TITLE: The notation $\prod_{g\in G} g$ for a finite group not well-defined.
QUESTION [12 upvotes]: 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.
http://www.math.fsu.edu/~aluffi/algebraerrata/Errata.html

REPLY [2 votes]: 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.