拉丁bitrades、等边三角形和阿贝尔群的剖析
Latin bitrades, dissections of equilateral triangles and abelian groups
摘要：
令$T = (T^{\textstyle \ast}, T^{\scriptscriptstyle \triangle})$为球形拉丁比特率。每个$a=(a_1,a_2,a_3)\in T^{\textstyle \ast}$与一组形式为$b_1+b_2=b_3$的线性方程$\eq(T,a)$相关联，其中$b = (b_1,b_2,b_3)$贯穿$T^{\textstyle \ast} \setminus \{a\}$。假设$a_1 = 0 = a_2$和$a_3 = 1$。那么$\eq(T,a)$在有理数中有唯一解$b_i = \bar b_i$。假设对于所有$b,c \in T^{\textstyle \ast}$，$\bar b_i \ne \bar c_i$使得$b_i \ne c_i$和$i \in \{1,2,3\}$。我们证明，然后$T^{\scriptscriptstyle \triangle}$可以被解释为一个等边三角形的解剖。我们还考虑拉丁bitrades组修改，并表明，用于生成的解剖的方法可以用于证明，$T^{\textstyle \ast}$可以嵌入到一个有限的阿贝尔群的操作表，为每个球形拉丁bitrade $T$。
英文摘要：
Let $T = (T^{\textstyle \ast}, T^{\scriptscriptstyle \triangle})$ be a spherical latin bitrade. With each $a=(a_1,a_2,a_3)\in T^{\textstyle \ast}$ associate a set of linear equations $\eq(T,a)$ of the form $b_1+b_2=b_3$, where $b = (b_1,b_2,b_3)$ runs through $T^{\textstyle \ast} \setminus \{a\}$. Assume $a_1 = 0 = a_2$ and $a_3 = 1$. Then $\eq(T,a)$ has in rational numbers a unique solution $b_i = \bar b_i$. Suppose that $\bar b_i \ne \bar c_i$ for all $b,c \in T^{\textstyle \ast}$ such that $b_i \ne c_i$ and $i \in \{1,2,3\}$. We prove that then $T^{\scriptscriptstyle \triangle}$ can be interpreted as a dissection of an equilateral triangle. We also consider group modifications of latin bitrades and show that the methods for generating the dissections can be used for a proof that $T^{\textstyle \ast}$ can be embedded into the operational table of a finite abelian group, for every spherical latin bitrade $T$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
