Product of all elements of a finite group with an 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

If the group is not abelian, the expression $$\prod_{g\in G} g$$ makes no sense. (It makes no sense at all if the group is infinite, but I assume that it isn't...)

Indeed, take any elements $x,y\in G$ such that $xy\neq yx$ and call the other elements $$g_1,g_2,\ldots,g_r$$ Then, $$g_1g_2\cdots g_rxy\neq g_1g_2\cdots g_ryx$$ so the product of the elements of $G$ depends on the chosen order.

So I think that there are two possibilities: perhaps you must assume that the prod expression makes sense, and you must show that, then, the group must be abelian (as I have just done), or the problem is wrongly set out.

Answer 2

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.

Answer 3

This is a mistake in textbook. In errata is written that it should be commutative group.

Answer 4

In addition to all the answers, there is also a very neat answer to the general 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.