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I want to solve the following exercise.

Let $G$ be a finite group, with exactly one element $f$ of order $2$. Prove that $\prod_{g\in G} g = f$.

I have a question regarding the notation, the expression $\prod_{g\in G} g$ stands for the product over all elements of $G$? But then it is not well-defined for non-abelian group, cause the order in which this product is evaluated matters? Or for what else can $\prod_{g\in G} g$ stand?

EDIT: I have taken the exercise from an algebra book, and I just looked up the homepage of the author and found an errata. And indeed, the exercise is wrongly stated! The group must be assumed to be abelian.

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+1: Even if the group is abelian, is the product of all its elements well defined? – Manos Feb 9 '13 at 22:28
@Manos: Yes, an inductive definition is used. – Martin Brandenburg Feb 9 '13 at 22:32
You may be interested in Wilson's Theorem, which asserts that if $p$ is prime then $(p-1)!\equiv -1\pmod{p}$. That is an important special case of the result you are proving. – André Nicolas Feb 9 '13 at 22:35
btw, this is a special case of – Martin Brandenburg Feb 10 '13 at 1:36
possible duplicate of Groups with only one element of order 2 – Code-Guru Feb 2 '14 at 22:20

5 Answers 5

up vote 5 down vote accepted

The statement is wrong, not well-defined, and in fact one should assume that $G$ is abelian. Then the proof is easy by using the equivalence relation $x \sim y \Leftrightarrow x=y \vee x=y^{-1}$.

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$@$Martin: could you elaborate a bit more on the proof you have in mind? – Pete L. Clark Feb 10 '13 at 0:54
Equivalent non-equal elements cancel in the product. The only equivalence classes with one element are $\{1\}$ and $\{f\}$. – Martin Brandenburg Feb 10 '13 at 1:34
Okay, fair enough. I guess I was thinking of the converse part of the result: in a finite commutative group which does not have exactly one element of order $2$, the product of all the elements is the identity. This is slightly harder... – Pete L. Clark Feb 10 '13 at 2:28

One might interpret the statement as saying that the product of all elements taken in any order gives the unique element of order $2$. But this fails for the quaternion group, which has $-1$ as unique element of order $2$, while the product $1ijk(-1)(-i)(-j)(-k)$ equals $1$, not $-1$.

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Moreover, the statement is false for any non-abelian group: if the value of $\prod_{g\in G} g$ doesn't depend on the order, then $G$ is abelian. – Yury Feb 9 '13 at 23:03
@Yuri Interesting, can you provide a short proof of this fact? I am lazy now... – Matemáticos Chibchas Feb 10 '13 at 0:11
@MatemáticosChibchas If $G=\{a,b\}\sqcup H$ with $a\ne b$, the assumption is that $ab\prod_{g\in H}g=ba\prod_{g\in H}g$, so $ab=ba$. By the way, surely we know each other, right? I'm curious. – Andrés Caicedo Feb 10 '13 at 0:40
@AndresCaicedo Thanks for the answer. Your reputation (in Colombia) precedes you, so in particular I know who you are. Regarding me, this is my answer (watch from 4:24 to 4:30) – Matemáticos Chibchas Feb 10 '13 at 0:53
Just a comment: because the quaternion group is the smallest nonabelian group with a unique element of order 2, it gives a minimal counterexample to the OP's original claim. – Pete L. Clark Feb 10 '13 at 0:56

Indeed, the group needs to be commutative for this result to hold. It might be nice for someone to post a minimal counterexample! (Added: never mind: Marc van Leeuwen's answer and the comments below it take care of this.)

Because of a request from a colleague, a little while back I had the occasion to write down an excruciatingly elementary proof of this fact. See here.

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In addition to all the answers, there is also a very neat answer to the 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. The theorem connects to the theory of Latin Squares and so-called complete maps.

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This is indeed a beautiful generalization. I think I learned it from you on this site before, but I was struggling to remember the exact statement, so thanks for bringing it up again here. – Pete L. Clark Feb 11 '13 at 16:48

I have taken the exercise from an algebra book, and I just looked up the homepage of the author and found an errata. And indeed, the exercise is wrongly stated! The group must be assumed to be abelian.

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Please make this a comment to the question instead of an answer. Better yet, edit it in to the question with a note that it has been added. – Ross Millikan Feb 10 '13 at 1:01

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