杨全李, 唐国平
中国科学院大学数学科学学院, 北京 100049
2019年1月16日 收稿; 2019年5月6日 收修改稿
基金项目: 国家自然科学基金（11771422）资助
通信作者: 杨全李, E-mail:yangquanli16@mails.ucas.ac.cn

摘要: 整群环是代数乃至许多数学分支中很重要的一类环，也是代数K理论主要的研究对象之一。对几类交换p（p为素数）群G的整群环$\mathbb{Z}$G，作为半单代数$\mathbb{Q}$G的一个$\mathbb{Z}$-序，通过将其嵌入到极大$\mathbb{Z}$-序Γ之中，然后利用核群的性质研究K₁（$\mathbb{Z}$G）在K₁（Γ）中的指数问题。主要结果有：首先对几类交换p群G给出[（Γ_p）^×：（$\mathbb{Z}$_pG）^×]的确切表达式，然后用来确定[K₁（Γ）：K₁（$\mathbb{Z}$G）]的具体数值。
关键词: 整群环$\mathbb{Z}$-序极大序核群K₁群
The K₁ group of integral group ring and its maximal order for a commutative p group
YANG Quanli, TANG Guoping
School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China


Abstract: Group rings are very important rings in algebra and many other branches of mathematics. It is also one of the main research subjects of algebraic K-theory. In this work, we mainly deal with integral group rings $\mathbb{Z}$G for some abelian p(p is prime) groups G. We can regard $\mathbb{Z}$G as a $\mathbb{Z}$-order of the semi-simple algebra $\mathbb{Q}$G and embed it into the maximal $\mathbb{Z}$-order Γ. Then we use the properties of the kernel group to study the exponential problem of K₁($\mathbb{Z}$G) in K₁(Γ). In this paper, there are two main results. First, the explicit formula of[(Γ_p)^×:($\mathbb{Z}$_pG)^×] is obtained for some abelian p groups. Secondly, by using the formula, we get the specific result of[K₁(Γ):K₁($\mathbb{Z}$G)] for some abelian p groups.
Keywords: integral group ring$\mathbb{Z}$-ordermaximal orderkernel groupK₁ group
当G为有限交换群时，K₁($\mathbb{Z}$G)$\cong $SK₁($\mathbb{Z}$G)$ \oplus $($\mathbb{Z}$G)^×是代数K理论中众所周知的结果。然而，通常情况下SK₁($\mathbb{Z}$G)的结构十分复杂。后面要列出的引理1.3说明在很多我们关心的情形下SK₁($\mathbb{Z}$G)平凡。通常将$\mathbb{Z}$G视为半单代数$\mathbb{Q}$G的一个$\mathbb{Z}$-序，由引理1.4，可将$\mathbb{Z}$G嵌入到一个极大$\mathbb{Z}$-序Γ中。定义：$\mathbb{Z}$G_p=$\mathbb{Z}$_pG=$\mathbb{Z}$G?_$\mathbb{Z}$$\mathbb{Z}$_p，Γ_p=Γ?_$\mathbb{Z}$$\mathbb{Z}$_p，其中p为素数，$\mathbb{Z}$_p为p-adic整数环。对几类特殊的有限交换p群G，本文给出指数[(Γ_p)^×:($\mathbb{Z}$_pG)^×]的确切公式。然后利用核群D($\mathbb{Z}$G)的相关性质得到当p=2时有关[K₁(Γ):K₁($\mathbb{Z}$G)]的部分结果。
1 预备知识定义1.1 ??对阶为素数p的幂的交换群G，由有限生成交换群的结构定理可将G唯一地表示为某些p^k_i阶循环群的直和，i=1, 2, …, n，且k₁≤k₂≤…≤k_n, 数组(k₁, k₂, …, k_n)_p称为群G的型。
引理1.1??令p是一个素数，群G的型为(k₁, k₂, …, k_n)_p。为方便起见记k₀=0，对任意整数h，0 < h≤k_n，记v_h是满足k_vh < h≤k_vh+1的非负整数，且记${a_h} = \sum\limits_{\mu = 0}^{{v_h}} {{k_\mu }} $。则G的阶为p^h的循环子群的个数是

$t({p^h}) = \frac{{{p^{n - {v_h}}} - 1}}{{p - 1}}{p^{(n - {v_h} - 1)(h - 1) + {a_h}}}.$

证明??见文献[1-3]。
引理1.2??对代数数域F的代数整数环O_F，有SK₁(O_F)=1。
证明??见文献[4]。
引理1.3??对任一有限交换群G，SK₁($\mathbb{Z}$G)=1当且仅当$G \cong {\left({{C_2}} \right)^n}$或者G的每一Sylow-p子群形如C_pⁿ或者C_p×C_pⁿ。
证明??见文献[5]。
引理1.4??对有限交换群G，$\mathbb{Z}$G是半单代数$\mathbb{Q}$G的一个$\mathbb{Z}$-序，且$\mathbb{Q}$G只有唯一极大$\mathbb{Z}$-序Γ，Γ?$\mathbb{Z}$G。如果$\mathbb{Q}G \cong \prod\nolimits_{i = 1}^r {\mathbb{Q}\left({{\xi _{{l_i}}}} \right)} $(其中ξ_li表l_i次本原单位根)，则Γ=∏_i=1^r$\mathbb{Z}$[ξ_li]。
证明??见文献[6]。
引理1.5??在上述引理条件下，($\mathbb{Z}$G)^×的挠部分为{±g|g∈G}，且($\mathbb{Z}$G)^×与Γ^×的自由部分的秩相同。特别有，[Γ^×:($\mathbb{Z}$G)^×] < ∞。
证明??见文献[7]。
定义1.2??对任意素数p，记$\mathbb{Z}$G_p=$\mathbb{Z}$_pG=$\mathbb{Z}$G?_$\mathbb{Z}$$\mathbb{Z}$_p，Γ_p=Γ?_$\mathbb{Z}$$\mathbb{Z}$_p，其中$\mathbb{Z}$_p为p-adic整数环。$\mathbb{Z}$中的素理想(p)称为$\mathbb{Z}$G-奇异的，若Γ_p≠$\mathbb{Z}$G_p。记S($\mathbb{Z}$G)为所有$\mathbb{Z}$G奇异素理想之集。
定义1.3??令G是有限群，对$\mathbb{Q}$G中任一$\mathbb{Z}$-序Λ及素数p，用Λ_(p)表示Λ在$\mathbb{Z}$的素理想(p)处的局部化，则有K₀(Λ)→K₀(Λ_(p)), [M]→[Λ_(p)?_ΛM]。局部自由类群CL(Λ)定义为映射K₀(Λ)→∏_pK₀(Λ_(p))的核，其中p取遍所有素数。
在上述定义的条件下，若$\mathbb{Q}$G中存在包含Λ的极大$\mathbb{Z}$-序，任取其中一个并记为Γ(注：当G非交换时Γ不唯一)。由文献[8]，映射[M]→[Γ?_ΛM]给出一个满同态CL(Λ)→CL(Γ)，该满同态的核称为核群, 记为D(Λ)。于是有交换群的正合列

$0 \to D(\Lambda ) \to CL(\Lambda ) \to CL(\Gamma ) \to 0.$

而由文献[9]可知(在同构意义下)D(Λ)与极大$\mathbb{Z}$-序Γ的选取无关。在上述意义下，特别当G是一个交换群时，取Λ=$\mathbb{Z}$G。由引理1.4知这种情形下上述意义下的极大序唯一，而且引理1.4给出了这个极大序。根据文献[10]有
引理1.6??有正合序列

$\begin{array}{*{20}{c}}{1 \to {\Gamma ^ \times }/{{(\mathbb{Z}G)}^ \times } \to \prod\limits_{S(\mathbb{Z}G)} {{{({\Gamma _{{p_0}}})}^ \times }} /{{(\mathbb{Z}{G_{{p_0}}})}^ \times }}\\{ \to CL(\mathbb{Z}G) \to CL(\Gamma ) \to 1.}\end{array}$

且有
$\begin{array}{l}\left( {\rm{i}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} |D(\mathbb{Z}G)| \cdot [{\Gamma ^ \times }:{(\mathbb{Z}G)^ \times }] = {\kern 1pt} \prod\limits_{S\left( {\mathbb{Z}G} \right)} {[{{({\Gamma _{{p_0}}})}^ \times }:{{(\mathbb{Z}{G_{{p_0}}})}^ \times }]} .\end{array}$
(ⅱ) $\left[\left(\Gamma_{p_{0}}\right)^{x}:\left(\mathbb{Z} G_{p_{0}}\right)^{\times}\right]=l_{p_{0}} \cdot m_{p_{0}}$,这里

$\begin{array}{*{20}{l}}{{l_{{p_0}}} = \frac{{|{{({\Gamma _{{p_0}}}/{J_{{p_0}}})}^ \times }|}}{{|{{\{ \mathbb{Z}{G_{{p_0}}}/({J_{{p_0}}} \cap \mathbb{Z}{G_{{p_0}}})\} }^ \times }|}}, }\\{{m_{{p_0}}} = [{J_{{p_0}}}:(\mathbb{Z}{G_{{p_0}}} \cap {J_{{p_0}}})]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = \frac{{[{\Gamma _{{p_0}}}:\mathbb{Z}{G_{{p_0}}}][\mathbb{Z}{G_{{p_0}}}:({J_{{p_0}}} \cap \mathbb{Z}{G_{{p_0}}})]}}{{[{\Gamma _{{p_0}}}:{J_{{p_0}}}]}}, }\end{array}$

其中J_p0是Γ_p0的Jacobson根。
证明??见文献[10]。
特别地，当G是阶为pⁿ的有限交换群时，S($\mathbb{Z}$G)={(p)}，引理1.6中的正合列变得简单了。设$\mathbb{Q}$G=∏_i=1^mF_i是分圆域的直和，记d_i是F_i的判别式，则引理1.6有如下的简单形式
引理1.7?? |D($\mathbb{Z}$G)|是p的幂，更确切地说有

$|D(\mathbb{Z}G)| = [{({\Gamma _p})^ \times }:{(\mathbb{Z}{G_p})^ \times }]/[{\Gamma ^ \times }:{(\mathbb{Z}G)^ \times }]$

且

${[{({\Gamma _p})^ \times }:{(\mathbb{Z}{G_p})^ \times }]^2} = {p^{n{p^n} - 2t}}/\prod\limits_{i = 1}^m | {d_i}|, $

其中t=t(G)=m-1是G的非平凡循环子群的个数。
证明??见文献[10]。
引理1.8??若G是指数为m的有限交换群, 则$\mathbb{Q}G = \mathop \oplus \limits_{d|m} t(d)\mathbb{Q}\left({{\xi _d}} \right)$(ξ_d表d次本原单位根)，其中t(d)是G的d阶循环子群的个数。
证明??见文献[11]。
2 主要结果我们的第一个主要结果是对G=C_pⁿ1×C_pⁿ2, 0≤n₁≤n₂，给出[(Γ_p)^×:($\mathbb{Z}$_pG)^×]的具体公式，为此分别考虑以下情形。
定理2.1??若G=C_pⁿ, (n≥0)，则

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {p^{\frac{{{p^n} - 1}}{{{p^{ - 1}}}} - n}}.$

证明??由引理1.1, G的型为(n)_p。这时，对任意0 < h≤n，G有唯一的阶为p^h的循环子群, 于是t=n。由引理1.8知$\mathbb{Q}$G=$\mathbb{Q}$$ \oplus $Q(ξ_p)$ \oplus $…$ \oplus $$\mathbb{Q}$(ξ_pⁿ)。文献[12]给出分圆域$\mathbb{Q}$(ξ_S)的判别式

$d(\mathbb{Q}({\xi _s})) = {( - 1)^{\varphi (s)/2}}{s^{\varphi (s)}}/\prod\limits_{p|s} {{p^{\varphi (s)/(p - 1)}}} .$

故|d_i|=|d($\mathbb{Q}$(ξ_pⁱ))|=p^{pi-1_(ip-i-1)}, 从而

$\begin{array}{l}\prod\nolimits_{i = 1}^n {\left| {{d_i}} \right|} = \prod\nolimits_{i = 1}^n {{p^{{p^{i - 1}}(ip - i - 1)}}} \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}}{ = {p^{\mathop \Sigma \limits_{i = 1}^n [{p^{i - 1}}(ip - i - 1)]}}}\\{ = {p^{n{p^n} - \frac{{2({p^n} - 1)}}{{p - 1}}}}.}\end{array}\end{array}$

于是由引理1.7得

$\begin{array}{*{20}{l}}{{{[{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]}^2} = {p^{n{p^n} - 2t}}/\prod\nolimits_{i = 1}^n {\left| {{d_i}} \right|} }\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {p^{\frac{{2({p^n} - 1)}}{{p - 1}} - 2n}}, }\end{array}$

故$\left[ {{{\left( {{\Gamma _p}} \right)}^ \times }:{{\left( {{\mathbb{Z}_p}G} \right)}^ \times }} \right] = {p^{\frac{{{p^n} - 1}}{{p - 1}} - n}}.$
当n=0时，上式显然成立。
定理2.2??当G=C_pⁿ×C_pⁿ，(0≤n)时，则

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {p^q}, $

其中$q = \frac{n}{2}{p^{2n}} + \frac{{(p + 2)\left({{p^{2n}} - 1} \right)}}{{2\left({{p^2} - 1} \right)}} - \frac{{(p + 1)\left({{p^n} - 1} \right)}}{{p - 1}}.$
证明??此时|G|=p²ⁿ, G的型为，(n, n)_p, 则当1≤h≤n时，${v_h} = 0, \; {a_h} = \sum\limits_{\mu = 0}^{{v_h}} {{k_\mu }} = 0$。由引理1.1有

$\begin{array}{*{20}{l}}{t({p^h}) = \frac{{{p^{2 - {v_h}}} - 1}}{{p - 1}}{p^{(2 - {v_h} - 1)(h - 1) + {a_h}}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = (p + 1){p^{h - 1}}.}\end{array}$

于是$t = t(G) = \sum\limits_{h = 1}^n {(p + 1)} {p^{h - 1}} = \frac{{(p + 1)\left({{p^n} - 1} \right)}}{{p - 1}}$, 再由引理1.8知t(pⁱ)(1≤i≤n)就是$\mathbb{Q}$G=$\mathbb{Q}$$ \oplus $t₁$\mathbb{Q}$(ξ_p)$ \oplus $…$ \oplus $t_n$\mathbb{Q}$(ξ_pⁿ)中t_i的值。于是有

$\begin{array}{*{20}{c}}{\mathbb{Q}G = \mathbb{Q} \oplus (p + 1)\mathbb{Q}({\xi _p}) \oplus (p + 1)p\mathbb{Q}({\xi _{{p^2}}}) \oplus }\\{ \cdots \oplus (p + 1){p^{n - 1}}\mathbb{Q}({\xi _{{p^n}}}).}\end{array}$

根据文献[12]有关分圆域的判别式的计算公式

$d(\mathbb{Q}({\xi _s})) = {( - 1)^{\varphi (s)/2}}{s^{\varphi (s)}}/\prod\limits_{\left. p \right|s} {{p^{\varphi (s)/(p - 1)}}} , $

于是在这种情形下有

$\begin{array}{*{20}{l}}{\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} = {p^{\mathop \Sigma \limits_{i = 1}^n (p + 1){p^{2i - 2}}(ip - i - 1)}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {p^{n{p^{2n}} - \frac{{(p + 2)({p^{2n}} - 1)}}{{{p^2} - 1}}}}.}\end{array}$

由引理1.7得

$\begin{array}{*{20}{c}}{{{[{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]}^2} = }\\{{p^{2n{p^{2n}} - \frac{{2(p + 1)({p^n} - 1)}}{{p - 1}}}}/{p^{n{p^{2n}} - \frac{{{p^{2n + 1}} + 2{p^{2n}} - p - 2}}{{{p^2} - 1}}}}}\\{ = {p^{2q}}, }\end{array}$

从而[(Γ_p)^×:($\mathbb{Z}$_pG)^×]=p^q.
上式对n=0情形显然成立。
定理2.3??当$G = {C_{{p^{{n_1}}}}} \times {C_{{p^{{n_2}}}}}, 0 < {n_1} < {n_2}$时，有

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {p^q}.$

其中

$\begin{array}{l}q = \frac{{{n_1}}}{2}{p^{{n_1} + {n_2}}} - ({n_2} - {n_1}){p^{{n_1}}} - \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{(p + 1)({p^{{n_1}}} - 1) + ({p^{{n_1}}} - {p^{{n_2}}}){p^{{n_1}}}}}{{p - 1}} + \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{(p + 2)({p^{2{n_1}}} - 1)}}{{2({p^2} - 1)}}.\end{array}$

证明??此时|G|=p^n₁+n₂, G的型为(n₁, n₂)_p。则当1≤h≤n₁时，${v_h} = 0, \; {a_h} = \sum\limits_{\mu = 0}^{{v_h}} {{k_\mu }} = 0, $, 于是此时有

$\begin{array}{*{20}{l}}{t({p^h}) = \frac{{{p^{2 - {v_h}}} - 1}}{{p - 1}}{p^{(2 - {v_h} - 1)(h - 1) + {a_h}}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = (p + 1){p^{h - 1}}.}\end{array}$

当n₁ < h≤n₂时，${v_h} = 1, \; {a_h} = \sum\limits_{\mu = 0}^{{v_h}} {{k_\mu }} = {n_1}$，此时有

$\begin{array}{*{20}{l}}{t({p^h}) = \frac{{{p^{2 - {v_h}}} - 1}}{{p - 1}}{p^{(2 - {v_h} - 1)(h - 1) + {a_h}}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {p^{{n_1}}}.}\end{array}$

综上所得有

$t({p^h}) = \left\{ {\begin{array}{*{20}{l}}{(p + 1){p^{h - 1}}, }&{{\rm{ 若 }}0 < h \le {n_1}}\\{{p^{{n_1}}}, }&{{\rm{ 若 }}{n_1} < h \le {n_2}}\end{array}} \right..$

于是

$\begin{array}{*{20}{c}}{t = t(G) = \mathop \Sigma \limits_{h = 1}^{{n_2}} t({p^h}) = \mathop \Sigma \limits_{h = 1}^{{n_1}} t({p^h}) + \mathop \Sigma \limits_{h = {n_1} + 1}^{{n_2}} t({p^h})}\\{ = \frac{{(p + 1)({p^{{n_1}}} - 1)}}{{p - 1}} + ({n_2} - {n_1}){p^{{n_1}}}.}\end{array}$

类似上面定理叙述可求得对应于引理1.7中的

$\begin{array}{l}\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} \\\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} = {p^{[{n_2}{p^{{n_2}}} - {n_1}{p^{{n_1}}} + \frac{{2({p^{{n_1}}} - {p^{{n_2}}})}}{{p - 1}}]{p^{{n_1}}} + {n_1}{p^{{2^n}1}} - \frac{{(p + 2)({p^{{2^n}1}} - 1)}}{{{p^2} - 1}}}}.\end{array}$

由引理1.7

$\begin{array}{*{20}{l}}{{{[{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]}^2} = {p^{({n_1} + {n_2}){p^{{n_1} + {n_2}}} - 2t}}/\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} }\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {p^{2q}}.}\end{array}$

于是[(Γ_p)^×:($\mathbb{Z}$_pG)^×]=p^q.
综合定理2.1~定理2.3的结果有
定理2.4??对$G = {C_{{p^{{n_1}}}}} \times {C_{{p^{{n_2}}}}}, 0 \le {n_1} \le {n_2}$，有

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {p^q}, $

其中

$\begin{array}{l}q = \frac{{{n_1}}}{2}{p^{{n_1} + {n_2}}} - ({n_2} - {n_1}){p^{{n_1}}} - \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{(p + 1)({p^{{n_1}}} - 1) + ({p^{{n_1}}} - {p^{{n_2}}}){p^{{n_1}}}}}{{p - 1}} + \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{(p + 2)({p^{2{n_1}}} - 1)}}{{2({p^2} - 1)}}.\end{array}$

对于任意(k₁, k₂, …, k_n)_p型的交换群，相应的计算公式非常难以给出，然而对下面特殊情形，有
定理2.5??当$G = {\left({{C_{{p^n}}}} \right)^l}, n \ge 0;\; l > 0$时有

$[{({\Gamma _p})^x}:{({\mathbb{Z}_p}G)^x}] = {p^q}, $

其中

$\begin{array}{l}q = \frac{{n(l - 1)}}{2}{p^{nl}} + \frac{{{p^{nl}} + p - 2}}{{2(p - 1)}} + \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{{p^{nl}} - {p^l}}}{{2({p^l} - 1)}} - \frac{{({p^l} - 1)({p^{n(l - 1)}} - 1)}}{{({p^{l - 1}} - 1)(p - 1)}}.\end{array}$

证明??此时|G|=p^nl, G的型为${\underbrace {\left({n, n, \cdots, n} \right)}_{共l项{\rm{ }}}}$_p。当1≤h≤n时，${v_h} = 0, {a_h} = \sum\limits_{\mu = 0}^{{v_h}} {{k_\mu }}=0 $。于是此时有

$\begin{array}{*{20}{l}}{t({p^h}) = \frac{{{p^{l - {v_h}}} - 1}}{{p - 1}}{p^{(l - {v_h} - 1)(h - 1) + {a_h}}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = \frac{{{p^l} - 1}}{{p - 1}}{p^{(l - 1)(h - 1)}}.}\end{array}$

故$t = t(G) = \sum\limits_{h = 1}^n t \left({{p^h}} \right) = \frac{{\left({{p^l} - 1} \right)\left({{p^{n(l - 1)}} - 1} \right)}}{{\left({{p^{l - 1}} - 1} \right)(p - 1)}}$.类似上面定理的叙述可求得

$\begin{array}{*{20}{c}}{\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} = \prod\nolimits_{h = 1}^n {{{({p^{{p^{h - 1}}(hp - h - 1)}})}^{t({p^h})}}} }\\{ = {p^{n{p^{nl}} - \frac{{pl}}{{p - 1}} + \frac{{{p^l} - {p^{nl}}}}{{{p^l} - 1}}}}.}\end{array}$

于是由引理1.7

$\begin{array}{*{20}{l}}{{{[{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]}^2} = {p^{nl{p^{nl}} - 2t}}/\prod\nolimits_{i = 1}^m {\left| {{d_i}} \right|} }\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {p^{2q}}, }\end{array}$

故[(Γ_p)^×:($\mathbb{Z}$_pG)^×]=p^q.易验证当n=0时，上式依然成立。
推论2.1?? (ⅰ)若G=C₂×C₂，则[K₁(Γ):K₁($\mathbb{Z}$G)]=2，
(ⅱ)若G=C₂×C_2²，则[K₁(Γ):K₁($\mathbb{Z}$G)]=2⁴，
(ⅲ)若G=C₂×C_2³，则[K₁(Γ):K₁($\mathbb{Z}$G)]=2¹²,
(ⅳ)若G=C₂, 则[K₁(Γ):K₁($\mathbb{Z}$G)]=1,
(ⅴ)若G=C_2²，则[K₁(Γ):K₁($\mathbb{Z}$G)]=2,
(ⅵ)若G=C_2³，则[K₁(Γ):K₁($\mathbb{Z}$)]=2⁴,
(ⅶ)若G=C_2⁴，则[K₁(Γ):K₁($\mathbb{Z}$G)]=2¹⁰。
证明??由文献[4]有

${{K_1}(\Gamma ) = {{(\Gamma )}^ \times } \oplus S{K_1}(\Gamma ), }$

${{K_1}(\mathbb{Z}G) = {{(\mathbb{Z}G)}^ \times } \oplus S{K_1}(\mathbb{Z}G).}$

由引理1.4知Γ是一些代数整数环的直和，于是由引理1.2知SK₁(Γ)=1, 而由引理1.3知在上述(i)-(vii)情况下均有SK₁($\mathbb{Z}$G)=1, 于是对于(i)-(vii)，下式总是成立的

$[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{(\Gamma )^ \times }:{(\mathbb{Z}G)^ \times }].$

(ⅰ)由文献[13]知，此时|D($\mathbb{Z}$G)|=1，由定理2.2得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = 2.}\end{array}$

(ⅱ)由文献[13]得知此时|D($\mathbb{Z}$G)|=2, 由定理2.3得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^4}.}\end{array}$

(ⅲ)由文献[13]知，此时|D($\mathbb{Z}$G)|=2³, 由定理2.3得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^{12}}.}\end{array}$

(ⅳ)当G=C₂时由文献[13]知|D($\mathbb{Z}$G)|=1，由定理2.1得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = 1.}\end{array}$

(ⅴ)当G=C_2²时由文献[13]知|D($\mathbb{Z}$G)|=1，由定理2.1得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = 2.}\end{array}$

(ⅵ)当G=C_2³时由文献[13]知|D($\mathbb{Z}$G)|=1，由定理2.1得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^4}.}\end{array}$

(ⅶ)当G=C_2⁴时由文献[13]知|D($\mathbb{Z}$G)|=2，由定理2.1得

$\begin{array}{*{20}{l}}{[{K_1}(\Gamma ):{K_1}(\mathbb{Z}G)] = [{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = [{{({\Gamma _2})}^ \times }:{{({\mathbb{Z}_2}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^{10}}.}\end{array}$

推论2.2?? (i)当G=C₂×C₂×C₂时，[K₁(Γ):K₁($\mathbb{Z}$G)]=2⁴，
(ii) 当G=C₂×C₂×C₂×C₂时，[K₁(Γ):K₁($\mathbb{Z}$G)]=2¹¹。
证明??由引理1.2~引理1.4知, 在(i), (ii)两种情况下均有SK₁(ZG)=1和SK₁(Γ)=1，于是均有[K₁(Γ):K₁($\mathbb{Z}$G)]=[(Γ)^×:($\mathbb{Z}$G)^×]。
(ⅰ)由定理2.5可得

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {2^5}.$

由文献[13]，此时|D($\mathbb{Z}$G)|=2，于是由引理1.7

$\begin{array}{*{20}{l}}{[{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }] = [{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^4}.}\end{array}$

(ⅱ)由定理2.5可得

$[{({\Gamma _p})^ \times }:{({\mathbb{Z}_p}G)^ \times }] = {2^{17}}.$

由文献[13]，此时|D($\mathbb{Z}$G)|=2⁶，于是由引理1.7

$\begin{array}{*{20}{l}}{[{{(\Gamma )}^ \times }:{{(\mathbb{Z}G)}^ \times }] = [{{({\Gamma _p})}^ \times }:{{({\mathbb{Z}_p}G)}^ \times }]/|D(\mathbb{Z}G)|}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {2^{11}}, }\end{array}$

故[K₁(Γ):K₁($\mathbb{Z}$G)]=2¹¹.

参考文献

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