在文献[1]中, 通过对群$ G $进行Bruck-Reilly扩张, Reilly获得了正则双单$ \omega $ -半群的$ BR(G, \theta) $结构. Warne研究了正则双单$ \omega^{n} $ -半群, 他在文献[2]中证明了正则双单$ \omega^{n} $ -半群具有$ (G\times C_{n}, \circ) $的结构, 其中$ G $为群, $ C_{n} $为2$ n $ -循环半群, “$ \circ $”是一种乘法.本文将用文献[1]的方法研究正则双单$ \omega^{2} $ -半群.在本节中, 引入了$ \omega^{2} $ -链; 在第2节中, 将引入幺半群$ T $的一种广义Bruck-Reilly扩张, 然后通过群$ G $的关于它的两个同态$ \beta, $ $ \gamma $及它的一个元$ u $的广义Bruck-Reilly扩张, 得到正则双单$ \omega^{2} $ -半群; 在第3节中, 将证明任意一个正则双单$ \omega^{2} $ -半群都可以这样构造.

我们将使用文献[3, 4]的概念及记号, 其余相关概念参见文献[3-16].在本文中, 映射作用在元素上都统一用映射写在元素的右侧来表示.设$ a, b $为半群$ S $的元, 若$ S^{1}a = S^{1}b $, 则称$ a, b $是$ \mathcal{L} $ -相关的.若$ aS^{1} = bS^{1} $, 则称$ a, b $是$ \mathcal{R} $ -相关的.规定$ \mathcal{H} = \mathcal{L}\cap \mathcal{R} $且$ \mathcal{D} = \mathcal{L}\vee\mathcal{R}. $可知$ \mathcal{L}, \mathcal{R}, \mathcal{H} $和$ \mathcal{D} $是$ S $上的等价关系且满足$ \mathcal{H}\subseteq \mathcal{L}\subseteq \mathcal{D}, $ $ \mathcal{H}\subseteq \mathcal{R}\subseteq \mathcal{D} $及$ \mathcal{D} = \mathcal{L}\circ \mathcal{R} = \mathcal{R}\circ \mathcal{L}. $为了避免混淆, 记$ S $上的关系$ \mathcal{K} $为$ \mathcal{K}(S). $用$ L_{a} $记$ S $上包含$ a $的$ \mathcal{L} $ -类, $ R_{a} $记$ S $上包含$ a $的$ \mathcal{R} $ -类.若半群$ S $只含一个$ \mathcal{D} $ -类, 则称它是双单的.若对半群$ S $的任一元$ a $, 存在唯一的$ S $的元$ x $, 满足$ axa = a $且$ xax = x $, 则称半群$ S $为逆半群, 称$ x $为$ a $的逆元, 记为$ a^{-1}. $用$ E_{S} $表示半群$ S $的幂等元的集合.在$ E_{S} $上定义偏序“$ \leq $”为$ e\leq f $当且仅当$ ef = fe = e. $设$ S $为半群且$ C_{\omega} = \{e_{0}, e_{1}, e_{2}, \cdot\cdot\cdot\}, $其中$ e_{0}, e_{1}, e_{2}, \cdot\cdot\cdot $为幂等元且$ e_{0}>e_{1}>e_{2}>\cdot\cdot\cdot, $若$ E_{S}\cong C_{\omega }, $则称$ S $为$ \omega $ -半群.用$ N^{0} $表示所有非负整数的集合, $ N $表示所有正整数的集合, $ T $为具有单位元$ e $的幺半群, $ \theta $为从$ T $到$ T $的单位的群$ H_{e} $的同态.在$ N^{0}\times T\times N^{0} $上规定乘法为

$ (m, a, n)(p, b, q) = (m-n+t, (a\theta^{t-n})(b\theta^{t-p}), q-p+t), $

其中$ t = {\rm{max}}\{n, p\} $, $ \theta^{0} $为$ T $的恒等映射, 则$ N^{0}\times T\times N^{0} $在上述乘法下构成一个半群, 称为$ T $的由$ \theta $决定的Bruck-Reilly扩张, 记为BR$ (T, \theta), $可知

$ (m, a, n)\mathcal{H}(p, b, q) \Leftrightarrow m = p, \; a\mathcal{H}b, \; n = q. $

引理 1.1 ^[3]  设$ G $为群, $ \theta $为$ G $的自同态, $ S = {\hbox{BR}}(G, \theta) $为$ G $的由$ \theta $决定的Bruck-Reilly扩张, 则$ S $为双单逆$ \omega $ -半群.反之, 任一个双单逆$ \omega $ -半群同构于某一个BR$ (G, \theta) $.

定义 1.2   在集合$ N^{0}\times N^{0} $上规定一个序为: $ (m, n)\leq (p, q) $当且仅当$ m> p, $或$ m = p $且$ n\geq q, $则称具有这样序的集合$ N^{0}\times N^{0} $为一个$ \omega^{2} $ -链, 记为$ C_{\omega^2}. $任一个序同构于$ C_{\omega^2} $的偏序集也称为$ \omega^{2} $ -链.

若半群$ S $的幂等元集$ E_{S} $序同构于$ C_{\omega ^{2}}, $则称$ S $为$ \omega^{2} $ -半群.因此, 若$ S $为$ \omega^{2} $ -半群, 则$ E_{S} = \{e_{m, n}: m, \; n\in N^{0}\}, $其中$ e_{m, n}\leq e_{p, q} $当且仅当$ (m, n)\leq (p, q). $用$ R_{m, n} $表示$ S $的包含幂等元$ e_{m, n} $的$ \mathcal{R} $ -类, 用$ L_{m, n} $表示$ S $的包含幂等元$ e_{m, n} $的$ \mathcal{L} $ -类, 用$ H_{(m, n), (q, p)} $表示$ R_{m, n}\cap L_{p, q}, $即

$ \begin{eqnarray*} &&R_{m, n} = \{a\in S: a\mathcal{R}e_{m, n}\}, \; \; L_{p, q} = \{a\in S: a\mathcal{L}e_{p, q}\}, \\ &&H_{(m, n), (q, p)} = \{a\in S:e_{m, n}\mathcal{R}a\mathcal{L}e_{p, q}\}. \end{eqnarray*} $

若$ H_{(m, n), (q, p)}\neq \emptyset $, 则$ H_{(m, n), (q, p)} $为$ S $的一个$ \mathcal{H} $ -类.

引理 1.3   ^[5]  设$ S $为$ \omega^{2} $ -逆半群, 则$ H_{(m, n), (q, p)}H_{(a, b), (d, c)}\subseteq H_{(i, j), (l, k)}, $其中

$ \begin{equation*} (i, j, l, k) = \begin{cases} (m, \max{\{q, b\}}-q+n, \max{\{q, b\}}-b+d, c), & \text{当 $p = a, $}\\ (m, n, q, p-a+c), & \text{当 $p>a, $}\\ (a-p+m, b, d, c), & \text{当 $p<a$}. \end{cases} \end{equation*} $

引理 1.4   设$ S $为正则$ \omega^{2} $ -半群, 则$ S $为具有单位元的$ \omega^{2} $ -逆半群.特别地, 双单$ \omega^{2} $ -半群是双单$ \omega^{2} $ -逆半群.

证  设$ S $为正则$ \omega^{2} $ -半群, $ e $和$ f $为$ S $的幂等元, 则$ e\leq f $或$ f\leq e, $从而$ ef = fe = e $或$ ef = fe = f. $不论何种情况, 都有$ ef = fe. $因此$ S $的幂等元可交换, 从而$ S $为逆半群.设$ E_{S} = \{e_{m, n}: m, \; n\in N^{0}\} $, 其中$ e_{m, n}\leq e_{p, q} $当且仅当$ (m, n)\leq(p, q) $.设$ a $为$ S $的任一元, 则存在$ S $的幂等元$ e_{m, n} $, 使得$ aa^{-1} = e_{m, n} $, 因此$ e_{0, 0}a = e_{0, 0}(e_{m, n}a) = (e_{0, 0}e_{m, n})a = e_{m, n}a = a $.类似有$ ae_{0, 0} = a $, 于是$ e_{0, 0} $是单位元, 故$ S $为具有单位元的$ \omega^{2} $ -逆半群.

在这一节中, 将引入一种广义Bruck-Reilly扩张, 对群和它的一对同态做这种扩张, 可得到正则双单$ \omega^{2} $ -半群.在下一节中, 将证明任一个正则双单$ \omega^{2} $ -半群都可以这样构造出来.

设$ T $为具有单位元$ e $的幺半群, $ H_{e} $为$ T $的包含$ e $作为单位元的极大子群, $ u $为$ H_{e} $的一个元, $ \tau_{u} $为$ H_{e} $的内部自同构, 即对任意$ g\in H_{e} $, 有$ g\tau_{u} = ugu^{-1} $.设$ \beta $, $ \gamma $为从$ T $到$ H_{e} $的两个同态, 且满足$ \gamma\tau _{u} = \beta \gamma, $其中$ \beta^{0} $, $ \gamma^{0} $为$ T $的恒等映射, $ u^{0} = e $.对任意$ (m, n, t, q, p), (m', n', t', q', p')\in N^{0}\times N^{0} \times T \times N^{0}\times N^{0} $, 在$ N^{0}\times N^{0} \times T \times N^{0}\times N^{0} $上规定乘法如下

$ \begin{equation} \begin{split} &(m, n, a, q, p)(m', n', a', q', p')\\ = &\begin{cases} (m, n-q+ {\rm{max}}\{q, n'\}, a \beta ^{ {\rm{max}}\{q, n'\}-q} a' \beta ^{ {\rm{max}}\{q, n'\}-n'}, \\ q'-n'+ {\rm{max}}\{q, n'\}, p'), & \text{当 $p = m', $}\\ (m, n, a(u^{-n'} a'\gamma u^{q'})\gamma ^{p-m'-1} \beta^{q}, q, p+p'-m'), & \text{当 $p>m', $}\\ (m+m'-p, n', (u^{-n} a\gamma u^{q})\gamma ^{m'-p-1} \beta^{n'} a', q', p'), & \text{当 $p<m'.$} \end{cases} \end{split} \end{equation} $

(2.1)

可以验证(2.1)式满足结合律, 由于证明过程是直接的和繁琐的, 因此省略了验证过程.规定从BR$ (T, \gamma) $到$ N^{0}\times N^{0} \times T \times N^{0}\times N^{0} $的映射$ \phi $为$ (m, a, n)\phi = (m, 0, a, 0, n). $显然$ \phi $为单射.对任意$ (m, 0, a, 0, n), (m', 0, a', 0, n') \in N^{0}\times N^{0} \times T \times N^{0}\times N^{0} $, 有

$ \begin{equation*} \begin{split} &(m, 0, a, 0, n)(m', 0, a', 0, n')\\ = &(m-n+ {\rm{max}}\{n, m'\}, 0 , a\gamma ^{ {\rm{max}}\{n, m'\}-n} a' \gamma ^{ {\rm{max}}\{n, m'\}-m'} , 0, n'-m'+ {\rm{max}}\{n, m'\}), \end{split} \end{equation*} $

从而$ \phi $为同态.在这个观点下, 称上述构造的半群为$ T $的由$ \beta, \gamma, u $所决定的广义Bruck-Reilly扩张, 记为GBR$ (T;\beta, \gamma;u). $可验证$ (m, n, a, q, p) $为GBR$ (T;\beta, \gamma;u) $的幂等元当且仅当$ m = p, $ $ n = q $且$ a $为$ T $的幂等元.

引理 2.1   设$ S = {\hbox{GBR}}(T;\beta, \gamma;u) $为幺半群$ T $的由$ \beta, \gamma, u $所决定的广义Bruck-Reilly扩张, $ (m, n, a, q, p), (m', n', a', q', p') $为$ S $的任意元, 则

(1)  $ (m, n, a, q, p)\mathcal{R}(S)(m', n', a', q', p') $当且仅当$ m = m', \; n = n' $及$ a\mathcal{R}(T)a' $.

(2)  $ (m, n, a, q, p)\mathcal{L}(S)(m', n', a', q', p') $当且仅当$ q = q', \; p = p' $及$ a\mathcal{L}(T)a' $.

证   (1)  设$ (m, n, a, q, p), \; (m', n', a', q', p') $为$ S $的两个元且$ (m, n, a, q, p) \mathcal{R}(S)(m', n', a', q', p'), $则存在$ S $中的元$ (x, y, b, z, w) $使得$ (m, n, a, q, p) (x, y, b, z, w) = (m', n', a', q', p'), $从而

$ \begin{equation*} \begin{split} &\begin{cases} (m, n-q+ {\rm{max}}\{q, y\}, a\beta ^{ {\rm{max}}\{q, y\}-q} b\beta ^{ {\rm{max}}\{q, y\}-y}, z-y+ {\rm{max}}\{q, y\}, w), & \text{当 $p = x, $}\\ (m, n, a (u^{-y} b\gamma u^{z})\gamma^{p-x-1}\beta^{q}, q, w-x+p), & \text{当 $p>x, $}\\ (m-p+x, y, (u^{-n} a\gamma u^{q})\gamma^{x-p-1}\beta^{y} b, z, w), & \text{当 $p<x$} \end{cases}\\ = &(m', n', a', q', p'). \end{split} \end{equation*} $

比较第一分量, 得$ m'\geq m. $对偶地, 有$ m\geq m' $, 从而$ m = m' $, 因此$ p\geq x. $比较第二分量, 得$ n\leq n'. $对偶地, 有$ n\geq n' $, 于是$ n = n' $.若$ p = x, $比较第二分量, 得$ {\rm{max}}\{q, y\} = q, $从而

$ a\beta^{ {\rm{max}}\{q, y\}-q} b\beta^{ {\rm{max}}\{q, y\}-y} = a b\beta^{q-y}. $

比较第三分量得$ a' = a b\beta^{q-y}, $从而在$ T $中有$ R_{a'}\leq R_{a} $.若$ p>x, $比较第三分量, 从而在$ T $中有$ R_{a'}\leq R_{a}. $对偶地, 有$ R_{a}\leq R_{a'} $.故$ a\mathcal{R}(T)a'. $

反之, 若$ a\mathcal{R}(T)a', $则存在$ T $的元$ c $, $ d $, 使得$ ac = a' $且$ a'd = d $, 从而

$ \begin{equation*} \begin{split} &(m, n, a, q, p)(p, q, c, q', p') = (m, n, a', q', p'), \\ &(m, n, a', q', p')(p', q', d, q, p) = (m, n, a, q, p). \end{split} \end{equation*} $

故$ (m, n, a, q, p)\mathcal{R}(S)(m, n, a', q', p'). $ (2)类似可证.

引理 2.2   设$ S = {\hbox{GBR}}(T;\beta, \gamma;u) $为幺半群$ T $的由$ \beta, \gamma, u $所决定的广义Bruck-Reilly扩张, 则$ S $为逆半群当且仅当$ T $是逆半群.

证  若$ T $为逆半群, 则对$ S $的任一元$ (m, n, a, q, p) $, 有

$ \begin{equation*} \begin{split} (m, n, a, q, p)(p, q, a^{-1}, n, m)(m, n, a, q, p)& = (m, n, aa^{-1}, n, m)(m, n, a, q, p)\\ & = (m, n, a, q, p) \end{split} \end{equation*} $

且

$ \begin{equation*} \begin{split} (p, q, a^{-1}, n, m)(m, n, a, q, p)(p, q, a^{-1}, n, m)& = (p, q, a^{-1}a, q, p)(p, q, a^{-1}, n, m)\\ & = (p, q, a^{-1}, n, m). \end{split} \end{equation*} $

因此$ (m, n, a, q, p) $有逆元$ (p, q, a^{-1}, n, m) $, 于是$ S $是正则的.若$ (m, n, e, n, m) $和$ (m', n', e', $ $ n', m') $为$ S $的两个幂等元, 则

$ \begin{equation*} \begin{split} &(m, n, e, n, m)(m', n', e', n', m')\\ = &\begin{cases} (m, {\rm{max}}\{n, n'\}, e\beta ^{ {\rm{max}}\{n, n'\}-n} e' \beta ^{ {\rm{max}}\{n, n'\}-n'}, {\rm{max}}\{n, n'\}, m'), & \text{当 $m = m', $}\\ (m, n, e (u^{-n'} e'\gamma u^{n'}) \gamma ^{m-m'-1}\beta ^{n}, n, m), & \text{当 $m>m', $}\\ (m', n', (u^{-n} e\gamma u^{n})\gamma ^{m'-m-1} \beta ^{n'} e', n', m'), & \text{当 $m<m'$} \end{cases} \end{split} \end{equation*} $

且

$ \begin{equation*} \begin{split} &(m', n', e', n', m')(m, n, e, n, m)\\ = &\begin{cases} (m', {\rm{max}}\{n, n'\}, e'\beta ^{ {\rm{max}}\{n, n'\}-n'} e \beta ^{ {\rm{max}}\{n, n'\}-n}, {\rm{max}}\{n, n'\}, m), & \text{当 $m' = m, $}\\ (m', n', e'(u^{-n} e\gamma u^{n}) \gamma ^{m'-m-1}\beta ^{n'}, n', m'), & \text{当 $m'>m, $}\\ (m, n, (u^{-n'} e'\gamma u^{n'}) \gamma ^{m-m'-1}\beta ^{n} e, n, m), & \text{当 $m'<m$}, \end{cases} \end{split} \end{equation*} $

由于$ T $为逆半群, 幂等元可交换, 从而$ S $的幂等元是可交换的.

反之, 若$ S $为逆半群, 设$ (m, n, a, q, p)^{-1} = (x, y, b, z, w), $则

$ \begin{equation*} \begin{split} &(m, n, a, q, p)(x, y, b, z, w)\\ = &\begin{cases} (m, n-q+ {\rm{max}}\{q, y\}, a\beta ^{ {\rm{max}}\{q, y\}-q} b \beta ^{ {\rm{max}}\{q, y\}-y}, z-y+ {\rm{max}}\{q, y\}, w), & \text{当 $p = x$, }\\ (m, n, a(u^{-y} b\gamma u^{z}) \gamma ^{p-x-1}\beta ^{q}, q, w-x+p), & \text{当 $p>x, $}\\ (m-p+x, y, (u^{-n} a\gamma u^{q})\gamma ^{x-p-1} \beta ^{y} b, z, w), & \text{当 $p<x$} \end{cases} \end{split} \end{equation*} $

是幂等元且$ \mathcal{R} $ -相关于$ (m, n, a, q, p) $且$ \mathcal{L} $ -相关于$ (x, y, b, z, w), $从而$ p = x. $由于

$ (m, n, a, q, p)(x, y, b, z, w) $

是幂等元, 从而$ m = w $且$ n = n-q+ {\rm{max}}\{q, y\} = z-y+ {\rm{max}}\{q, y\} = z, $因此$ q = y, $于是

$ \begin{equation*} \begin{split} (m, n, a, q, p)& = (m, n, a, q, p)(p, q, b, n, m)(m, n, a, q, p)\\ & = (m, n, ab, n, m)(m, n, a, q, p) = (m, n, aba, q, p) \end{split} \end{equation*} $

且

$ \begin{equation*} \begin{split} (p, q, b, n, m)& = (p, q, b, n, m)(m, n, a, q, p)(p, q, b, n, m)\\ & = (p, q, ba, q, p)(p, q, b, n, m) = (p, q, bab, n, m). \end{split} \end{equation*} $

故在$ T $中有$ a = aba $且$ b = bab $, 从而$ T $是正则的.若$ e $, $ f $为$ T $的两个幂等元, 由于

$ (0, 0, e, 0, 0)(0, 0, f, 0, 0) = (0, 0, f, 0, 0)(0, 0, e, 0, 0), $

从而$ ef = fe. $因此$ T $为逆半群.

定理 2.3   设$ G $为群且单位元为$ e $, $ u $为$ G $的元, $ \beta $和$ \gamma $为$ G $的同态, $ S = {\hbox{GBR}}(G; \beta ;\gamma;u) $为$ G $的由$ \beta , \gamma , u $决定的广义Bruck-Reilly扩张, 则$ S = {\hbox{GBR}} (G;\beta, \gamma;u) $为正则双单$ \omega ^{2} $ -半群.

证  由引理2.2知, $ S $为正则半群.设$ (m, n, a, q, p), $ $ (m', n', a', q', p') $为$ S $的任意两个元, 则

$ (m, n, a, q, p)\mathcal{R}(m, n, a, q', p')\mathcal{L} (m', n', a', q', p'), $

从而$ (m, n, a, q, p)\mathcal{D}(m', n', a', q', p'), $因此$ S $是双单的.设$ (m, n, e, n, m) $, $ (m, n', e, n', m) $为$ S $的任意两个幂等元且$ n<n', $则

$ (m, n', e, n', m)< (m, n, e, n, m). $

若$ (m, n, e, n, m) $, $ (m', n', e, n', m') $为$ S $的两个幂等元且$ m< m', $则

$ (m, n, e, n, m)(m', n', e, n', m') = (m', n', e, n', m') = (m', n', e, n', m')(m, n, e, n, m), $

从而$ (m', n', e, n', m')<(m, n, e, n, m) $, 因此$ S $的幂等元构成一个链

$ \begin{equation*} \begin{split} &\; (0, 0, e, 0, 0)>(0, 1, e, 1, 0)>(0, 2, e, 2, 0)>\cdots\\ >&\; (1, 0, e, 0, 1)>(1, 1, e, 1, 1)>(1, 2, e, 2, 1)>\cdots\\ >&\; (2, 0, e, 0, 2)>(2, 1, e, 1, 2)>(2, 2, e, 2, 2)>\cdots\\ & \quad \vdots \end{split} \end{equation*} $

对任意$ m, n\in N^{0}, $规定$ e_{m, n} = (m, n, e, n, m), $从而$ S $的幂等元的集合为$ \{e_{m, n}:m, \; n\in N^{0}\} $且$ e_{m, n}\leq e_{p, q} $当且仅当$ (m, n)\leq (p, q). $因此$ S = GBR \, (G;\beta, \gamma ;u) $为正则双单$ \omega^2 $ -半群.

在本节中, 我们将证明任一个正则双单$ \omega^2 $ -半群同构于群$ G $的由$ \beta, \gamma, u $决定的Bruck-Reilly扩张GBR$ (G;\beta, \gamma;u). $

设$ S $为正则双单$ \omega ^{2} $ -半群, $ E_{S} $为$ S $的幂等元的集合, 则

$ E_{S} = \{ e_{m, n}: m, \; n \in N^{0}\} $

构成$ \omega^{2} $ -链且$ e_{m, n}<e_{p, q} $当且仅当$ (m, n)<(p, q) $且$ e_{0, 0} $为$ S $的单位元.设$ a\in H_{(0, 0), (1, 0)}, $ $ b\in H_{(0, 0), (0, 1)}. $规定$ a^{0} = e_{0, 0} $且$ b^{0} = e_{0, 0}, $则$ a^{0}\in H_{(0, 0), (0, 0)}. $假设$ a^{n-1}\in H_{(0, 0), (n-1, 0)}, $则

$ a^{n} = a^{n-1} a\in H_{(0, 0), (n-1, 0)} H_{(0, 0), (1, 0)} \subseteq H_{(0, 0), (n, 0)}. $

从而对任意非负整数$ n $, 有$ a^{n} \in H_{(0, 0), (n, 0)}. $类似地, 设$ m, n $为非负整数, 则

$ a^{-n} \in H_{(0, n), (0, 0)}, \; \; b^{m} \in H_{(0, 0), (0, m)}, \; \; b^{-m} \in H_{(m, 0), (0, 0)} $

且

$ a^{n}a^{-n} = e_{0, 0}, \; \; a^{-n}a^{n} = e_{0, n}, \; \; b^{m}b^{-m} = e_{0, 0}, \; \; b^{-m}b^{m} = e_{m, 0}. $

由引理1.3, 对$ H_{(0, 0), (0, 0)} $的任意元$ g, $有$ b^{-m}a^{-n}ga^{q}b^{p} \in H_{(m, n), (q, p)}. $对任意$ m, n , p, q \in N^{0}, $规定从$ H_{(0, 0), (0, 0)} $到$ H_{(m, n), (q, p)} $的对应法则$ \sigma $为$ g\sigma = b^{-m}a^{-n}ga^{q}b^{p}. $若$ b^{-m}a^{-n}g_{1}a^{q}b^{p} = b^{-m}a^{-n}g_{2}a^{q}b^{p} $, 则

$ b^{m}b^{-m}a^{-n}g_{1}a^{q}b^{p}b^{-p} = b^{m}b^{-m}a^{-n}g_{2}a^{q}b^{p}b^{-p}, $

从而$ a^{-n}g_{1}a^{q} = a^{-n}g_{2}a^{q}, $因此

$ a^{n}a^{-n}g_{1}a^{q}a^{-q} = a^{n}a^{-n}g_{2}a^{q}a^{-q}, $

于是$ g_{1} = g_{2}. $设$ x $为$ H_{(m, n), (q, p)} $的任意元, 则

$ b^{m}xb^{-p}\in H_{(0, n), (q, 0)}, \; \; a^{n}b^{m}xb^{-p}a^{-q} \in H_{(0, 0), (0, 0)}, $

从而$ (a^{n}b^{m}xb^{-p}a^{-q})\sigma = x, $因此$ \sigma $为双射.

由于$ S = \bigcup \{H_{(m, n), (q, p)}|\; m, n, p, q \in N^{0}\}, $从而由上述论证可知下列引理成立.

引理 3.1   设$ a\in H_{(0, 0), (1, 0)} $且$ b\in H_{(0, 0), (0, 1)}, $则$ S $的任一元都可唯一的表为$ b^{-m}a^{-n}ga^{q}b^{p} $的形式, 其中$ m, n, p, q \in N^{0}, g\in H_{(0, 0), (0, 0)}. $

设$ H_{(0, 0), (0, 0)} = G, g $为$ G $的任一元, 则$ ag \in H_{(0, 0), (1, 0)} H_{(0, 0), (0, 0)}\subseteq H_{(0, 0), (1, 0)} $且$ bg \in H_{(0, 0), (0, 1)} H_{(0, 0), (0, 0)}\subseteq H_{(0, 0), (0, 1)}, $从而由引理3.1知$ ag $可唯一的表示为$ b^{0}a^{0}g'a^{1}b^{0} = g'a $; $ bg $可唯一的表示为$ b^{0}a^{0}g''a^{0}b^{1} = g''b $, 其中$ g', g''\in G. $设$ \beta $, $ \gamma $分别为$ G $的按如下条件所决定的映射

$ ag = (g\beta)a, \; \; \; \; bg = (g\gamma)b, $

则

$ \begin{equation*} \begin{split} &[(g_{1}g_{2})\beta]a = a(g_{1}g_{2}) = (ag_{1})g_{2} = [(g_{1}\beta)a]g_{2} = (g_{1}\beta)(ag_{2}) = (g_{1}\beta)(g_{2}\beta)a, \\ &[(g_{1}g_{2})\gamma]b = b(g_{1}g_{2}) = (bg_{1})g_{2} = [(g_{1}\gamma)b]g_{2} = (g_{1}\gamma)(bg_{2}) = (g_{1}\gamma)(g_{2}\gamma)b. \end{split} \end{equation*} $

由于$ aa^{-1} = e_{0, 0} $且$ bb^{-1} = e_{0, 0}, $从而

$ (g_{1}g_{2})\beta = (g_{1}\beta)(g_{2}\beta), \; \; \; \; (g_{1}g_{2})\gamma = (g_{1}\gamma)(g_{2}\gamma). $

因此$ \beta $, $ \gamma $为$ G $的自同态.于是对任意$ n\in N^{0} $, 有

$ \begin{equation} a^{n}g = a^{n-1}ag = a^{n-1}(g\beta)a = a^{n-2} (g\beta^{2})a^{2} = \cdot\cdot\cdot = (g\beta^{n})a^{n}. \end{equation} $

(3.1)

进一步地, 由于$ ga^{-n}\in H_{(0, n), (0, 0)} $且$ a^{n}a^{-n} = e_{0, 0} $为$ S $的单位元, 从而

$ \begin{equation} ga^{-n} = e_{0, n}ga^{-n} = a^{-n}(a^{n}g)a^{-n} = a^{-n}(g\beta^{n})a^{n}a^{-n} = a^{-n}(g\beta^{n}). \end{equation} $

(3.2)

类似地, 对任意$ n\in N^{0} $, 有

$ \begin{equation} b^{n}g = (g\gamma^{n})b^{n}, \; \; \; \; gb^{-n} = b^{-n}(g\gamma^{n}). \end{equation} $

(3.3)

令$ u = bab^{-1}, $则$ u^{-1} = ba^{-1}b^{-1}. $规定$ u^{0} = e_{0, 0}. $由于

$ ba^{-1}\in H_{(0, 0), (0, 1)} H_{(0, 1), (0, 0)} \subseteq H_{(0, 0), (0, 1)}, $

从而$ ba^{-1} = u^{-1}b. $类似有

$ ba = ub, \quad ab^{-1} = b^{-1}u, \quad a^{-1}b^{-1} = b^{-1}u^{-1}. $

因此, 对任意$ m, n, q, p\in N^{0} $, 有

$ \begin{equation} ba^{-m} = u^{-m}b, \quad ba^{n} = u^{n}b, \quad a^{q}b^{-1} = b^{-1}u^{q}, \quad a^{-p}b^{-1} = b^{-1}u^{-p}. \end{equation} $

(3.4)

引理 3.2   设$ S $为正则双单$ \omega ^{2} $-半群, $ G $为$ S $的单位的群, $ a \in H_{(0, 0), (1, 0)} $, $ b \in H_{(0, 0), (0, 1)} $, 则$ u = bab^{-1}\in G $且$ \gamma \tau _{u} = \beta \gamma $, 其中$ \tau _{u} $为$ G $的内部自同构, 即对任意$ g\in G, g\tau_{u} = ugu^{-1}. $

证  由$ H_{(0, 0), (0, 1)} H_{(0, 0), (1, 0)} H_{(1, 0), (0, 0)}\subseteq H_{(0, 0), (0, 0)} $知$ u = bab^{-1} \in G $且对任意$ g\in G, $有

$ \begin{equation*} \begin{split} u (g\gamma)& = (bab^{-1})g\gamma = ba(b^{-1}g\gamma) = ba(gb^{-1}) = b(ag)b^{-1} = b (g\beta) a b^{-1}\\ & = (b (g\beta)) a b^{-1} = ((g\beta \gamma) b)ab^{-1} = g\beta \gamma (bab^{-1}) = g\beta \gamma u, \end{split} \end{equation*} $

因此对任意$ g\in G, $有$ g\gamma\tau_{u} = ug\gamma u^{-1} = g\beta\gamma $, 于是$ \gamma \tau _{u} = \beta \gamma $.

定理 3.3   设$ S $为正则双单$ \omega^{2} $ -半群, 则$ S\simeq {\hbox{GBR}}(G;\beta, \gamma;u), $其中$ G $为群, $ \beta, \gamma $都为$ G $的自同态, $ u\in G $.

证  设$ G, \beta, \gamma, u $如前述所规定.由引理3.1知, $ S $的每一元都可唯一表示为$ b^{-m}a^{-n}ga^{q}b^{p} $的形式, 其中$ m, n, p, q \in N^{0}, g \in G. $设$ x = b^{-m}a^{-n}ga^{q}b^{p}, \; y = b^{-m'}a^{-n'}g'a^{q'}b^{p'} $, 其中$ g, \; g'\in G. $分以下三种情形.

情形 1  若$ p = m', $则

$ xy = b^{-m}a^{-n}ga^{q}b^{p} b^{-m'} a^{-n'}g'a^{q'}b^{p'} = b^{-m}a^{-n}ga^{q-n'}g' a^{q'}b^{p'}; $

若$ q = n', $则

$ xy = b^{-m}a^{-n}gg'a^{q'}b^{p'}; $

若$ q>n', $则由$ (3.1) $式知

$ xy = b^{-m}a^{-n}g(g'\beta^{q-n'})a^{q+q'-n'}b^{p'}; $

若$ q<n', $则由$ (3.2) $式知

$ xy = b^{-m}a^{-(n-q+n')}(g\beta^{n'-q})g'a^{q'}b^{p'}. $

情形 2  若$ p> m' $, 则由$ (3.1), (3.3), (3.4) $式知

$ \begin{align*} xy& = b^{-m}a^{-n}ga^{q}b^{p-m'-1}ba^{-n'}g'a^{q'}b^{p'}\\ & = b^{-m}a^{-n}ga^{q}b^{p-m'-1}u^{-n'}bg'a^{q'}b^{p'} & \\ & = b^{-m}a^{-n}ga^{q}(u^{-n'}\gamma^{p-m'-1})b^{p-m'-1}bg'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})a^{q}b^{p-m'}g'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})a^{q}(g'\gamma^{p-m'})b^{p-m'}a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})(g'\gamma^{p-m'}\beta^{q})a^{q}b^{p-m'}a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})(g'\gamma^{p-m'}\beta^{q})a^{q}b^{p-m'-1}u^{q'}b^{p'+1} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})(g'\gamma^{p-m'}\beta^{q})a^{q} (u^{q'}\gamma^{p-m'-1}) b^{p-m'+p'} &\\ & = b^{-m}a^{-n}g(u^{-n'}\gamma^{p-m'-1}\beta^{q})(g'\gamma^{p-m'}\beta^{q}) (u^{q'}\gamma^{p-m'-1}\beta^{q})a^{q}b^{p-m'+p'} &\\ & = b^{-m}a^{-n}g(u^{-n'} g'\gamma u^{q'})\gamma^{p-m'-1}\beta ^{q} a^{q} b^{p-m'+p'}. \end{align*} $

情形 3  若$ p< m' $, 则由(3.2)–(3.4)式知

$ \begin{align*} xy& = b^{-m}a^{-n}ga^{q}b^{-1}b^{p-m'+1}a^{-n'} g'a^{q'}b^{p'}\\ & = b^{-m}a^{-n}gb^{-1}u^{q}b^{p-m'+1}a^{-n'} g'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}gb^{-1}b^{p-m'+1}(u^{q}\gamma^{m'-p-1})a^{-n'}g'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}gb^{p-m'}a^{-n'}(u^{q}\gamma^{m'-p-1}\beta^{n'}) g'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}b^{p-m'}(g\gamma^{m'-p})a^{-n'}(u^{q}\gamma^{m'-p-1}\beta^{n'})g'a^{q'}b^{p'} &\\ & = b^{-m}a^{-n}b^{p-m'}a^{-n'}(g\gamma^{m'-p}\beta^{n'})(u^{q}\gamma^{m'-p-1}\beta^{n'})g'a^{q'}b^{p'} &\\ & = b^{-m}b^{-1}u^{-n}b^{p-m'+1}a^{-n'}(g\gamma^{m'-p}\beta^{n'})(u^{q}\gamma^{m'-p-1}\beta^{n'})g'a^{q'}b^{p'} &\\ & = b^{-m-1}b^{p-m'+1}(u^{-n}\gamma^{m'-p-1})a^{-n'}(g\gamma^{m'-p}\beta^{n'})(u^{q}\gamma^{m'-p-1}\beta^{n'})g'a^{q'}b^{p'} &\\ & = b^{p-m-m'}a^{-n'}(u^{-n}\gamma^{m'-p-1}\beta^{n'})(g\gamma^{m'-p}\beta^{n'})(u^{q}\gamma^{m'-p-1}\beta^{n'})g'a^{q'}b^{p'} &\\ & = b^{-(m-p+m')}a^{-n'}(u^{-n} g\gamma u^{q})\gamma^{m'-p-1}\beta^{n'} g'a^{q'}b^{p'}. \end{align*} $

因此从$ S $到GBR$ (G;\beta, \gamma;u) $的映射

$ \psi:\; \; (b^{-m}a^{-n}ga^{q}b^{p})\psi = (m, n, g, q, p) $

为同构映射.故$ S\simeq {\hbox{GBR}}(G;\beta, \gamma;u). $