Why is $\prod_{a\in G}a = e_G$ if there exist several elements with $\operatorname{ord}(u) = 2$

abelian-groupsabstract-algebragroup-theory

I'm currently reading Algebra by Karpfinger, Christian, and Meyberg, Kurt. There I found the following exercise:

Prove that for any abelian finite group $G$ it holds true that: If $G$ has more than one element $u$ with $\operatorname{ord}(u) = 2$ then: $$\prod_{a\in G}a = e_G$$ where $e_G$ is the neutral element in $G$.

Idea: Since $G$ is abelian we can rewrite the product $\prod_{a\in G}a = a_1\cdot a_2 \cdot \ldots$ by placing any $a_i$ next to its inverse and they'll cancel out. Then we are only left with all selfinverse elements $u$. Further I know that there must be an odd number of those elements $u$, because:

  • the number of elements with order greater than $2$ is even
  • there is one neutral element
  • the order of $G$ is even, because of Lagrange's theorem

Question: I'd like to get a hint on how to solve this. Is my progress until now correct or leading to the solution? In particular, I don't understand how we get the neutral element through the multiplication of the selfinverse elements, because I have no intuition on how they are connected.

Best Answer

Hint: You are doing great. Now, get all the elements of order $2$, notice that they form a group when you add $e$ to them. Moreover, this group should contain a subset of elements (why?), say $X$, that generate the group. The group, then, is obtained by doing all possibilities of elements on $X$. Notice that every element on $X$ is in $2^{n-1}$ elements of the group where $|X|=n$. What happens, then, when you multiply them all together?

Related Question