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 $e$?


2 Answers 2


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 $H_{k+1}$.

  • $\begingroup$ 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? $\endgroup$
    – Junglemath
    Sep 30, 2019 at 3:06
  • $\begingroup$ 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? $\endgroup$ 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 $2$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.