In Aluffi's Algebra Chapter 0, there's an error in the 2009 publication that took me a while to find:
Let $G$ be a finite abelian group with exactly one element $f$ of order 2. Prove that $\prod_{g\in G}g = f$.
With the addition of the abelian condition, the problem is much simpler. However, I could not find a counterexample or proof for the non-abelian case. How can I go about proving/finding a counterexample to the statement without the abelian condition?
Consider the quaternion group of order $8$, $Q_8$.
That has a unique element of order $2$, namely $-1$.
Of course there is an ordering for the product that gives the desired result but we could also have the ordering:
$$-1\times i\times j\times -i\times k\times -k\times -j=1$$