论文标题
纯粹的循环基团的单数分裂
Purely singular splittings of cyclic groups
论文作者
论文摘要
让$ g $为有限的阿贝尔集团。 我们说，如果每个非零元素$ g $的$ g $ of $ g $的$ m $和$ s $ a \ textsl {spriptsl {sprapting}的$ g $的独特表示，则$ g = ms $，in m $ in m $ in m $，而$ 0 $ in s $，而$ 0 $则没有这样的表示。如果对于每个Prime Divisor $ p $ $ | g | $，则称为\ textit {纯singular}，至少有$ m $的一个元素可除以$ p $。 在本文中，我们主要研究循环基团的纯粹分裂。我们首先证明，如果$ k \ ge3 $是一个正整数，以至于$ [-k+1，\，k]^*$拆分一个环状组$ \ mathbb {z} _m $，则$ m = 2k $。接下来，我们有以下一般结果。假设$ m = [ - k_1，\，k_2]^*$ splits $ \ mathbb {z} _ {n（k_1+k_2）+1} $，$ 1 \ leq k_1 <k_2 $。如果$ n \ geq 2 $，则$ k_1 \ leq n-2 $和$ k_2 \ leq 2n-5 $。应用此结果，我们证明，如果$ m = [ - k_1，\，k_2]^*$ spling $ \ mathbb {z} _m $纯粹是纯粹的，而$（i）$ $ $ \ gcd（s，\，\，m）= 1 $ in s $ s $ s $ s $或$ $ $ m = 2^2^2^2^^^^$或$ 2^^^$或$α\ geq 0 $，$β\ geq 1 $和$ p $，$ p_1 $，$ p_2 $奇数，然后$ m = k_1+k_2+k_2+1 $或$ k_1 = 0 $和$ m = k_2+1 $或$ 2k_2+1 $。
Let $G$ be a finite abelian group. We say that $M$ and $S$ form a \textsl{splitting} of $G$ if every nonzero element $g$ of $G$ has a unique representation of the form $g=ms$ with $m\in M$ and $s\in S$, while $0$ has no such representation. The splitting is called \textit{purely singular} if for each prime divisor $p$ of $|G|$, there is at least one element of $M$ is divisible by $p$. In this paper, we mainly study the purely singular splittings of cyclic groups. We first prove that if $k\ge3$ is a positive integer such that $[-k+1, \,k]^*$ splits a cyclic group $\mathbb{Z}_m$, then $m=2k$. Next, we have the following general result. Suppose $M=[-k_1, \,k_2]^*$ splits $\mathbb{Z}_{n(k_1+k_2)+1}$ with $1\leq k_1< k_2$. If $n\geq 2$, then $k_1\leq n-2$ and $k_2\leq 2n-5$. Applying this result, we prove that if $M=[-k_1, \,k_2]^*$ splits $\mathbb{Z}_m$ purely singularly, and either $(i)$ $\gcd(s, \,m)=1$ for all $s\in S$ or $(ii)$ $m=2^αp^β$ or $2^αp_1p_2$ with $α\geq 0$, $β\geq 1$ and $p$, $p_1$, $p_2$ odd primes, then $m=k_1+k_2+1$ or $k_1=0$ and $m=k_2+1$ or $2k_2+1$.
