 27. This incidence geometry is called the projective plane of order 2 or the Fano Plane,and will be denoted by $\mathcal{F}$ . (Generally,a projective plane of “order” $N$ has ${N}^{2} + N + 1$ points,and the same number of lines.) Note that at this point the projective plane $\mathcal{F}$ does exist—we have explicitly exhibited points and lines satisfying (11) to (13) even though the group $G$ is not yet known to exist.

（这种点和线的配置因此满足了一个被称为射影平面的公理。）现在可以很容易地证明，这种 incidence 几何是唯一确定的，并且可以用图1中的图来表示，其中点是顶点，线是三角形的六条边和中线以及内切圆——见练习27。这种 incidence 几何称为阶为2的射影平面或法诺平面，并将用 $\mathcal{F}$ 表示。（一般地，一个“阶”为 $N$ 的射影平面有 ${N}^{2} + N + 1$ 个点，以及相同数量的线。）请注意，此刻射影平面 $\mathcal{F}$ 确实存在——我们已经明确展示了满足（11）到（13）的点和平线，尽管群 $G$ 尚未知道是否存在。

Figure 1

图1

An automorphism of this plane is any permutation of points and lines that preserves the incidence relation. For example, any of the six symmetries of the triangle in Figure 1 give automorphisms of $\mathcal{F}$ ,but we shall see that $\mathcal{F}$ has many more automorphisms than these.

该平面的自同构是任何保持 incidence 关系的点和线的置换。例如，图1中三角形的六个对称性中的任何一个都给出了 $\mathcal{F}$ 的自同构，但我们将会看到 $\mathcal{F}$ 有比这些更多的自同构。

Each $g \in G$ acts by conjugation on the set of points and lines,and this action preserves the incidence relation. Only the identity element in $G$ fixes all points and so via this action the group $G$ would be isomorphic to a subgroup of the group of $\operatorname{Aut}\left( \mathcal{F}\right)$ , the group of all automorphisms of $\mathcal{F}$ .

每个 $g \in G$ 通过共轭作用于点集和线集，这种作用保留了 incidence 关系。只有在 $G$ 中的单位元素才能固定所有点，因此通过这种作用，群 $G$ 将与 $\operatorname{Aut}\left( \mathcal{F}\right)$ 的子群同构，$\operatorname{Aut}\left( \mathcal{F}\right)$ 是所有 $\mathcal{F}$ 自同构的群。

Any automorphism of $\mathcal{F}$ that fixes two points on a line as well as a third point not on that line is easily seen to fix all points. Thus any automorphism of $\mathcal{F}$ is uniquely determined by its action on any three noncollinear points. Since one easily computes that there are 168 such triples, $\mathcal{F}$ has at most 168 automorphisms. This proves

任何固定直线上两点以及该直线外第三点的 $\mathcal{F}$ 自同构都可以轻易看出固定了所有点。因此，任何 $\mathcal{F}$ 的自同构都是由其作用于任意三个不共线点的行为唯一确定的。由于可以轻易计算出存在 168 个这样的三元组，$\mathcal{F}$ 至多有 168 个自同构。这证明了

if the simple group $G$ exists it is unique and $G \cong \operatorname{Aut}\left( \mathcal{F}\right)$ .

如果简单群 $G$ 存在，它是唯一的且 $G \cong \operatorname{Aut}\left( \mathcal{F}\right)$。

Two steps in the classification process yet remain: to prove that $\mathcal{F}$ does have 168 automorphisms and to prove $\operatorname{Aut}\left( \mathcal{F}\right)$ is indeed a simple group. Although one can do these graph-theoretically, we adopt an approach following ideas from the theory of “algebraic groups.” Let $V$ be a 3-dimensional vector space over the field of 2 elements, ${\mathbb{F}}_{2}$ ,so $V$ is the elementary abelian 2-group ${Z}_{2} \times {Z}_{2} \times {Z}_{2}$ of order 8. By Proposition 17 in Section 4.4, $\operatorname{Aut}\left( V\right) = {GL}\left( V\right) \cong G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ has order 168. Call the seven 1- dimensional subspaces (i.e.,the nontrivial cyclic subgroups) of $V$ points,call the seven 2-dimensional subspaces (i.e.,the subgroups of order 4) lines,and say the point $p$ is incident to the line $L$ if $p \subset L$ . Then the points and lines are easily seen to satisfy the same axioms (11) to (13) above,hence to represent the Fano Plane. Since ${GL}\left( V\right)$ acts faithfully on these points and lines preserving incidence, $\mathrm{{Aut}}\left( \mathcal{F}\right)$ has order at least 168. In light of the established upper bound for $\left| {\operatorname{Aut}\left( \mathcal{F}\right) }\right|$ this proves

分类过程中仍有两步需要完成：证明 $\mathcal{F}$ 确实有 168 个自同构，并证明 $\operatorname{Aut}\left( \mathcal{F}\right)$ 确实是一个简单群。尽管可以通过图论方法完成这些，我们采用了一种遵循“代数群”理论思想的途径。设 $V$ 是一个定义在包含两个元素的域上的三维向量空间，${\mathbb{F}}_{2}$，因此 $V$ 是一个阶为 8 的初等阿贝尔 2-群 ${Z}_{2} \times {Z}_{2} \times {Z}_{2}$。根据 4.4 节命题 17，$\operatorname{Aut}\left( V\right) = {GL}\left( V\right) \cong G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ 的阶为 168。将 $V$ 的七个一维子空间（即非平凡循环子群）称为点，将七个二维子空间（即阶为 4 的子群）称为线，如果 $p$ ，则说点 $L$ 与线 $p \subset L$ 相交。那么点和线很容易被看出满足上述的公理 (11) 到 (13)，因此表示了法诺平面。由于 ${GL}\left( V\right)$ 在这些点和线上的作用是忠实的，并保持交点，$\mathrm{{Aut}}\left( \mathcal{F}\right)$ 的阶至少为 168。鉴于对 $\left| {\operatorname{Aut}\left( \mathcal{F}\right) }\right|$ 已建立的 upper bound，这证明了

Finally we prove that ${GL}\left( V\right)$ is a simple group. By way of contradiction assume $H$ is a proper nontrivial normal subgroup of ${GL}\left( V\right)$ . Let $\Omega$ be the 7 points and let $N$ be the stabilizer in ${GL}\left( V\right)$ of some point in $\Omega$ . Since ${GL}\left( V\right)$ acts transitively on $\Omega ,N$ has index 7. Since the intersection of all conjugates of $N$ fixes all points,this intersection is the identity. Thus $H \nleqslant N$ ,and so ${GL}\left( V\right) = {HN}$ . Since $\left| {H : H \cap N}\right| = \left| {{HN} : N}\right|$ we have 7 $\left| H\right|$ . Since ${GL}\left( V\right)$ is isomorphic to a subgroup of ${S}_{7}$ and since Sylow 7-subgroups of ${S}_{7}$ have normalizers of order 42, ${GL}\left( V\right)$ does not have a normal Sylow 7-subgroup,so by Sylow’s Theorem ${n}_{7}\left( {{GL}\left( V\right) }\right) = 8.$ A normal Sylow 7-subgroup of $H$ would be characteristic in $H$ ,hence normal in ${GL}\left( V\right)$ ,so also $H$ does not have a unique Sylow 7-subgroup. Since ${n}_{7}\left( H\right) \equiv 1$ mod7and ${n}_{7}\left( H\right) \leq {n}_{7}\left( {{GL}\left( V\right) }\right) = 8$ we must have ${n}_{7}\left( H\right) = 8$ . This implies $\left| H\right|$ is divisible by 8,so 56 | $\left| H\right|$ ,and since $H$ is proper we must have $\left| H\right| = {56}$ . By usual counting arguments (cf. Exercise 7(b) of Section 5.5) $\begin{matrix} H \end{matrix}$ has a normal,hence characteristic,Sylow 2-subgroup,which is therefore normal in ${GL}\left( V\right)$ . But then ${GL}\left( V\right)$ would have a unique Sylow 2-subgroup. Since the set of upper triangular matrices and the set of lower triangular matrices are two subgroups of $G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ each of order 8,we have a contradiction. In summary we have now proven the following theorem.

最终我们证明了 ${GL}\left( V\right)$ 是一个简单群。通过反证法假设 $H$ 是 ${GL}\left( V\right)$ 的一个适当的非平凡正规子群。设 $\Omega$ 为 7 个点，$N$ 为 ${GL}\left( V\right)$ 中某个点的稳定子。由于 ${GL}\left( V\right)$ 在 $\Omega ,N$ 上作用传递，$N$ 的指数为 7。由于所有共轭的 $N$ 的交集固定所有点，这个交集是单位元。因此 $H \nleqslant N$，所以 ${GL}\left( V\right) = {HN}$。由于 $\left| {H : H \cap N}\right| = \left| {{HN} : N}\right|$，我们有 7 $\left| H\right|$。由于 ${GL}\left( V\right)$ 同构于 ${S}_{7}$ 的一个子群，且由于 ${S}_{7}$ 的 Sylow 7-子群的正规化阶为 42，${GL}\left( V\right)$ 没有正规 Sylow 7-子群，所以根据 Sylow 定理 ${n}_{7}\left( {{GL}\left( V\right) }\right) = 8.$ $H$ 的一个正规 Sylow 7-子群将在 $H$ 中是特征子群，因此在 ${GL}\left( V\right)$ 中也是正规的，所以 $H$ 也没有唯一的 Sylow 7-子群。由于 ${n}_{7}\left( H\right) \equiv 1$ 模 7 同余和 ${n}_{7}\left( H\right) \leq {n}_{7}\left( {{GL}\left( V\right) }\right) = 8$，我们必须有 ${n}_{7}\left( H\right) = 8$。这意味着 $\left| H\right|$ 可被 8 整除，所以 56 | $\left| H\right|$，由于 $H$ 是适当的，我们必须有 $\left| H\right| = {56}$。通过通常的计数论证（参见第 5.5 节练习题 7(b)）$\begin{matrix} H \end{matrix}$ 有一个正规子群，因此是特征子群，Sylow 2-子群，因此也在 ${GL}\left( V\right)$ 中是正规的。但是这样 ${GL}\left( V\right)$ 将有一个唯一的 Sylow 2-子群。由于上三角矩阵集合和下三角矩阵集合是 $G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ 的两个子群，每个阶数为 8，我们得到了矛盾。总之，我们现在已经证明了以下定理。

Theorem 15. Up to isomorphism there is a unique simple group of order ${168},G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ , which is also the automorphism group of the projective plane $\mathcal{F}$ .

定理15. 在同构意义下，存在唯一的阶为 ${168},G{L}_{3}\left( {\mathbb{F}}_{2}\right)$ 的简单群，它也是射影平面的自同构群 $\mathcal{F}$。

Note that we might just as well have called the ${W}_{j}$ points and the ${U}_{i}$ lines. This “duality” between points and lines together with the uniqueness of a simple group of order 168 may be used to prove the existence of an outer automorphism of $G$ that interchanges points and lines i.e.,conjugates $U$ to $W$ .

注意，我们同样也可以将 ${W}_{j}$ 称为点，将 ${U}_{i}$ 称为线。点与线之间的这种“对偶性”以及168阶简单群的唯一性可以用来证明存在一个 $G$ 的外自同构，它交换点与线，即，将 $U$ 共轭为 $W$。

Many families of finite simple groups can be classified by analogous methods. In more general settings geometric structures known as buildings play the role of the projective plane (which is a special case of a building of type ${\mathcal{A}}_{2}$ ). In this context the subgroups ${N}_{G}\left( U\right)$ and ${N}_{G}\left( W\right)$ are parabolic subgroups of $G,$ and $U,W$ are their unipotent radicals respectively. In particular, all the simple linear groups (cf. Section 3.4) are characterized by the structure and intersections of their parabolic subgroups, or equivalently, by their action on an associated building.

许多有限简单群的族可以通过类似的方法进行分类。在更一般的设置中，被称为建筑的几何结构起着射影平面（它是类型 ${\mathcal{A}}_{2}$ 的建筑的特殊情况）的作用。在这种背景下，子群 ${N}_{G}\left( U\right)$ 和 ${N}_{G}\left( W\right)$ 是 $G,$ 的抛物子群，$U,W$ 是它们的无扭核分别。特别是，所有简单的线性群（参见第3.4节）都是由它们的抛物子群的结构和交集，或者等价地，由它们在相关建筑上的作用来表征的。

Remarks on the Existence Problem for Groups

关于群存在问题的备注

As in other areas of mathematics (such as the theory of differential equations) one may hypothesize the existence of a mathematical system (e.g., solution to an equation) and derive a great deal of information about this proposed system. In general, if after considerable effort no contradiction is reached based on the initial hypothesis one begins to suspect that there actually is a system which does satisfy the conditions hypothesized. However, no amount of consistent data will prove existence. Suppose we carried out an analysis of a hypothetical simple group $G$ of order ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ analogous to our analysis of a simple group of order 168 (which we showed to exist). After a certain amount of effort we could show that there are unique possible Sylow numbers:

正如在数学的其他领域（如微分方程理论）中一样，人们可以假设存在一个数学系统（例如，方程的解）并推导出关于这个假设系统的许多信息。通常，如果在经过大量努力后，基于初始假设没有达到矛盾，人们开始怀疑实际上确实存在一个满足假设条件的系统。然而，无论多少一致的数据都无法证明存在性。假设我们对一个假设的简单群 $G$ 的阶数 ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ 进行了分析，类似于我们对168阶简单群的分析（我们证明了其存在）。在经过一定量的努力后，我们可以证明存在唯一的可能的Sylow数：

We could further show that such a $G$ would have no elements of order ${pq},p$ and $q$ distinct primes,no elements of order 9,and that distinct Sylow subgroups would intersect in the identity. We could then count the elements in Sylow $p$ -subgroups for all primes $p$ and we would find that these would total to exactly $\left| G\right|$ . At this point we would have the complete subgroup structure and class equation for $G$ . We might then guess that there is a simple group of this order, but the Feit-Thompson Theorem asserts that there are no simple groups of odd composite order. (Note, however, that the configuration for a possible simple group of order ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ is among the cases that must be dealt with in the proof of the Feit-Thompson Theorem, so quoting this result in this instance is actually circular. We prove no simple group of this order exists in Section 19.3; see also Exercise 29.) The point is that even though we have as much data in this case as we had in the order 168 situation (i.e., Proposition 14), we cannot prove existence without some new techniques.

我们可以进一步证明这样一个 $G$ 将不含有阶为 ${pq},p$ 和 $q$ 的不同素数阶的元素，也不含有阶为9的元素，并且不同的 Sylow 子群在单位元处相交。然后我们可以计算所有素数 $p$ 的 Sylow $p$ -子群中的元素数量，我们会发现这些元素总数恰好为 $\left| G\right|$ 。在这一点上，我们将得到 $G$ 的完整子群结构和类方程。我们可能会猜测这个阶存在一个简单群，但 Feit-Thompson 定理断言不存在奇数合数阶的简单群。（注意，然而，一个可能的阶为 ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ 的简单群的配置是必须在 Feit-Thompson 定理证明中处理的情况之一，所以在这个情况下引用这个结果实际上是有问题的。我们在第19.3节证明了不存在这个阶的简单群；也见练习29。）关键在于，尽管在这种情况下我们拥有与168阶情况（即命题14）一样多的数据，但如果没有一些新技术，我们无法证明其存在性。

When we are dealing with nonsimple groups we have at least one method of building larger groups from smaller ones: semidirect products. Even though this method is fairly restrictive it conveys the notion that nonsimple groups may be built up from smaller groups in some constructive fashion. This process breaks down completely for simple groups; and so this demarcation of techniques reinforces our appreciation for the Hölder Program: determining the simple groups,and finding how these groups are put together to form larger groups.

当我们处理非简单群时，我们至少有一种方法可以从较小的群构建较大的群：半直积。尽管这种方法相当有限制，但它传达了一个概念，即非简单群可能以某种构造性的方式由较小的群构建而成。这个过程在简单群上完全崩溃；因此这种技术划分强化了我们对 Hölder 计划的认可：确定简单群，并找到这些群如何组合形成较大群的方法。

The study of simple groups, as illustrated in the preceding discussion of groups of order 168, uses many of the same tools as the study of nonsimple groups (to unravel their subgroup structures, etc.) but also requires other techniques for their construction. As we mentioned at the end of that discussion, these often involve algebraic or geometric methods which construct simple groups as automorphisms of mathematical structures that have intrinsic interest, and thereby link group theory to other areas of mathematics and science in fascinating ways. Thus while we have come a long way in the analysis of finite groups, there are a number of different areas in this branch of mathematics on which we have just touched.

对简单群的研究，如前面对168阶群的研究所示，使用了与研究中简单群（解开它们的子群结构等）相同的许多工具，但也需要其他技术来构造它们。正如我们在那次讨论的结尾提到的，这些通常涉及代数或几何方法，这些方法将简单群构造为具有内在兴趣的数学结构的自同构，并通过引人入胜的方式将群论与其他数学和科学领域联系起来。因此，虽然我们在有限群的解析上取得了长足的进步，但在这门数学分支的许多不同领域，我们只是触及了表面。

The analysis of infinite groups generally involves quite different methods, and in the next section we introduce some of these.

无限群的分析通常涉及相当不同的方法，在下一节中，我们将介绍其中的一些。

EXERCISES

练习题

Counting elements:

元素计数：

1. Prove that for fixed $P \in {\operatorname{Syl}}_{p}\left( G\right)$ if $P \cap R = 1$ for all $R \in {\operatorname{Syl}}_{p}\left( G\right) - \{ P\}$ ,then ${P}_{1} \cap {P}_{2} = 1$ whenever ${P}_{1}$ and ${P}_{2}$ are distinct Sylow $p$ -subgroups of $G$ . Deduce in this case that the number of nonidentity elements of $p$ -power order in $G$ is $\left( {\left| P\right| - 1}\right) \left| {G : {N}_{G}\left( P\right) }\right|$ .

2. 证明对于固定的 $P \in {\operatorname{Syl}}_{p}\left( G\right)$ 如果 $P \cap R = 1$ 对于所有 $R \in {\operatorname{Syl}}_{p}\left( G\right) - \{ P\}$ 成立，那么 ${P}_{1} \cap {P}_{2} = 1$ 当 ${P}_{1}$ 和 ${P}_{2}$ 是 $G$ 的不同 Sylow $p$ 子群时。在这种情况下推断 $G$ 中 $p$ 次幂阶的非单位元素的个数是 $\left( {\left| P\right| - 1}\right) \left| {G : {N}_{G}\left( P\right) }\right|$。

3. In the group ${S}_{3} \times {S}_{3}$ exhibit a pair of Sylow 2-subgroups that intersect in the identity and exhibit another pair that intersect in a group of order 2 .

4. 在群 ${S}_{3} \times {S}_{3}$ 中展示一对交于单位元的 Sylow 2-子群，并展示另一对交于阶为2的群的 Sylow 2-子群。

5. Prove that if $\left| G\right| = {380}$ then $G$ is not simple. [Just count elements of odd prime order.]

6. 证明如果 $\left| G\right| = {380}$ 那么 $G$ 不是简单群。[只需计算奇素数阶的元素数量。]

7. Prove that there are no simple groups of order 80,351,3875 or 5313.

8. 证明不存在阶为80,351,3875或5313的简单群。

9. Let $G$ be a solvable group of order ${pm}$ ,where $p$ is a prime not dividing $m$ ,and let $P \in {\operatorname{Syl}}_{p}\left( G\right)$ . If ${N}_{G}\left( P\right) = P$ ,prove that $G$ has a normal subgroup of order $m$ . Where was the solvability of $G$ needed in the proof? (This result is true for nonsolvable groups as well - it is a special case of Burnside’s N/C-Theorem.)

10. 设 $G$ 是一个阶为 ${pm}$ 的可解群，其中 $p$ 是一个不整除 $m$ 的素数，并且设 $P \in {\operatorname{Syl}}_{p}\left( G\right)$ 。如果 ${N}_{G}\left( P\right) = P$ ，证明 $G$ 有一个阶为 $m$ 的正规子群。在证明中哪里需要用到 $G$ 的可解性？（这个结果对于非可解群也成立 - 它是伯恩赛德 N/C-定理的一个特例。）

Exploiting subgroups of small index:

利用小指数的子群：

6. Prove that there are no simple groups of order 2205, 4125, 5103, 6545 or 6435.

7. 证明阶为 2205、4125、5103、6545 或 6435 的简单群不存在。

Permutation representations:

排列表示：

7. Prove that there are no simple groups of order 1755 or 5265. [Use Sylow 3-subgroups to show $G \leq {S}_{13}$ and look at the normalizer of a Sylow 13-subgroup.]

8. 证明阶为 1755 或 5265 的简单群不存在。[使用 Sylow 3-子群来证明 $G \leq {S}_{13}$ 并观察 Sylow 13-子群的正规化子。]

9. Prove that there are no simple groups of order 792 or 918.

10. 证明阶为 792 或 918 的简单群不存在。

11. Prove that there are no simple groups of order 336.

12. 证明阶为 336 的简单群不存在。

Playing $p$ -subgroups off against each other:

玩 $p$ -子群相互对抗：

10. Prove that there are no simple groups of order 4095, 4389, 5313 or 6669.

11. 证明阶为 4095、4389、5313 或 6669 的简单群不存在。

12. Prove that there are no simple groups of order 4851 or 5145.

13. 证明阶为 4851 或 5145 的简单群不存在。

14. Prove that there are no simple groups of order 9555. [Let $Q \in {\operatorname{Syl}}_{13}\left( G\right)$ and let $P \in$ ${\operatorname{Syl}}_{7}\left( {{N}_{G}\left( Q\right) }\right)$ . Argue that $Q \trianglelefteq {N}_{G}\left( P\right)$ - why is this a contradiction?]

15. 证明阶为 9555 的简单群不存在。[设 $Q \in {\operatorname{Syl}}_{13}\left( G\right)$ 并且设 $P \in$ ${\operatorname{Syl}}_{7}\left( {{N}_{G}\left( Q\right) }\right)$ 。论证 $Q \trianglelefteq {N}_{G}\left( P\right)$ - 这为什么是一个矛盾？]

Normalizers of Sylow intersections:

Sylow 交集的正规化子：

13. Let $G$ be a group with more than one Sylow $p$ -subgroup. Over all pairs of distinct Sylow $p$ -subgroups let $P$ and $Q$ be chosen so that $\left| {P \cap Q}\right|$ is maximal. Show that ${N}_{G}\left( {P \cap Q}\right)$ has more than one Sylow $p$ -subgroup and that any two distinct Sylow $p$ -subgroups of ${N}_{G}\left( {P \cap Q}\right)$ intersect in the subgroup $P \cap Q$ . (Thus $\left| {{N}_{G}\left( {P \cap Q}\right) }\right|$ is divisible by $p \cdot \left| {P \cap Q}\right|$ and by some prime other than $p$ . Note that Sylow $p$ -subgroups of ${N}_{G}\left( {P \cap Q}\right)$ need not be Sylow in $G$ .)

14. 设 $G$ 是一个具有多个西罗 $p$ 子群的群。在所有不同的西罗 $p$ 子群对中，选取 $P$ 和 $Q$，使得 $\left| {P \cap Q}\right|$ 是最大的。证明 ${N}_{G}\left( {P \cap Q}\right)$ 有多个西罗 $p$ 子群，并且 ${N}_{G}\left( {P \cap Q}\right)$ 的任意两个不同的西罗 $p$ 子群在子群 $P \cap Q$ 中相交。（因此 $\left| {{N}_{G}\left( {P \cap Q}\right) }\right|$ 可以被 $p \cdot \left| {P \cap Q}\right|$ 整除，也可以被不同于 $p$ 的某个素数整除。注意 ${N}_{G}\left( {P \cap Q}\right)$ 的西罗 $p$ 子群不一定是 $G$ 的西罗子群。）

15. Prove that there are no simple groups of order 144,525,2025 or 3159.

16. 证明不存在阶数为 144,525,2025 或 3159 的简单群。

General exercises:

一般练习：

15. Classify groups of order 105.

16. Prove that there are no non-abelian simple groups of odd order $< {10000}$ .

17. (a) Prove that there is no simple group of order 420 .

(b) Prove that there are no simple groups of even order $< {500}$ except for orders2,60, 168 and 360.

18. Prove that if $G$ is a group of order 36 then $G$ has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.

19. Show that a group of order 12 with no subgroup of order 6 is isomorphic to ${A}_{4}$ .

20. Show that a group of order 24 with no element of order 6 is isomorphic to ${S}_{4}$ .

21. Generalize Lemma 13 by proving that if ${n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{k}}\right)$ then there are distinct Sylow $p$ -subgroups $P$ and $R$ of $G$ such that $P \cap R$ is of index $\leq {p}^{k - 1}$ in both $P$ and $R$ .

22. Suppose over all pairs of distinct Sylow $p$ -subgroups of $G,P$ and $R$ are chosen with $\left| {P \cap R}\right|$ maximal. Prove that ${N}_{G}\left( {P \cap R}\right)$ is not a $p$ -group.

23. Let $A$ and $B$ be normal subsets of a Sylow $p$ -subgroup $P$ of $G$ . Prove that if $A$ and $B$ are conjugate in $G$ then they are conjugate in ${N}_{G}\left( P\right)$ .

24. Let $G$ be a group of order ${pqr}$ where $p,q$ and $r$ are primes with $p < q < r$ . Prove that a Sylow $r$ -subgroup of $G$ is normal.

25. Let $G$ be a simple group of order ${p}^{2}{qr}$ where $p,q$ and $r$ are primes. Prove that $\left| G\right| = {60}$ .

26. Prove or construct a counterexample to the assertion: if $G$ is a group of order 168 with more than one Sylow 7-subgroup then $G$ is simple.

27. Show that if $\mathcal{F}$ is any set of points and lines satisfying properties (11) to (13) in the subsection on simple groups of order 168 then the graph of incidences for $\mathcal{F}$ is uniquely determined and is the same as Figure 1 (up to relabeling points and lines). [Take a line and any point not on this line. Depict the line as the base of an equilateral triangle and the point as the vertex of this triangle not on the base. Use the axioms to show that the incidences of the remaining points and lines are then uniquely determined as in Figure 1.]

28. Let $G$ be a simple group of order ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ . Compute all permissible values of ${n}_{p}$ for each $p \in \{ 3,7,{13},{409}\}$ and reduce to the case where there is a unique possible value for each ${n}_{p}$ .

29. Given the information on the Sylow numbers for a hypothetical simple group of order ${3}^{3} \cdot 7 \cdot {13} \cdot {409}$ ,prove that there is no such group. [Work with the permutation representation of degree 819.]

30. Suppose $G$ is a simple group of order 720 . Find as many properties of $G$ as you can (Sylow numbers, isomorphism type of Sylow subgroups, conjugacy classes, etc.). Is there such a group?