Product of all elements in a finite, boolean group So I've seen similar questions here that are solved using heavy weapons, such as Wilson's theorem, but my question is about a proof I saw that I don't quite understand.
So suppose $G$ is a finite, abelian group, $|G| \geq 3$, and every nonidentity element has order 2. We want to prove that the product of all the elements of $G$ is the identity.
Here's the construction they offer:
Take two distinct nonidentity elements $a$ and $b$ (which exist because $|G| \geq 3$), and form the set $H_1 = \{e, a, b, ab\}$. Then $H_1$ is a subgroup and the product of its elements is $e$. If $H_1 = G$, then we are done. Otherwise, there is some element $c$ in $G$ that has not yet been used. Form the subgroup $H_2 = H \, \cup cH_1 = \{e, a, b, ab, c, ca, cb, cab\}$. If we have exhausted all the elements, great. Otherwise continue in this manner, at each stage forming a subgroup $H_k$ in which the product of all its elements is $e$.
The question I have is the following: how can one show definitively, without making a huge multiplication table, that at each step $H_k$ is a subgroup? And how do you convince yourself that the product of all the elements of $H_k$ is $e$?
 A: Suppose $H$ is a subgroup of $G$, and $c\in G$. Since $G$ is abelian, the subgroup $\langle H,c\rangle$, which in general is the set of all finite products of elements of $H\cup \{c\}$ and its inverses, will just reduce to products of an element of $H$ and a power of $c$. Since $c$ has order $2$, the power of $c$ is either trivial or equal to $c$. Thus, every element of $\langle H,c\rangle$ is of the form $h$ or $ch$, with $h\in H$. That is, $\langle H,c\rangle = H\cup cH$. 
When $c\notin H$, $cH\cap H=\varnothing$, so this is a disjoint union. Otherwise, $cH=H$ and you just get $H$ back.
More generally, if $G$ is abelian, $H$ is a subgroup, and $g\in G$ has order $n$, then $\langle H,g\rangle = H\cup gH\cup g^2H\cup\cdots\cup g^{n-1}H$. (This also holds if $g$ centralizes $H$, regardless of whether $G$ is abelian).
As to why the product of the elements of $H_k$ is trivial, suppose those of $H_k$ are known to be trivial. Note that $|H_k|$ is even. Then $H_{k+1}$ consists of the elements of $H_k$ (whose product is trivial), and the elements of $cH_{k}$. The product of the latter elements is equal to $c^{|H_k|}\prod_{h\in H_k}h$. The product of the elements of $H_k$ is already known to be trivial, and since $|H_k|$ is even, the factor $c^{|H_k|}$ is also trivial. Thus, the product of the elements of $cH_k$ is trivial, and hence so is the product of all elements of $H_{k+1}$. 
A: Let the integer $k$ be greater than $1$, and regard $G$ additively, as a $k$-dimensional vectorspace over $\mathbb{F}_2=\{0, 1\}$. Write 0 for the zero vector and 1 for the $k$-tupel $(1, 1,\cdots, 1)$. With each vector $v \in G$, we associate the vector $1 − v$, when
summing all vectors, i.e. all elements of $G$. Since there are $2^{k−1}$ such pairs, the sum equals $2^{k−1} \cdot$ 1 $\equiv$ 0 (mod $2$).
