Non-abelian groups in which every subgroup is abelian
Non-abelian groups in which every subgroup is abelian
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DOI：
10.1090/s0002-9947-1903-1500650-9
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发表时间：
1903-04
影响因子：
1.3
通讯作者：
G. A. Miller;H. X. Moreno
中科院分区：
文献类型：
--
作者：
G. A. Miller;H. X. Moreno
几年前戴德金和其他人调查的团体，其中每一个小组是不变的，并发现理论，这些团体提出了显着的困难，除了如涉及阿贝尔群。每个子群都是阿贝尔群的非阿贝尔群给出了一个简单而一般的结果的平行例子。以下是一些最重要的：所有这些群都是可解的。它们的序不能被两个以上不同的素数整除。每个交换子都是素数阶的。当阶为p a，（p和q是素数; a，/3> 0）时，只存在q个p a阶子群，且只存在一个q a阶子群。前者是循环的，后者是（1，1，1，···）型的。当阶为pa时，只有p + 1个p°-l阶子群，且它们都不包含超过3个不变量.如果有三个不变量，其中至少有一个必须是p阶的。设G表示任何非阿贝尔群，其中每个子群都是阿贝尔群。首先证明G是可解的。如果G被表示为一个可迁置换群，则它要么是本原的，要么是非本原的。在后一种情况下，它将与某个本原群P同构。f G的对应于P中单位元的子群是阿贝尔的，并且P的每个子群都是阿贝尔的。群G是可解的，只要P是可解的。因此，它仍然是要证明，一个非阿贝尔本原群P，其中每个子群是阿贝尔的总是可解的。设A* 是P的子群，它由所有省略给定字母的替换组成。因为P是非正则的，所以J P除了单位元之外还包括至少一个替换。如果Px的两个共轭除了单位元之外还有一个公共的置换，这个置换在P下是不变的，因为Px是P的一个极大子群。因此P必须是n - 1类的，n是A的次数。因此P包含一个阶为w，§的不变子群，关于这个不变子群，商群与Px简单同构。作为
Several years ago Dedekind and others investigated the groups in which every subgroup is invariant, and found that the theory of these groups presents remarkably few difficulties except such as are involved in abelian groups. The non-abelian groups in which every subgroup is abelian present a parallel example of simple and general results. The following are some of the most important ones : All such groups are solvable. Their orders cannot be divided by more than two distinct primes. Every commutator is of prime order. When the order ispaqß, (p and q being prime; a, /3> 0), there are just qß subgroups of order pa and there is only one subgroup of order qß . The former are cyclic and the latter is of type (1, 1, 1, •••). When the order is pa, there are just p + 1 subgroups of order p°-~l and none of them involves more than three invariants. If there are three invariants at least one of them must be of order p. Let G represent any non-abelian group in which every subgroup is abelian. We shall first prove that G is solvable. If G is represented as a transitive substitution group it will be either primitive or imprimitive. In the latter case it will be isomorphic with some primitive group P.-f The subgroup of G which corresponds to identity in P is abelian and every subgroup of P is abelian. The group G is solvable whenever P is solvable. Hence it remains to prove that a non-abelian primitive group P in which every subgroup is abelian is always solvable. Let A*, be the subgroup of P which is composed of all the substitutions omitting a given letter. Since P is non-regular,J P includes at least one substitution besides the identity. If two conjugates of Px had a common substitution besides the identity, this substitution would be invariant under P, since Px is a maximal subgroup of P. Hence P must be of class n — 1, n being the degree ofA\ Therefore P contains an invariant subgroup of order w, § with respect to which the quotient group is simply isomorphic with Px. As