Let $G$ be an abelian group of order $n$, and $a_1,a_2,...a_n$ its elements. If $G$ has more than one nontrivial elements with order $2$, how to show that $\prod^n_{x=1}a_x=1$?
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pxc3110
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$\newcommand{\ZZ}{\mathbb{Z}}$ This follows from the structure theorem for finite abelian groups. Namely, $G$ must have the form $$\ZZ/m_1\ZZ\oplus\cdots\oplus\ZZ/m_n\ZZ$$ When given such a group, checking to see if the sum of every element of the group is 0, amounts to checking that each coordinate of the sum is 0, ie, you want to check that for every $i\in\{1,\ldots,n\}$, $$(0 + 1 + 2 + \cdots + m_i)\prod_{j\ne i}m_j \equiv 0\mod m_i$$ or equivalently, if for every $i$, $$\frac{m_i(m_i+1)}{2}\prod_{j\ne i}m_j \equiv 0\mod m_i$$ Here the product counts the number of times you have to sum over the possible values of that coordinate.
The congruence would easily be true if it weren't for the 2 in the denominator. However, the fact that you have at least two elements of order 2 means that for any $i$, there is a $j\ne i$ such that $m_j$ is even, so the product $\prod_{j\ne i}m_j$ is even, which cancels out the 2 in the denominator, and the congruence above is true.
oxeimon
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