PARTIALLY ORDERED ABELIAN SEMIGROUPS II. ON THE 2019-03-18آ  PARTIALLY ORDERED ABELIAN SEMIGROUPS II

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    Title PARTIALLY ORDERED ABELIAN SEMIGROUPS II. ON THE STRONGNESS OF THE LINEAR ORDERDEFINED ON ABELIAN SEMIGROUPS

    Author(s) Nakada, Osamu

    Citation Journal of the Faculty of Science Hokkaido University. Ser. 1 Mathematics, 12(2), 073-086

    Issue Date 1952

    Doc URL http://hdl.handle.net/2115/55977

    Type bulletin (article)

    File Information JFSHIU_12_N2_073-086.pdf

    Hokkaido University Collection of Scholarly and Academic Papers : HUSCAP

    https://eprints.lib.hokudai.ac.jp/dspace/about.en.jsp

  • PARTIALLY ORDERED ABELIAN SEMIGROUPS II. ON THE STRONGNESS OF THE LINEAR ORDER

    DEFINED ON ABELIAN SEMIGROUPS

    BY

    Osamu NAKADA

    In my previous paper, I note that a strong linearly ordered abelian semigroup (strong 1. $0$ . semigroup) is always normal, its elements are of infinite order except the unit element (if there exists) and the product cancellation law is held in it. In this Part, I shall seek for the other conditions which are exchangeable for the cancellation law. But the some theorems in this note will be expressed in the form of a partially ordered abelian semigroup.

    Definition 1. Let $a$ be any element of an abelian semigroup $S$ . We denote the set of all positive powers of $a$ , in other words, the sub- semigroup of $S$ generated by $a$ , by $S(a)$ , which is called the sector by $a$ . And let $T(\backslash a)$ be the set of all elements of $S$ whose some positive powers belong to $S(a)$ , that is, the set of all elements $x$ of $S$ such that $x^{m}=a^{n}$ for some positive integers $m$ and $n$ . $T(a)$ is called the complete sector by $a$ .

    Let $a$ and $b$ be any two elements of $S$ . Then the two complete sectors $T(a)$ and $\mathcal{I}^{\prime}(b)$ are either disjoint or identical. Hence, $S$ is the union of mutually disjoint complete sectors, and clearly $T(a)$ is a sub- semigroup of $S$ .

    Definition 2 Let $a$ be any element of an abelian semigroup $S$ , and we shall consider the sector $S(a)$ by $a$ . There are two possible cases.

    (1) Partially ordered abelian semigroups. I. On the extension of the strong partial order defined on abelian semigroups. Journal of the Faculty of Science, Hokkaido University, Series I, vol. XI, No. 4 (1951) pp’ 181-189; this is referred to hereafter as 0. $I$.

    A set $S$ is said to be a partiauy ordered $abel\dot{w}n$ semigroup (p.o. ssmigroup), when $S$ is (I) an abelian semigroup (not necessarily contains th2 unit element), (II) a partially ordered set, and satisfies (III) the homogeneity: $a\geqq b$ implies $ac\geqq bc$ for any $c$ of $S$. When a partial order $P$ of an abelian semigroup $S$ satisfies the condition (III), we call $P$ a partial or&r defined on an abelian semigroup S. (Definition 1, $0$ I.)

    (2) Cf. D. $R\underline{q}oe$ : On semi-group3, Proc. Cambridge Phil. Soc, vol. 36 (1940), pp. 387-400.

  • 74 O. Nakada

    First, $a$ may satisfy no equation of the form $a^{m}=a^{n}(m\neq n)$ . In this case, the element $a$ is said to be of infinite order.

    Secondly, $a$ may satisfy a relation of the form $a^{m}=a^{n}(n>m)$ . If $a$ satisfies such a relation, choose one for which $n$ is a minimum. Then clearly $a,$ $a^{\varphi}\rightarrow,$ $\cdots a^{n-1}$ are distinct elements of $S(a)$ . We write $n=m+r$ , $r\underline{\geq_{-}}1$ . Then, if $q$ is any positive integer, by simple induction we have immediately $a^{m}=a^{m+qr}$ . Now, if $N$ is any integer not less than $n$ , we may write it in the form $N=m+qr+s(1\leqq q, 0\leqq s

  • Partially Ordered Abelian Semigroups II 78

    and $c\cdot T(b)$ are disjoint for any $c$ of $S$ . Moreover, we consider the following conditions for a p.o. semi-

    group $S$ : $(A)$ : If $a$ is positive or negative, then $ab\geqq b$ or $b\geqq ab$ for any

    $b$ of $S$ respectively. $(B)$ : If $a$ ( $-\neq$ the unit element) is positive or negative, then $\phi>b$

    or $b>d$ for any $b$ of $S$ respectively.

    Theorem 1. If an abelian semigroup $S$ satisfies the condition $(a)$ , then the element of $S$ is idempotent or of infinite order.

    Proof. Let $a$ be an element of finite order, $m$ and $r$ be the length and the period of $a$ respectively. And let $Z=\{a^{m}, \cdots, a^{m+r-1}\}$ , then $Z$ is a cyclic group. Hence there exists a positive integer $q$ such that $a^{q}=(a^{2})^{q}=y$ , where $y$ is the identity of the cyclic group $Z$. Therefore, by the condition $(a)$ we have $ a=a^{9}\lrcorner$ that is, $a$ is idempotent.

    Corollary. An element of a normal(5) p.o. semigroup $S$ is idempotent or of infinite order.

    Proof. For, $S$ satisfies the condition $(\alpha)$ by Theorem 12, $0$.I. Example 1. Let $S_{1}$ be a free abelian semigroup generated by one

    element $a$ and $S_{1}‘‘=$ { $b,$ $e=the$ identity} be a group of order 2, which does not satisfy the conditions $(a)$ and $(\beta)$ . And let $S_{1}$ be a direct product of $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ in the usual sense. Then $S_{1}$ becomes the abelian semi- group which satisfies the conditions $(\beta),$ $(r)$ and $(\delta)$ , but, since $(xe)^{o}\rightarrow=$ $(xb)^{\Omega}\lrcorner$ in spite of $xe\neq xb(x\in S_{1}^{\prime})$ , does not sat-sfy the condition $(\alpha)$ .

    Clearly an abelian semigroup, all elements of which are idempotent, satisfies the condition $(a)$ . For instance, see Example 8, which does not satisfy the conditions $(\beta),$ $(r)$ and $(\delta)$ .

    The conditions $(a)$ and $(\beta)$ are equivalent to each other in abelian groups but not in semigroups.

    Theorem 2. If an abelian semigroup $S$ satisfies the condition $(\alpha)$, then any complete sector $T(a)$ is isomorphic with a sub-semigroup of the additive semigroup of the non-negative rational numbers.

    Proof. By Theorem 1, an element of $S$ is idempotent or of infinite order.

    (4) An element $a$ of a p.o. semigroup $S$ is called $posit\dot{w}e$ or negative, when $a^{B}\geqq a$ or $a\geqq a^{2}$ respectively. (Definition 4, 0.I.)

    (5) A partial order defined on an abelian semigroup $S$ is called normal, when the following condition is satisfied:

    $a^{n}\geqq b^{n}$ for some positive integer $n$ implies $a_{=}^{\approx}b$ . (Defimition 6, 0.I.)

  • 76 O. Nakada

    i) $a$ is idempotent. Let $x$ be any element of $T(a)$ , then $x$ is of finite order, and hence $x$ is idempotent. Therefore, $x=a$ . Then the complete sector $\mathcal{I}’(a)$ consists of only $a$ . Consider the correspondence $a\leftarrow\rightarrow 0$ .

    ii) $a$ is of infinite order. Let $x$ be any element of the complete sector $T(a)$ . Then $x$ is of infinite order and there exist positive integers $n$ and $m$ such that $x^{m}=a^{n}$ . We consider the correspondence $x-n/m$ .

    First, let $x^{m}=a^{n}$ and $x^{m^{\prime}}=a^{n^{\prime}}$ , then $x^{mn^{\prime}}=a^{nn^{\prime}}=x^{nm^{\prime}}$ . Hence $mn^{\prime}=$ $nm^{f}$ because $x$ is of infinite order. Therefore we have $n/m=n^{\prime}/m^{\prime},$ $i.e.$ , the correspondence is unique.

    Now, let $y\in T(a),$ $y-j/i$ , that is, $y^{i}=a^{j}$ . If $n/m=j/i$ , then $x^{mi}=$ $a^{ni}=a^{mj}=y^{mi}$ , and hence $x=y$ by the condition $(\alpha),$ $i.e.$, this corres- pondence is one-to-one.

    Finally, since $x^{mi}=a^{ni}$ and $y^{mi}=a^{mj}$ , we have $(xy)^{mi}=a^{ni+mj}$, there- fore, $xy\rightarrow(ni+mj)/mi=n/m+j/i$ .

    Consequently, the correspondence $x\leftarrow\rightarrow n/m$ is the isomorphism of $T(a)$ into the additive semigroup of the positive rational numbers.

    $C\sigma rolhry1$ . If an abelian semigroup $S$ satisfies the condition $(\alpha)$ , then the product cancellation law is held in a (complete) sector by any element of $S$ .

    Corollary 2. If an abelian semigroup $S$ satisfies the condition $(a)$ , then a strong linear order may be defined on a (complete) sector by any element of $S^{(7)}$ .

    Definition 4. Let $S$ be an abelian semigroup and $P$ be a partial order defined on $S$ . And let $T$ be a subset of $S$ . If any two elements of $T$ are comparable in $P$, then $T$ is called a cmnparable subset (in $P$) of $S$ , on the contrary if any two distinct elements of $T$ are non-com- parable in $P$, then $T$ is called a non-copmarable subset (in $P$) of $S$ .

    Next, let $U$ and $V$ be two disjoint subsets of $S$ . If any elements $u\in U$ and $v\in V$ are comparable in $P$, then $U$ and $V$ are called comparable subsets to each other (in $P$) of $S$ , and if any elements $u$ and $v$ are non- comparable in $P$, then they are called non-comparable subsets to each other (in $P$) of $S$ (or we say that $U$ is comparable (or non-comparable) to $V$ in $P$).

    If $U$ and $V$ are comparable subsets to each other in $P$ and $u>v$ in $P$ for any $u\in U$ and $v\in V$, then we denote $U>V$ in $P$ . $\overline{(6)}$A $p:\overline{o.}$semigroup $S$ is called strong, when the following condition is satisfied in $S$: $ac\geqq bc$ implles $a\geqq c$ . (Definition 2, 0.I.)

    (7) Cf. Theorem 15, O.I.

  • Parttally Ordered Abelian &migroups $\Pi$ 77

    If a complete sector by an element of infinite order contains a posi- tive element, then it does not contain a negative element. Accordingly, we shall call such a complete sector positive. Similarly, we may define a negative complete sector.

    Theorem 3. Let $S$ be a normal $p.0$ . semigroup. Then the positive or negative complete sector is the comparable subset of $S$ .

    Proof. Let $qa$) be a positive complete sector, where we assume that $a$ is positive without loss of generality. Then clearly the sector $S(a)$ is the comparable subset. Now let $u$ and $v$ be any two elements of the complete sector $T(\backslash a)$