How to determine $\prod_{g\in G}g$ in all generality

Question

This question is based on another question that is closed as a duplicate:How to determine $\prod_{g\in G}g$?, which was in the reopening queue but is removed again, so I decided to ask myself. ("This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.") So the question was

Let $G$ be a finite Abelian group, then determine $\prod\limits_{g\in G}g.$

This question has answers in the original post and here. I reasoned as follows:

• If there is no element of order 2, every element and its inverse appear in the product, and the identity $e$ appears once, such that the product equals $e$. By Cauchy's theorem, this is the case when $\mathrm{order}(G)$ is odd.

• If $\mathrm{order}(G)$ is even, again by Cauchy's theorem, there is an element of order $2$. Suppose there are $k$ elements of order $2$ and denote these by $g_1,g_2,\ldots,g_k$. Then, because $G$ is abelian by assumption, $\{e,g_1,g_2,\ldots,g_k\}\subset G$ is a subgroup (verify this). Then by Cauchy's theorem again, the order of this subgroup must be even, thus $k$ is odd. If we write out $\prod_{g\in G}g$ now, we observe that for all $g\not\in\{e,g_1,g_2,\ldots,g_k\}$, both the element and its inverse appear exactly once in the product, thereby yielding the identity element. Thus the product reduces to $\prod_{i=1}^{k}g_i$. For $k=1$, it's simple, $\prod_{g\in G}g=g_1$, for $k=3$ also: $\prod_{g\in G}g=g_1\circ g_2\circ g_3=g_3^2=e$, since $g_1\circ g_2\not\in \{e,g_1,g_2\}$. For any $k>3$ (the general case) the product reduces to exactly one element in $\{e,g_1,g_2,\ldots,g_k\}$ but to be honest I don't see how to infer anything about this.

So this is not a duplicate of the linked questions, the way I see it, because you are asked to determine the product in all generality. Maybe someone else knows how to calculate the product for $k>3$.

Question: for the case described above, how do I determine $\prod\limits_{g\in G}g$? This case is not treated in the other answers. Obviously the product reduces to exactly one element in $\{e,g_1,g_2,\ldots,g_k\}$, but can we say for which values of $k$ the product equals $e$ and when it equals a non-identity element from $\{e,g_1,g_2,\ldots,g_k\}$?

And please, don't close again unless there is a good reason. Also, if I'm missing something and I'm asking something that is totally trivial, please explain! Thanks in advance!

Answer 1

In the general case you always get the identity. First note that the elements of order $2$ don't just have the structure of an abelian group, but also of a vector space over $\mathbb{F}_2$.

You are then just asking what the sum of every element in a finite dimensional vector space over $\mathbb{F}_2$ is. The answer is of $0$ unless the dimension is $1$ in which case it's the nonzero element. To see this either use induction starting at the 2-dimensional case or just note that the answer must be invariant under the action of $GL_n(\mathbb{F}_2)$, which acts transitively on the non-zero vectors.

Answer 2

In addition to all the answers, there is also a very neat, but non-trivial answer to the general question What is the set of all different products of all the elements of a finite group $G$? So $G$ not necessarily being abelian.

Well, if a $2$-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$.

If a $2$-Sylow subgroup of $G$ is cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution (=element of order $2$) of a $2$-Sylow subgroup.

See also J. Dénes and P. Hermann, `On the product of all elements in a finite group', Ann. Discrete Math. 15 (1982) 105-109. The theorem connects to the theory of Latin Squares and so-called complete maps.

Answer 3

Here's a proof based on the fundamental theorem of finitely generated abelian groups. Because you're using abelian groups, I will use the additive notation.

Any finite abelian group $G$ is isomorphic to a direct sum of the form $$ G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \mathbb{Z}_{p_2^{k_2}} \oplus \cdots \oplus \mathbb{Z}_{p_n^{k_n}}$$ An element $x = (x_1, x_2, \dots, x_n)$ has order $2$ if and only if for every $i$, $x_i$ is either the neutral element $0$ or has order $2$ in $\mathbb{Z}_{p_i^{k_i}}$. The latter case is only possible when $p_i = 2$, in which case $x_i = 2^{k_i-1}$.

In other words, we only care about those direct summands with $p_i = 2$, so let us ignore the other summands: $$ G \cong \mathbb{Z}_{2^{k_1}} \oplus \mathbb{Z}_{2^{k_2}} \oplus \cdots \oplus \mathbb{Z}_{2^{k_n}}$$

There are $2^n$ elements $x$ with order $2$: exactly half ($2^{n-1}$) of those have the $i$-th component equal to $0$, and the other half have the $i$-th component equal to $2^{k_i-1}$ (this is some easy combinatorics).

So the sum $y$ of all $2^n$ elements of order $2$ is $$y = \sum_{x^2 = 1} x = \left( 2^{n-1} \cdot 2^{k_1-1}, 2^{n-1} \cdot 2^{k_2-1}, \ldots, 2^{n-1} \cdot 2^{k_n-1}\right)$$ If $n \geq 2$, then $2^{n-1} \cdot 2^{k_i-1} = 2^{n-2} \cdot 2^{k_i} \equiv 0$ in $\mathbb{Z}_{2^{k_i}}$.

Answer 4

I have a new idea. I’d like to write more for the sake of completeness, although it might be quite a simple one to think. Please have a look.

Suppose $G$ to be a finite Abelian group.

$$G_2:=\{x\in G\mid |x|=2\}\cup \{1\}$$

It’s easy to show $G_2$ is a subgroup of $G$; and obviously $G_2$ is a $2$-group, hence $G_2\cong C_2\times \cdots\times C_2$.

Therefore, for some $n\in\mathbb{N}_+$, $G_2=\langle a_1\rangle\times\cdots\times \langle a_n\rangle$, where $a_i\in G_2$, $i=1,...,n$.

Then an element $a$ is in $G_2$ iff it can be written in a unique way as product of $1$ and $a_i$ for some $i\in\{1,...,n\}$ (not repeatedly).

Thus $\prod_{a\in G_2}a$ is actually product of $a_i$ (repeatable). We can count exactly that how many times $a_i$ repeats; and that is $$\binom{n-1}{0}+ \binom{n-1}{1}+ \cdots+ \binom{n-1}{n-1}=2^{n-1}=\left\{\begin{array}{ll} 1,&n=1;\\\text{an even number}, &n>1.\end{array}\right. $$

Since $G_2\leq G$ is Abelian, we are done.