关于Abel群的一个等和问题

An equal sum question about Abel group

设G是一个Abel加群,假设G不含奇数阶元,我们证明了如果a1,a2,…,a2k+1是由G的元构成的一个序列满足下面性质:从2k+1项中任去掉一项,其余2k项总可分成和相等的两组,每组k项,则必有a1=a2=…=a2k+1。当G含奇数阶元时,我们举例说明上述结论不成立。

Set up G as an Abel add group,suppose G doesn't include step unit of odd number,we have identified if a1,a2,...,a(2k+1) is an array formed by the G unit,which meets the nature followed:remove any item from 2k+1,the other 2k items can be always divided into two groups with the same sum,there must be a1=a2=...=a(2k+1) in each group of K item.When G includes the step unit of odd number,it proves above-mentioned conclusions are untenable.