Radicable Nilpotent Groups of Finite Rank with Indecomposable Abelian Commutator Subgroups

DOI：10.16205/j.cnki.cama.2018.0025

 作者 单位 E-mail 刘合国 湖北大学数学系, 武汉 430062. ghliu@hubu.edu.cn 张继平 北京大学数学科学学院, 北京 100871. jzhang@pku.edu.cn 廖　军 通讯作者. 湖北大学数学系, 武汉 430062. jliao@hubu.edu.cn

完整地确定了换位子群是不可分Abel群的有限秩可除幂零群的结构, 证明了下面的定理. 设$G$是有限秩的可除幂零群, 则$G$的换位子群是不可分Abel群当且仅当$G'=\mathbb{Q}$或$\mathbb{Q}_p/\mathbb{Z}$且$G$可以分解为$G=S\times D$, 其中当$G'=\mathbb{Q}$时, $$S=\left\{\left[ \begin{array}{ccccccc} 1&a_{12}&a_{13}&a_{14}&\cdots&a_{1~r+1}&a_{1~r+2}\0&1&0&0&\cdots&0&a_{2~r+2}\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\0&0&0&\cdots&1&0&a_{r~r+2}\0&0&0&\cdots&0&1&a_{r+1~r+2}\0&0&0&\cdots&0&0&1 \end{array} \right]\left| \begin{array}{c} ~~\~~\~~\a_{ij}\in \mathbb{Q} \~~\~~\~\ \end{array} \right. \right\},$$ 当$G'=\mathbb{Q}_p/\mathbb{Z}$时, $S$有中心积分解$S=S_1\ast S_2\ast\cdots \ast S_r$, 并且可以将$S$形式化地写成 $$S=\left\{\left[ \begin{array}{ccccccc} 1&a_{12}&a_{13}&a_{14}&\cdots&a_{1~r+1}&b_{1~r+2}\0&1&0&0&\cdots&0&a_{2~r+2}\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\0&0&0&\cdots&1&0&a_{r~r+2}\0&0&0&\cdots&0&1&a_{r+1~r+2}\0&0&0&\cdots&0&0&1 \end{array} \right]\left| \begin{array}{c} ~~\~~\~~\a_{ij}\in \mathbb{Q} \b_{1~r+2}\in \mathbb{Q}_p/\mathbb{Z}\~~\~~\ \end{array} \right. \right\},$$ 其中$D\cong \underbrace{\mathbb{Q}\oplus\mathbb{Q}\oplus\cdots\oplus\mathbb{Q}}_s \bigoplus\bigoplus\limits_{k=1}^t(\mathbb{Q}_{\pi_k}/\mathbb{Z})$, 式中$s$, $t$都是非负整数, $\mathbb{Q}$是有理数加群, $\pi_k$ $(k=1, 2,\cdots,t)$是某些素数的集合, 满足$\pi_1\subseteq\pi_2\subseteq \cdots \subseteq \pi_t$, $\mathbb{Q}_{\pi_k}=\{\frac mn\mid(m,n)=1, m\in \mathbb{Z}, n\mbox{ 为正的$\pi_k${-}数 }\}$. 进一步地, 当$G'=\mathbb{Q}$时, $(r; s; \pi_1, \pi_2, \cdots, \pi_t)$是群$G$的同构不变量; 当$G'=\mathbb{Q}_p/\mathbb{Z}$时, $(p, r; s; \pi_1, \pi_2, \cdots, \pi_t)$是群$G$的同构不变量. 即若群$H$也是有限秩的可除幂零群, 它的换位子群是不可分Abel群, 那么$G$同构于$H$的充分必要条件是它们有相同的不变量.

The structure of the radicable nilpotent groups of finite rank with indecomposable abelian commutator subgroups is completely determined. More exactly, the following theorem is proved. Let $G$ be a radicable nilpotent group of finite rank. Then the commutator subgroup of $G$ is indecomposable and abelian if and only if $G'=\mathbb{Q}$ or $\mathbb{Q}_p/\mathbb{Z}$ and $G$ has a decomposition $G=S\times D$, where $$S=\left\{~~\left[ \begin{array}{ccccccc} 1&a_{12}&a_{13}&a_{14}&\cdots&a_{1~r+1}&a_{1~r+2}\0&1&0&0&\cdots&0&a_{2~r+2}\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\0&0&0&\cdots&1&0&a_{r~r+2}\0&0&0&\cdots&0&1&a_{r+1~r+2}\0&0&0&\cdots&0&0&1 \end{array} \right]~~\left| \begin{array}{c} ~~\~~\~~\a_{ij}\in \mathbb{Q} \~~\~~\~\ \end{array} \right. \right\},$$ if $G'=\mathbb{Q}$ and $S=S_1\ast S_2\ast\cdots\ast S_r$, $S_i\cong S_p$ if $G'=\mathbb{Q}_p/\mathbb{Z}$. Write $S$ formally as $S=\left\{~~\left[ \begin{array}{ccccccc} 1&a_{12}&a_{13}&a_{14}&\cdots&a_{1~r+1}&b_{1~r+2}\0&1&0&0&\cdots&0&a_{2~r+2}\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\0&0&0&\cdots&1&0&a_{r~r+2}\0&0&0&\cdots&0&1&a_{r+1~r+2}\0&0&0&\cdots&0&0&1 \end{array} \right]~~\left| \begin{array}{c} ~~\~~\~~\a_{ij}\in \mathbb{Q} \b_{1~r+2}\in \mathbb{Q}_p/\mathbb{Z}\~~\~\ \end{array} \right. \right\}.$ Here $D$ is a divisible abelian group such that $D \cong \underbrace{\mathbb{Q}\oplus\mathbb{Q}\oplus\cdots\oplus\mathbb{Q}}_s\bigoplus\bigoplus\limits_{k=1}^t(\mathbb{Q}_{\pi_k}/\mathbb{Z})$, where $s$ and $t$ are nonnegative integers, and $\mathbb{Q}$ is the additive group of the rational number field, where $\pi_k$ $(k=1, 2,\cdots,t)$ are the sets of some prime numbers such that $\pi_1\subseteq\pi_2\subseteq \cdots \subseteq \pi_t$, and $\mathbb{Q}_{\pi_k}=\{\frac mn\mid(m,n)=1,\ m\in \mathbb{Z},\ n\mbox{ is a positive$\pi_k$-number}\}$. Moreover, $(p, r; s; \pi_1, \pi_2, \cdots, \pi_t)$ is an isomorphic invariant of $G$, that is to say, if $H$ is also a radicable nilpotent group of finite rank with indecomposable abelian commutator subgroup, then $G$ is isomorphic to $H$ if and only if they have the same invariants.