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Let $G$ be a finite group. By indexing $G = \{g_1,\ldots,g_n\}$ arbitrarily, we can make sense of the product $$ \prod_{i = 1}^n g_i. $$ If $G$ is abelian, the result is clearly the product of all elements of order 2, which is $1$ unless there is a unique element of order 2. If $G$ is not abelian, the ordering of the group (or, the product) matters, and there are at least two possible outcomes (just exchange two adjacent non-commuting elements). Can something be said in general about which outcomes are possible?

Mees de Vries
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  • Out of curiosity, must there always be two adjacent, non-commuting elements? – Arthur Sep 07 '16 at 20:32
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    @Arthur, if the group is non-abelian, then there is an ordering of the elements in which there are always two adjacent, non-commuting elements. Taking that ordering and the one with the elements exchanged gives two different ones. – Mees de Vries Sep 07 '16 at 20:33
  • There are no more than $n$ outcomes, but there are $n!$ possible arrangements. Quite interesting. – ajotatxe Sep 07 '16 at 20:35
  • Ahh, yes, of course. I was rather thinking about a different question: can you find an ordering so that any two adjacent elements commute? It's possible for some groups (take $S_3\times \Bbb Z_n$ for some large $n$), but is it always possible? – Arthur Sep 07 '16 at 20:38
  • @Arthur, that is also a good question, but perhaps better for a new post than the comment section of this one. :-) Edit: it is certainly not possible for $S_3$, because the 2-cycles commute only with the identity. – Mees de Vries Sep 07 '16 at 20:39
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    The related question Product of all elements in finite group has an answer (by Nicky Heckster) for this one, too. – ypercubeᵀᴹ Sep 07 '16 at 20:59
  • @ypercubeᵀᴹ, thank you! I had seen the question but not read beyond the accepted answer. What is protocol here? – Mees de Vries Sep 07 '16 at 21:24
  • Not sure. You could either write your own answer or invite NickyHeckster (comment in his answer) to provide his answer here. His answer is certainly relevant here and not exactly perfect fit there. One other option would be to close this as duplicate but I don't like it, as it is not really a duplicate. – ypercubeᵀᴹ Sep 07 '16 at 21:32
  • I think my answer here http://math.stackexchange.com/questions/1353829/product-of-all-elements-in-finite-group is the correct one. It is a very elegant result that is depending on the Sylow $2$-subgroup. I have researched the question myself (including a lot of GAP calculations) and wrote a paper on it then went into the Liber Amicorum of my former thesis. advisor Rob van der Waall. – Nicky Hekster Sep 07 '16 at 21:39

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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.

Nicky Hekster
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