# 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 $$e$$?

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 $$H_{k+1}$$.
• Should the first instance of $H_k$ in the last paragaraph be $H_{k + 1}$? Also, I agree that if $c \not \in H$ then all the elements $ch$ are not in $h$ and distinct from each other; but why is $H \cup cH$ a subgroup? Sep 30, 2019 at 3:06
• No; I’m doing an induction argument. The statement to be proven is “The product of the elements of $H_k$ is trivial.” So, assuming that the result holds for $k$, I show it holds for $H_{k+1}$. As to your second question, I prove that in the first paragraph. What are you not seeing? Sep 30, 2019 at 3:11
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 $$2$$).