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Clearly for $G \approx \Bbb{Z}_2$ this is not true since $G = \{1, a\}$ and so the product equals $a$. Was wondering what sufficient conditions are such that the product of all the group elements amounts to $1$.

For $G \approx \Bbb{Z}_3$ it's true since $G = \{1, a, b\}$ with $ab = 1$. I'm a little lost as to how to proceed.

I think if each element pairs with an inverse distinct from it, then it's true that the product equals $1$. But is there another way to state that, and is that a neccessary condition as well?

In addition, is it possible to cover all finite abelian groups by taking the product of the squares of all elements?

Daniel Donnelly
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    Are you assuming your group is abelian? Because if not, then it is not clear what “the product of all elements” means. What is the product of all elements of the dihedral group? In what order? – Arturo Magidin Aug 31 '19 at 03:38
  • @ArturoMagidin you're right. Let's say that $G$ is indeed abelian. – Daniel Donnelly Aug 31 '19 at 03:38
  • For the nonabelian case see this question. – nomadictype Aug 31 '19 at 03:41
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    If you take the product of the squares, that’s the same as squaring the product of all elements. Since the product of all elements is either trivial or of order $2$, the square will always be trivial. If you mean, first square every element, and then take the product of all the resulting elements without multiplicity, then you are taking the product of all elements in $G^2$ and the original result applies: if $G^2$ has a unique element of order $2$ then you get that element, otherwise the product is trivial. – Arturo Magidin Aug 31 '19 at 03:48
  • @ArturoMagidin thank you! I am excited to move onto the kernel of that hom, in my other post. – Daniel Donnelly Aug 31 '19 at 03:49
  • A very neat 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 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 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. – Nicky Hekster Sep 01 '19 at 13:35

3 Answers3

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Wilson's theorem for finite Abelian groups:

The product of all elements in a finite Abelian group is either $1$ or the element of order $2$ if there is only one such element.

lhf
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  • Would that imply that the product is then equal to the product of all order 2 elements, or 1 if there are none? – Daniel Donnelly Aug 31 '19 at 03:41
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    @ShineOnYouCrazyDiamond: Yes. Because all other elements cancel out with their inverses. But this product could be trivial, e.g., in the Klein $4$-group. – Arturo Magidin Aug 31 '19 at 03:41
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    @ShineOnYouCrazyDiamond: It says even more: as soon as there are two or more elements of order 2, the product of all elements is 1. So the only case where the product is not 1 is when there is exactly one element of order 2. – nomadictype Aug 31 '19 at 03:43
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    @ShineOnYouCrazyDiamond: Right; what this says is: if $G$ has exactly one element of order $2$, then the product will equal that element. Otherwise, the product will be trivial. – Arturo Magidin Aug 31 '19 at 03:45
  • See my edit to the question. What if we decided to square all elements then take product? I assume that because of your answer, that it is indeed always 1. – Daniel Donnelly Aug 31 '19 at 03:46
  • @ShineOnYouCrazyDiamond, Wilson's theorem implies that the product of the squares is always 1 – lhf Aug 31 '19 at 03:56
  • You used to meticulously link earlier incarnations of this question to the earliest ones. Why the change of heart here? – Jyrki Lahtonen Aug 31 '19 at 05:09
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If the order of $G$ is odd and $G$ is abelian. Then no element in $G$ except identity is self inverse. So, in this case the product of all elements of $G$ turn out to be $1$.

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Regarding the question about taking the product of the squares of all elements: Yes, this product is 1 for every finite abelian group, because as per lhf's answer, if the product of all elements is not 1 then it is $x$ for the only element $x$ of order 2, whose square is $1$.

nomadictype
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