The notation $\prod_{g\in G} g$ for a finite group not well-defined. 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
 A: 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 $x \sim y \Leftrightarrow x=y \vee x=y^{-1}$.
A: 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.
A: 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.
A: 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 $-1$.
A: 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
