TITLE: Product of all elements in finite group
QUESTION [16 upvotes]: Question: If $G$ is a finite group such that the product of its elements (each chosen only once) is always $1$, independent of the ordering in the product, what can we say about $G$?
I was trying to conclude that $G$ must be abelian (I am not sure about this). I proceed as follows: if $|G|\leq 4$, then obviously, $G$ is abelian. Suppose $|G|>4$. Then consider products of all elements in the order $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$. This, being identity, implies that $a^{-1}b^{-1}ab=1$, i.e. $G$ is abelian.
BUT, the problem is in $a^{-1}b^{-1}ab$, the elements $a$ and $a^{-1}$ should be distinct, and similarly for $b$. Otherwise, the product $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$ is not in the prescribed form as in question.
Also, note that the cyclic group of order $4$ doesn't satisfy that $abcd=1$ where $\{a,b,c,d\}\cong \mathbb{Z}_4$.

REPLY [11 votes]: 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.