换位子群是不可分Abel群的有限秩可除幂零群

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'=Q或Q_p/Z且G可以分解为G=S×D,其中当G'=Q时,■当G'=Q_p/Z时,S有中心积分解S=S_1*S_2*…*S_r,并且可以将S形式化地写成■其中■,式中s,t都是非负整数,Q是有理数加群,π_κ(k=1,2,…,t)是某些素数的集合,满足π_1■Cπ_2■…■π_t,Q_π_k={m/n|(m,n)=1,m∈Z,n为正的π_k-数}.进一步地,当G'=Q时,(r;s;π_1,π_2,…,π_t)是群G的同构不变量;当G'=Q_p/Z时,(p,r;s;π_1,π_2,…,πt)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.

Radicable Nilpotent Groups of Finite Rank with Indecomposable Abelian Commutator Subgroups

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.