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 $\{a,b,c,d\}\cong \mathbb{Z}_4$.
Best Answer
(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 $ab = ba$.