| 刘合国,张继平,廖 军.换位子群是不可分Abel群的有限秩可除幂零群[J].数学年刊A辑,2018,39(3):287~296 |
| 换位子群是不可分Abel群的有限秩可除幂零群 |
| Radicable Nilpotent Groups of Finite Rank with Indecomposable Abelian Commutator Subgroups |
| Received:February 03, 2016 Revised:April 28, 2016 |
| DOI：10.16205/j.cnki.cama.2018.0025 |
| 中文关键词: 幂零群, 局部循环群, 中心, 换位子群, 可除群 |
| 英文关键词:Nilpotent group, Locally cyclic group, center, Commutator subgroup, Radicable group |
| 基金项目:本文受到国家自然科学基金(No.11131001, No.11371124, No.11401186)的资助. |
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| 中文摘要: |
| 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. |
| 英文摘要: |
| 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. |
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