# Spring 2018, problem 63

For which commutative finite groups is the product of all elements equal to the unit element?

Claudio
3 years ago

Claim The condition holds true for all groups but two cases:

(a) the cyclic groups of even order, $\mathbb Z\diagup(2p)\mathbb Z$; with additive notations the sum of all elements is $p$.

(b) the direct sum of cyclic groups, all but one having odd order

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The easiest examples and counterexamples are given by the cyclic groups $\mathbb Z\diagup n\mathbb Z$, that satisfy the wanted property if and only if $n$ is odd. In general, denoting by $n$ the order of the group:

(i) because of the Klein-four group, the oddness of $n$ is not a necessary condition;

(ii) however, as we will show later on, the oddness of $n$ is a sufficient condition.

In order to find a general necessary and sufficient condition, let us fix some notations. $\mathbb G$ will denote our finite commutative group, and $e$ will be its unit element. We will call special any $s\in\mathbb G$ such that $s*s=e$; $\mathbb S$ will denote the family of special elements. We want to prove:

Theorem $\mathbb S$ is a subgroup, whose order $k$ can be 1,2,4,8,... The product of all elements of $\mathbb G$ coincides with the product of all elements of $\mathbb S$; and such product is $e$ if and only if $k\neq2$.

The proof will be splitted in many parts.

The group-structure of $\mathbb S$ is obvious; its order will be evaluated later on. In the product $\prod\limits_{g\in\mathbb G}g$ we can suppress any couple of factors like $g*g^{-1}$ provided $g\not\in\mathbb S$; thus the product reduces to $\prod\limits_{g\in\mathbb S}g$. Of course the elements of $\mathbb S$ can not be coupled because $g^{-1}\equiv g$.

Concerning the order $k$ of $\mathbb S$, we already encountered examples with $k=1$ (say $\mathbb S=\{e\}$; this appens e.g. for $\mathbb G=\mathbb Z\diagup n\mathbb Z$ with odd $n$) and examples with $k=2$, as in $\mathbb Z\diagup(2p)\mathbb Z$ where, with additive notations, $\mathbb S=\{0,p\}$.

Now let us assume $k>2$; if $e,a,b$ are elements of $\mathbb S$, also $a*b\in\mathbb S$; and it is a fourth element because it is obviously different from $e,a,b$. Remark that the product of these four elements is $e$; in fact each non-unit element appears twice in the product. Now we work in a similar way: if a fifth element $c$ belongs to $\mathbb S$, the further elements $a*c,b*c,a*b*c$ are all special and different from the previous four $a,b,a*b,c$; thus the order could be 8. Remark that again the product of the 8 elements is $e$, because each non-unit element appears in the product an even number of times.

A trivial induction argument now suffices for the proof of the Theorem; in particular: $$\left\{\prod\limits_{g\in\mathbb G}g=e\right\}\ \Longleftrightarrow\ \left\{k\neq2\right\}$$

Final remark Till now we just used elementary results about groups. However, by using the Kronecker decomposition theorem, we can prove that the condition $k=2$ holds true if and only if the group is a cyclic group of even order or, more generally, a direct sum of cyclic groups having all but one odd order.

naomievans
1 year ago

Let G be the ultimate Abelian band. Well, then: 1) a work of all elements of group G, the orders of which are different from 2, equal to a single item; 2) if group G contains a part of order 2, the work of all elements of group G shall be equal to the work of all elements of rule 2 of group G. Proof. If e≠x ∈ G, then O(x) = 2 then and only when x=x^(-1). If O(x) > 2, then O(x^(-1)) = O(x) > 2, then x ≠ x-1. Since G is an Abel band, then: