Abstract Algebra – Product of All Elements of a Finite Group with a Unique Element of Order 2

Question

I have the following problem from Aluffi's Algebra:
given a finite group $G$ with an unique element $f$ of order $2$, show that
\begin{equation}
\prod_{g\in G}g=f
\end{equation}
My reasoning is the following. Since $f$ is the unique element of order $2$, for each $g\in G$ such that $g\neq e$ and $g\neq f$, it must be that $g^{-1}\neq g$. So we can concoct a particular product coupling the elements with their inverses:
\begin{equation}
e\cdot f\cdot (g_1\cdot g_1^{-1})\cdot(g_2\cdot g_2^{-1})\cdots (g_n\cdot g_n^{-1})
\end{equation}
which is really just $f$ in the end.
My question is: how can I show that the order of the factors in this product really does not matter? I cannot see a way out.

Answer 1

As noted above, the expression $\prod_{g\in G}g$ does not make sense. To give a concrete example, consider the quaternionic group $G=Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ (under quaternion multiplication). It has only one element of order two, namely $-1$. Now $$ 1\cdot(-1)\cdot i\cdot(-i)\cdot j\cdot(-j)\cdot k\cdot(-k)=-1 $$ but $$ 1\cdot(-1)\cdot i\cdot j\cdot(-i)\cdot(-j)\cdot k\cdot(-k)=1. $$ Therefore one indeed has to specify an order for the product. It is true that there is an order so that the result holds, but it doesn't hold for all orders.