Product of all elements in finite group

Solution 1:

(as per my earlier comment): let $a$ and $b$ be arbitrary elements of your group and let $P$ be the product of all the other elements, in any order. Then, by assumption, $abP = baP$ so $ab = ba$.

Solution 2:

The group must be Abelian.

Fix an ordering of the elements of $G$: $g_1, g_2, \dots, g_n$.

Then $g_1 g_2 g_3 \cdots g_n = 1 = g_2 g_1 g_3 \cdots g_n$

Now cancel $g=g_3 \cdots g_n$ to get $g_1 g_2 = g_2 g_1$.

You don't need that the product is always $1$, just that it does not depend on the order. Moreover, this is the right hypothesis because of Wilson's theorem for finite Abelian groups:

The product of all elements in a finite Abelian group is either $1$ or the element of order $2$ if there is only one such element.

Solution 3:

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.