Why is $\prod_{a\in G}a = e_G$ if there exist several elements with $\operatorname{ord}(u) = 2$? 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.
 A: You almost solved the problem.
Hints.
Denote by $A$ the set of all elements of order 2. Then $A$ is a subgroup of $G$ with each element of order 2. The group $A$ can be viewed as a vector space over a field of two elements.  By hypothesis the dimension of this vector space are greater than 1. It remains to prove that the sum of all vectors of such space is 0.
A: 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?
A: Since $G$ is abelian, the product of two elements of order $2$ is also of order $2$ (unless the two elements are the same when it's $e$), so the elements of order $2$, together with $e$ form a subgroup $S$ and $$\prod_{s\in S}s=\prod_{g\in G}g$$
Consider a minimal set of generators $s_1,\dots s_k$ of $S$.  Then each $s\in S$ is of the form $s=s_1^{a_1}s_2^{a_2\cdots}s_k^{a_k}$ where $a_i\in\{0,1\}$, for $i=1,\dots,k$.  Then $s_i$ occurs $2^{k-1}$ times in $${\prod_{s\in S}s}$$
A: The elements of $G$ of order 2 together with $e$ form a group $H$. So to finish solving, it suffices to show that $\left(\prod_{a \in H} a \right)= e$ for any abelian group $H$ of order at least 4 where every element besides $e$ has order 2.
HINT: Let $H$ be an abelian group of order at least 4 where every element besides $e$ has order 2.
If $|H|=4$ then $H$ can be written $H=\{e,a_1,a_2, a_1a_2\}$ where $a_1,a_2$ generate $H$, and it is easy to see that $\prod_{a \in H} a = e$.
Otherwise let us assume that $|H|=2^n$ for some $n \ge 3$, and let $H'$ be a subgroup of $H'$ of index 2 in $H$, so $|H'|=2^{n-1} \ge 4$, and let $a_0$ be an element in $H \setminus H'$. Then let us assume by induction that $\left(\prod_{a \in H'}a\right) = e$.
Then $H\setminus H'=\{a_0a; a \in H'\}$ and $|H \setminus H'| =2^{n-1}$ and thus:
$$\prod_{a \in H}a \  = \ \left(\prod_{a \in H'}a \right) \times \left(\prod_{a \in H'}a_0a\right) = a_0^{|H'|} \times \left(\prod_{a \in H'} a \right)^2= e \times e^2 = e$$
this following because $H$ is abelian and $a_0$ has order 2 and $|H'|$ is even.
