本文涉及的群都是有限群, 采用的符号和术语都是标准的, 参照[1].

设$ \alpha $是群$ G $的自同构, 如果对于任意的$ x \in G $, 存在整数$ \epsilon_{1}, \epsilon_{2}, \cdots, \epsilon_{k} $和$ G $中元素$ v_{0}, v_{1}, \cdots, v_{k} $, 使得

$ \alpha(x)= v_{0}x^{\epsilon_{1}}v_{1}\cdots v_{k-1}x^{\epsilon_{k}}v_{k}, $

那么称$ \alpha $是$ G $的多项式自同构. 因为$ \alpha(1)=1 $, 所以多项式自同构可以写成如下形式

$ \alpha(x)=(u_{1}^{-1}x^{\epsilon_{1}}u_{1})\cdots (u_{k}^{-1}x^{\epsilon_{k}}u_{k}). $

不难发现内自同构是多项式自同构. 有限群$ G $的所有的多项式自同构构成的集合记为$ \mathrm{PAut}(G) $. 可以验证$ \mathrm{PAut}(G) $是$ G $的自同构群$ \mathrm{Aut}(G) $的正规子群, 且有下面的正规群列

$ 1\unlhd\mathrm{Inn}(G)\unlhd \mathrm{PAut}(G)\unlhd \mathrm{Aut}(G). $

群的多项式自同构的研究主要有两方面. 一方面, 研究多项式自同构与其它自同构的联系和区别. 如Frohlich ^[2]给出了非交换有限单群的自同构是多项式自同构, Endimioni ^[3]指出了$ n $次对称群$ S_{n} $的自同构是多项式自同构; 另一方面, 确定给定群的$ \mbox{PAut}(G) $的结构. 如Schweigert在[4] 中证明了如果$ G $是有限可解群, 那么$ \mbox{PAut}(G) $是可解群.

众所周知, 如果群$ G $的正规子群$ G_{1}, G_{2}, \cdots, G_{n} $, 满足ⅰ) $ G=G_{1}G_{2}\cdots G_{n} $, ⅱ) $ [G_{i}, G_{j}]=1 \; (i\neq j) $, ⅲ) $ G_{i}\cap \prod_{j\neq i} G_{j}=\zeta G \; (1\leq i\leq n) $, 那么称$ G $是$ G_{1}, G_{2}, \cdots, G_{n} $的中心积, 记为

$ G=G_{1}\circ G_{2}\circ\cdots \circ G_{n}. $

Bornand在[5] 里用中心积给出了一类中心循环的有限$ p $- 群, 即

$ \underbrace{X_{3}(p^{m})\circ X_{3}(p^{m})\circ \cdots \circ X_{3}(p^{m})}_{n}\circ\mathbb{Z}_{p^{m+r}}, $

其中$ n\geq 1 $, $ m\geq 1 $, $ r\geq 0 $且

$ X_{3}(p^{m})=\langle x, y \; | x^{p^{m}}=y^{p^{m}}=1, [x, y]^{p^{m}}=1, [x, [x, y]]=[y, [x, y]]=1\rangle. $

王玉雷等在[6] 中确定了上述群的自同构群. 本文研究了这类中心循环的有限$ p $- 群的多项式自同构, 得到了下面的结果.

定理1.1   设$ G=\underbrace{X_{3}(p^{m})\circ X_{3}(p^{m})\circ \cdots \circ X_{3}(p^{m})}_{n}\circ\mathbb{Z}_{p^{m+r}} $, 其中$ p $是奇素数, $ n\geq 1 $, $ m\geq 1 $, $ r\geq 0 $, 则$ \mathrm{PAut}(G)/\mathrm{Inn}(G)\cong \mathbb{Z}_{p^{r}} $.

定理1.2   设$ G=\underbrace{X_{3}(2^{m})\circ X_{3}(2^{m})\circ \cdots \circ X_{3}(2^{m})}_{n}\circ\mathbb{Z}_{2^{m+r}} $, 其中$ n\geq 1 $, $ m\geq 1 $, $ r\geq 0 $, 则

(ⅰ) 当$ r=0 $或$ 1 $时, $ \mathrm{PAut}(G)=\mathrm{Inn}(G) $.

(ⅱ) 当$ r\geq 2 $时, $ \mathrm{PAut}(G)/\mathrm{Inn}(G)\cong \mathbb{Z}_{2^{r-1}} $.

由[1] 的定理5.3.5容易得到下面的引理2.1.

引理2.1   设$ G $是幂零类$ \leq 2 $的幂零群, 则对于$ x, y \in G $, 有$ (xy)^{n}=x^{n}y^{n}[y, x]^{\frac{n(n-1)}{2}}, $其中$ n $是整数.

引理2.2   设$ G=\underbrace{X_{3}(p^{m})\circ X_{3}(p^{m})\circ \cdots \circ X_{3}(p^{m})}_{n}\circ\mathbb{Z}_{p^{m+r}} $, 其中$ n\geq 1 $, $ m\geq 1 $, $ r\geq 0 $且

$ X_{3}(p^{m})=\langle x, y \; | x^{p^{m}}=y^{p^{m}}=1, [x, y]^{p^{m}}=1, [x, [x, y]]=[y, [x, y]]=1\rangle. $

如果$ \alpha $是$ G $的一个如下形式的多项式自同构

$ \alpha(x)=(u_{1}^{-1}x^{\epsilon_{1}}u_{1})(u_{2}^{-1}x^{\epsilon_{2}}u_{2})\cdots (u_{k}^{-1}x^{\epsilon_{k}}u_{k}), $

那么

(ⅰ) 当$ p $是奇素数时, $ \epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}\equiv 1\; (\mathrm{mod} \; p^{m}) $.

(ⅱ) 当$ p=2 $时, $ \epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}\equiv 1\; (\mathrm{mod} \; 2^{m+1}) $.

证   因为$ G $的幂零类是2, 所以对于$ x \in G $, 有

$ \begin{eqnarray*} \alpha(x)&=&(u_{1}^{-1}x^{\epsilon_{1}}u_{1})(u_{2}^{-1}x^{\epsilon_{2}}u_{2})\cdots (u_{k}^{-1}x^{\epsilon_{k}}u_{k})\\ &=&x^{\epsilon_{1}}[x^{\epsilon_{1}}, u_{1}]x^{\epsilon_{2}}[x^{\epsilon_{2}}, u_{2}]\cdots x^{\epsilon_{k}}[x^{\epsilon_{k}}, u_{k}]\\ &=&x^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}[x^{\epsilon_{1}}, u_{1}][x^{\epsilon_{2}}, u_{2}]\cdots[x^{\epsilon_{k}}, u_{k}]\\ &=&x^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}[x, u_{1}^{\epsilon_{1}}u_{2}^{\epsilon_{2}}\cdots u_{k}^{\epsilon_{k}}]. \end{eqnarray*} $

因此$ \alpha(x)=x^{\epsilon}[x, u] $, 其中$ \epsilon=\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k} $, $ u=u_{1}^{\epsilon_{1}}u_{2}^{\epsilon_{2}}\cdots u_{k}^{\epsilon_{k}} $.

注意到$ \alpha(xy)=\alpha(x)\alpha(y) $, 我们有$ (xy)^{\epsilon}=x^{\epsilon}y^{\epsilon} $. 再根据引理2.1可知

$ x^{\epsilon}y^{\epsilon}[y, x]^{\frac{\epsilon(\epsilon-1)}{2}}=x^{\epsilon}y^{\epsilon}. $

从而

$ y^{-\epsilon}x^{-\epsilon}x^{\epsilon}y^{\epsilon}[y, x]^{\frac{\epsilon(\epsilon-1)}{2}}=1. $

进而

$ [y, x]^{\frac{\epsilon(\epsilon-1)}{2}}=1. $

因为$ G^{'}=\langle y^{p^{r}}\rangle $的阶数是$ p^{m} $, 所以

$ \frac{\epsilon(\epsilon-1)}{2}\equiv0 \; (\mathrm{mod} \; p^{m}). $

当$ p $是奇素数时, $ \epsilon\equiv1 \; (\mathrm{mod} \; p^{m}) $或者$ \epsilon \equiv0 \; (\mathrm{mod} \; p^{m}) $. 如果$ \epsilon\equiv0 \; (\mathrm{mod} \; p^{m}) $, 那么对于任意的$ 1 \neq g \in G^{'}\leq \zeta G $,

$ \alpha(g)=g^{\epsilon}[g, u]=[g, u]=1, $

矛盾于$ \alpha $的核是1. 因此$ \epsilon\equiv1 \; (\mathrm{mod} \; p^{m}) $.

当$ p=2 $时, $ \epsilon\equiv1 \; (\mathrm{mod} \; 2^{m+1}) $或者$ \epsilon\equiv0 \; (\mathrm{mod} \; 2^{m+1}) $. 如果$ \epsilon\equiv0 \; (\mathrm{mod} \; 2^{m+1}) $, 不妨设$ \epsilon=2^{m+1}k $, 其中$ k $是整数, 则对于任意的$ 1 \neq h \in G^{'} $,

$ \alpha(h)=h^{\epsilon}[h, u]=h^{2^{m}2k}[h, u]=(h^{2^{m}})^{2k}[h, u]=[h, u]=1. $

矛盾, 舍去$ \epsilon\equiv0 \; (\mathrm{mod} \; 2^{m+1}) $. 因此$ \epsilon\equiv1 \; (\mathrm{mod} \; 2^{m+1}) $.

由[6] 可知$ G $有生成元$ x_{1}, x_{2}, \cdots , x_{2n}, y $, 且$ x_{i}^{p^{m}}=1(1\leq i\leq 2n) $, $ y^{p^{m+r}}=1 $, 并满足

$ \; \; \; \; \; [x_{2i-1}, x_{2i}]=y^{p^{r}}, \; \; 1\leq i\leq n $

$ [x_{2i-1}, x_{j}]=1, \; \; j\neq 2i\; \; \; $

$ \; \; \; \; \; \; [x_{2i}, x_{l}]=1, \; \; l\neq 2i-1 $

任取$ \alpha\in \mathrm{PAut}(G) $, 存在整数$ \epsilon_{1}, \epsilon_{2}, \cdots, \epsilon_{k} $和$ G $中元素$ u_{t}=(\Pi_{j=1}^{2n} x_{j}^{a_{t, j}})y^{b_{t}} $ $ (1\leq t\leq k, 0\leq a_{t, j}< p^{m}, 0\leq b_{t}< p^{m+r} ) $, 使得

$ \begin{eqnarray*} \alpha(x_{2i-1})&=&(u_{1}^{-1}x_{2i-1}^{\epsilon_{1}}u_{1})(u_{2}^{-1}x_{2i-1}^{\epsilon_{2}}u_{2})\cdots(u_{k}^{-1}x_{2i-1}^{\epsilon_{k}}u_{k})\\ &=&x_{2i-1}^{\epsilon_{1}}[x_{2i-1}^{\epsilon_{1}}, u_{1}]x_{2i-1}^{\epsilon_{2}}[x_{2i-1}^{\epsilon_{2}}, u_{2}]\cdots x_{2i-1}^{\epsilon_{k}}[x_{2i-1}^{\epsilon_{k}}, u_{k}]\\ &=&x_{2i-1}^{\epsilon_{1}}[x_{2i-1}^{\epsilon_{1}}, (\Pi_{j=1}^{2n} x_{j}^{a_{1, j}}) y^{b_{1}}]\cdots x_{2i-1}^{\epsilon_{k}}[x_{2i-1}^{\epsilon_{k}}, (\Pi_{j=1}^{2n} x_{j}^{a_{k, j}}) y^{b_{k}}]\\ &=&x_{2i-1}^{\epsilon_{1}}[x_{2i-1}^{\epsilon_{1}}, \Pi_{j=1}^{2n} x_{j}^{a_{1, j}}]\cdots x_{2i-1}^{\epsilon_{k}}[x_{2i-1}^{\epsilon_{k}}, \Pi_{j=1}^{2n} x_{j}^{a_{k, j}}]\\ &=&x_{2i-1}^{\epsilon_{1}}[x_{2i-1}^{\epsilon_{1}}, x_{2i}^{a_{1, 2i}}]\cdots x_{2i-1}^{\epsilon_{k}}[x_{2i-1}^{\epsilon_{k}}, x_{2i}^{a_{k, 2i}}]\\ &=&x_{2i-1}^{\epsilon_{1}}y^{\epsilon_{1}a_{1, 2i}p^{r}}x_{2i-1}^{\epsilon_{2}}y^{\epsilon_{2}a_{2, 2i}p^{r}}\cdots x_{2i-1}^{\epsilon_{k}}y^{\epsilon_{k}a_{k, 2i}p^{r}}\\ &=&x_{2i-1}^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}y^{(\epsilon_{1}a_{1, 2i}+\epsilon_{2}a_{2, 2i}+\cdots+\epsilon_{k}a_{k, 2i})p^{r}}, \end{eqnarray*} $

$ \begin{eqnarray*} \alpha(x_{2i})&=&(u_{1}^{-1}x_{2i}^{\epsilon_{1}}u_{1})(u_{2}^{-1}x_{2i}^{\epsilon_{2}}u_{2})\cdots(u_{k}^{-1}x_{2i}^{\epsilon_{k}}u_{k})\\ &=&x_{2i}^{\epsilon_{1}}[x_{2i}^{\epsilon_{1}}, u_{1}]x_{2i}^{\epsilon_{2}}[x_{2i}^{\epsilon_{2}}, u_{2}]\cdots x_{2i}^{\epsilon_{k}}[x_{2i}^{\epsilon_{k}}, u_{k}]\\ &=&x_{2i}^{\epsilon_{1}}[x_{2i}^{\epsilon_{1}}, (\Pi_{j=1}^{2n} x_{j}^{a_{1, j}}) y^{b_{1}}]\cdots x_{2i}^{\epsilon_{k}}[x_{2i}^{\epsilon_{k}}, (\Pi_{j=1}^{2n} x_{j}^{a_{k, j}}) y^{b_{k}}]\\ &=&x_{2i}^{\epsilon_{1}}[x_{2i}^{\epsilon_{1}}, \Pi_{j=1}^{2n} x_{j}^{a_{1, j}}]\cdots x_{2i}^{\epsilon_{k}}[x_{2i}^{\epsilon_{k}}, \Pi_{j=1}^{2n} x_{j}^{a_{k, j}}]\\ &=&x_{2i}^{\epsilon_{1}}[x_{2i}^{\epsilon_{1}}, x_{2i-1}^{a_{1, 2i-1}}]\cdots x_{2i}^{\epsilon_{k}}[x_{2i}^{\epsilon_{k}}, x_{2i-1}^{a_{k, 2i-1}}]\\ &=&x_{2i}^{\epsilon_{1}}y^{-\epsilon_{1}a_{1, 2i-1}p^{r}}x_{2i}^{\epsilon_{2}}y^{-\epsilon_{2}a_{2, 2i-1}p^{r}}\cdots x_{2i}^{\epsilon_{k}}y^{-\epsilon_{k}a_{k, 2i-1}p^{r}}\\ &=&x_{2i}^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}y^{-(\epsilon_{1}a_{1, 2i-1}+\epsilon_{2}a_{2, 2i-1}+\cdots+\epsilon_{k}a_{k, 2i-1})p^{r}}, \end{eqnarray*} $

和

$ \; \; \; \; \; \; \alpha(y)=(u_{1}^{-1}y^{\epsilon_{1}}u_{1})(u_{2}^{-1}y^{\epsilon_{2}}u_{2})\cdots (u_{k}^{-1}y^{\epsilon_{k}}u_{k})=y^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}. $

根据引理2.2得到

$ \alpha(x_{2i-1})=x_{2i-1}y^{(\epsilon_{1}a_{1, 2i}+\epsilon_{2}a_{2, 2i}+\cdots+\epsilon_{k}a_{k, 2i})p^{r}}, $

$ \; \; \; \; \; \; \; \; \; \; \; \; \alpha(x_{2i})=x_{2i}y^{-(\epsilon_{1}a_{1, 2i-1}+\epsilon_{2}a_{2, 2i-1}+\cdots+\epsilon_{k}a_{k, 2i-1})p^{r}}, $

$ \alpha(y)=y^{\mu}, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; $

其中$ 1\leq \mu\leq (p^{r}-1)p^{m}+1 $. 又$ \zeta G $的阶数是$ p^{m+r} $, 故

$ \alpha(x_{2i-1})=x_{2i-1}y^{\mu_{2i-1}p^{r}}, $

$ \alpha(x_{2i})=x_{2i}y^{\mu_{2i}p^{r}}, \; \; \; $

$ \alpha(y)=y^{\mu}, \; \; \; \; \; \; \; \; \; \; $

其中$ 0\leq \mu_{2i-1}, \mu_{2i}< p^{m}, 1\leq \mu\leq (p^{r}-1)p^{m}+1 $.

设集合

$ A= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ a_{1}&a_{2}&\cdots&a_{2n}&a \end{array}\right)\right\vert 0\leq a_{i}< p^{m}, a \; \epsilon\; \langle p^{m}+1\rangle \text{且} a \; \epsilon\; \mathbb{Z}_{p^{m+r}} \right\}. $

下证$ A \cong (\underbrace{\mathbb{Z}_{p^{m}}\oplus\mathbb{Z}_{p^{m}}\oplus\cdots\oplus\mathbb{Z}_{p^{m}}}_{2n})\rtimes \mathbb{Z}_{p^{r}} $. 记

$ A_{1}= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ a_{1}&a_{2}&\cdots&a_{2n}&1 \end{array}\right)\right\vert 0\leq a_{i}< p^{m} \right\}, $

则$ A_{1} $是$ \mathrm{GL}(2n+1, \mathbb{Z}_{p^{m}}) $的一个子群. 通过定义

$ \Upsilon(\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ a_{1}&a_{2}&\cdots&a_{2n}&1 \end{array}\right))=(a_{1}, a_{2}, \cdots, a_{2n}), $

构造$ A_{1} $到$ \underbrace{\mathbb{Z}_{p^{m}}\oplus\mathbb{Z}_{p^{m}}\oplus\cdots\oplus\mathbb{Z}_{p^{m}}}_{2n} $的群同态$ \Upsilon $, 可知$ A_{1}\cong \underbrace{\mathbb{Z}_{p^{m}}\oplus\mathbb{Z}_{p^{m}}\oplus\cdots\oplus\mathbb{Z}_{p^{m}}}_{2n}. $记

$ A_{2}= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ 0&0&\cdots&0&a \end{array}\right)\right\vert a \; \epsilon\; \langle p^{m}+1\rangle \text{且} a \; \epsilon\; \mathbb{Z}_{p^{m+r}} \right\}. $

容易验证$ A_{2}\cong \mathbb{Z}_{p^{r}} $. 又因为$ A=A_{1}A_{2}, A_{1}\unlhd A, A_{1}\cap A_{2}=1 $, 所以$ A=A_{1}\rtimes A_{2} $, 即

$ A \cong (\underbrace{\mathbb{Z}_{p^{m}}\oplus\mathbb{Z}_{p^{m}}\oplus\cdots\oplus\mathbb{Z}_{p^{m}}}_{2n})\rtimes \mathbb{Z}_{p^{r}}. $

注意到多项式自同构$ \alpha $在基$ \{x_{1}, x_{2}, \cdots, x_{2n}, y\} $下的$ 2n+1 $阶方阵是

$ \left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \mu_{1}&\mu_{2}&\cdots&\mu_{2n}&\mu \end{array}\right), $

显然此方阵属于$ A $. 因此考虑群同态

$ \Psi : \; \mathrm{PAut}(G)\rightarrow A\; \; \; \; \; \; \; \; \; \; \; \; \; \; $

$ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \sigma \mapsto\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \varepsilon_{1}&\varepsilon_{2}&\cdots&\varepsilon_{2n}&\varepsilon \end{array}\right) $

其中$ \left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \varepsilon_{1}&\varepsilon_{2}&\cdots&\varepsilon_{2n}&\varepsilon \end{array}\right) $是$ \sigma $在基$ \{x_{1}, x_{2}, \cdots, x_{2n}, y\} $下的方阵. 由同态定理可得

$ \mathrm{PAut}(G)\cong A, $

进而$ \mathrm{PAut}(G)/\mathrm{Inn}(G)\cong \mathbb{Z}_{p^{r}}. $

由[6] 可设$ G $有生成元$ x_{1}, x_{2}, \cdots , x_{2n}, y $, 且$ x_{i}^{2^{m}}=1(1\leq i\leq 2n) $, $ y^{2^{m+r}}=1 $, 同时有

$ \; \; \; \; \; [x_{2i-1}, x_{2i}]=y^{2^{r}}, \; \; 1\leq i\leq n $

$ [x_{2i-1}, x_{j}]=1, \; \; j\neq 2i\; \; \; $

$ \; \; \; \; \; \; [x_{2i}, x_{l}]=1, \; \; l\neq 2i-1 $

任取$ \alpha\in \mathrm{PAut}(G) $, 存在整数$ \epsilon_{1}, \epsilon_{2}, \cdots, \epsilon_{k} $和$ G $中元素$ u_{t}=(\Pi_{j=1}^{2n} x_{j}^{a_{t, j}})y^{b_{t}} $ $ (1\leq t\leq k, 0\leq a_{t, j}< 2^{m}, 0\leq b_{t}< 2^{m+r} ) $, 使得

$ \begin{eqnarray*} \alpha(x_{2i-1})&=&(u_{1}^{-1}x_{2i-1}^{\epsilon_{1}}u_{1})(u_{2}^{-1}x_{2i-1}^{\epsilon_{2}}u_{2})\cdots(u_{k}^{-1}x_{2i-1}^{\epsilon_{k}}u_{k})\\ &=&x_{2i-1}^{\epsilon_{1}}[x_{2i-1}^{\epsilon_{1}}, x_{2i}^{a_{1, 2i}}]\cdots x_{2i-1}^{\epsilon_{k}}[x_{2i-1}^{\epsilon_{k}}, x_{2i}^{a_{k, 2i}}]\\ &=&x_{2i-1}^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}y^{(\epsilon_{1}a_{1, 2i}+\epsilon_{2}a_{2, 2i}+\cdots+\epsilon_{k}a_{k, 2i})2^{r}}, \end{eqnarray*} $

$ \begin{eqnarray*} \alpha(x_{2i})&=&(u_{1}^{-1}x_{2i}^{\epsilon_{1}}u_{1})(u_{2}^{-1}x_{2i}^{\epsilon_{2}}u_{2})\cdots(u_{k}^{-1}x_{2i}^{\epsilon_{k}}u_{k})\\ &=&x_{2i}^{\epsilon_{1}}[x_{2i}^{\epsilon_{1}}, x_{2i-1}^{a_{1, 2i-1}}]\cdots x_{2i}^{\epsilon_{k}}[x_{2i}^{\epsilon_{k}}, x_{2i-1}^{a_{k, 2i-1}}]\\ &=&x_{2i}^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}y^{-(\epsilon_{1}a_{1, 2i-1}+\epsilon_{2}a_{2, 2i-1}+\cdots+\epsilon_{k}a_{k, 2i-1})2^{r}}, \end{eqnarray*} $

和

$ \; \; \; \; \; \; \alpha(y)=(u_{1}^{-1}y^{\epsilon_{1}}u_{1})(u_{2}^{-1}y^{\epsilon_{2}}u_{2})\cdots (u_{k}^{-1}y^{\epsilon_{k}}u_{k})=y^{\epsilon_{1}+\epsilon_{2}+\cdots+\epsilon_{k}}. $

当$ r=0 $时, 由引理2.2知

$ \alpha(x_{2i-1})=x_{2i-1}y^{(\epsilon_{1}a_{1, 2i}+\epsilon_{2}a_{2, 2i}+\cdots+\epsilon_{k}a_{k, 2i})}, $

$ \; \; \; \; \; \; \; \; \; \; \; \; \alpha(x_{2i})=x_{2i}y^{-(\epsilon_{1}a_{1, 2i-1}+\epsilon_{2}a_{2, 2i-1}+\cdots+\epsilon_{k}a_{k, 2i-1})}, $

$ \alpha(y)=y.\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; $

又$ \zeta G $的阶数是$ 2^{m} $, 故

$ \alpha(x_{2i-1})=x_{2i-1}y^{\eta_{2i-1}}, $

$ \alpha(x_{2i})=x_{2i}y^{\eta_{2i}}, \; \; \; $

$ \alpha(y)=y, \; \; \; \; \; \; \; \; \; \; $

其中$ 0\leq \eta_{2i-1}, \eta_{2i}< 2^{m} $.

考虑$ \mathrm{GL}(2n+1, \mathbb{Z}_{2^{m}}) $的一个子群

$ B= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ b_{1}&b_{2}&\cdots&b_{2n}&1 \end{array}\right)\right\vert 0\leq b_{i}< 2^{m} \right\}. $

不难证明$ B\cong \underbrace{\mathbb{Z}_{2^{m}}\oplus\mathbb{Z}_{2^{m}}\oplus\cdots\oplus\mathbb{Z}_{2^{m}}}_{2n}. $因为多项式自同构$ \alpha $在基$ \{x_{1}, x_{2}, \cdots, x_{2n}, y\} $下的方阵属于$ B $, 所以构造群同态

$ \Gamma : \; \mathrm{PAut}(G) \rightarrow B\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; $

$ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \tau \mapsto\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \lambda_{1}&\lambda_{2}&\cdots&\lambda_{2n}&1 \end{array}\right) $

其中$ \left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \lambda_{1}&\lambda_{2}&\cdots&\lambda_{2n}&1 \end{array}\right) $是$ \tau $在基$ \{x_{1}, x_{2}, \cdots, x_{2n}, y\} $下的方阵. 由同态定理可得

$ \mathrm{PAut}(G)\cong B\cong\mathrm{Inn}(G). $

当$ r\geq 1 $时, 根据引理2.2可得

$ \alpha(x_{2i-1})=x_{2i-1}y^{(\epsilon_{1}a_{1, 2i}+\epsilon_{2}a_{2, 2i}+\cdots+\epsilon_{k}a_{k, 2i})2^{r}}, $

$ \; \; \; \; \; \; \; \; \; \; \; \; \alpha(x_{2i})=x_{2i}y^{-(\epsilon_{1}a_{1, 2i-1}+\epsilon_{2}a_{2, 2i-1}+\cdots+\epsilon_{k}a_{k, 2i-1})2^{r}}, $

$ \alpha(y)=y^{\vartheta}, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; $

其中$ 1\leq \vartheta\leq (2^{r-1}-1)2^{m+1}+1 $. 又$ \zeta G $的阶数是$ 2^{m+r} $, 故

$ \alpha(x_{2i-1})=x_{2i-1}y^{\vartheta_{2i-1}2^{r}}, $

$ \alpha(x_{2i})=x_{2i}y^{\vartheta_{2i}2^{r}}, \; \; \; $

$ \alpha(y)=y^{\vartheta}, \; \; \; \; \; \; \; \; \; \; $

其中$ 0\leq \vartheta_{2i-1}, \vartheta_{2i}< 2^{m}, 1\leq \vartheta\leq (2^{r-1}-1)2^{m+1}+1 $. 由此可以看到当$ r=1 $时, $ \vartheta=1 $, 这与$ r=0 $的情形一样. 因此$ r=1 $时, $ \mathrm{PAut}(G)=\mathrm{Inn}(G) $. 下面只需考虑$ r\geq 2 $的情形即可.

记$ \mathrm{GL}(2n+1, \mathbb{Z}_{2^{m}}) $的一个子群

$ C_{1}= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ c_{1}&c_{2}&\cdots&c_{2n}&1 \end{array}\right)\right\vert 0\leq c_{i}< 2^{m} \right\}. $

可知$ C_{1}\cong \underbrace{\mathbb{Z}_{2^{m}}\oplus\mathbb{Z}_{2^{m}}\oplus\cdots\oplus\mathbb{Z}_{2^{m}}}_{2n}. $再考虑集合

$ C_{2}= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ 0&0&\cdots&0&c \end{array}\right)\right\vert c \; \epsilon\; \langle 2^{m+1}+1\rangle \text{且} c \; \epsilon\; \mathbb{Z}_{2^{m+r}} \right\}. $

易证$ C_{2}\cong \mathbb{Z}_{2^{r-1}} $. 记

$ C= \left\{\left.\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ c_{1}&c_{2}&\cdots&c_{2n}&c \end{array}\right)\right\vert 0\leq c_{i}< 2^{m}, c \; \epsilon\; \langle 2^{m+1}+1\rangle \text{且} c \; \epsilon\; \mathbb{Z}_{2^{m+r}} \right\}, $

则$ C=C_{1}\rtimes C_{2} $, 从而$ C \cong (\underbrace{\mathbb{Z}_{2^{m}}\oplus\mathbb{Z}_{2^{m}}\oplus\cdots\oplus\mathbb{Z}_{2^{m}}}_{2n})\rtimes \mathbb{Z}_{2^{r-1}} $.

构造群同态

$ \Phi : \; \mathrm{PAut}(G) \rightarrow C\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; $

$ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \theta \mapsto\left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \xi_{1}&\xi_{2}&\cdots&\xi_{2n}&\xi \end{array}\right) $

其中$ \left(\begin{array}{ccccc} 1 & 0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\\ \xi_{1}&\xi_{2}&\cdots&\xi_{2n}&\xi \end{array}\right) $是$ \theta $在基$ \{x_{1}, x_{2}, \cdots, x_{2n}, y\} $下的方阵. 由同态定理可得

$ \mathrm{PAut}(G)\cong C, $

从而$ \mathrm{PAut}(G)/\mathrm{Inn}(G)\cong \mathbb{Z}_{2^{r-1}} $.