# Numbers of distinct products obtained by permuting 3 factors

Let $G$ be a group, and assume that there exist $a, b, c \in G$ such that $abc$, $acb$, $bac$, $bca$, $cab$ and $cba$ are precisely 5 distinct elements (i.e. that precisely two of the products are equal).

Question 1: Does it follow that there exist $d, e, f \in G$ such that $def$, $dfe$, $edf$, $efd$, $fde$ and $fed$ are precisely 4 distinct elements? And if not -- does the non-existence of such $d, e, f \in G$ at least imply that $G$ is infinite?

Remark: When one replaces 5 by 6, the answer is no. -- The smallest group which can be taken as an example here has order 54. It is $$G_{54,8} := \langle (1,4,7)(2,5,8)(3,6,9), (3,4,5)(6,8,7), (3,6)(4,7)(5,8) \rangle.$$

Question 2: Let $G$ be as above, and assume further that there are no $d, e, f \in G$ such that $def$, $dfe$, $edf$, $efd$, $fde$ and $fed$ are pairwise distinct. If $G$ is finite, does it follow that the order of $G$ is a multiple of 5?

Remark: The groups of order up to 625 which fulfill the conditions have orders 20, 40, 60, 80, 100, 120, 125, 140, 160, 180, 200, 220, 240, 250, 260, 280, 300, 320, 340, 360, 375, 380, 400, 420, 440, 460, 480, 500, 520, 540, 560, 580, 600, 620 and 625, respectively.

Side note: A related earlier question of mine remains unsolved so far.

Added on Aug 21, 2018: Given a group $G$, put $${\rm P}_3(G) := \left\{ |\{abc, acb, bac, bca, cab, cba\}| \ \big| \ a,b,c \in G \right\}.$$ Then clearly we have ${\rm P}_3(G) = \{1\}$ if and only if $G$ is abelian. Computational investigations further suggest that ${\rm P}_3(G)$ is always one of $\{1\}$, $\{1,2\}$, $\{1,2,3\}$, $\{1,2,3,4\}$, $\{1,2,3,4,5\}$, $\{1,2,3,4,5,6\}$, $\{1,2,3,4,6\}$, $\{1,2,3,6\}$ and $\{1,2,6\}$ -- where $\{1,2,3,4\}$, $\{1,2,3,4,5,6\}$ and $\{1,2,3,4,6\}$ are all very common, while $\{1,2,3,6\}$ and $\{1,2,3,4,5\}$ impose more-or-less severe restrictions on the structure of the group.

• Is there much literature on the three generated group with relation ABC=CBA ? Gerhard "Most Other Cases Don't Work" Paseman, 2018.08.18. – Gerhard Paseman Aug 18 '18 at 14:39
• @GerhardPaseman Probably not much literature is needed: the group $<A,B,C|ABC=CBA>$ is isomorphic to $Z\ast Z^2$. Indeed writing $b=AB$ and $c=CA^{-1}$ reduces to the presentation $<A,b,c|bc=cb>$. – YCor Aug 21 '18 at 11:01
• Suppose one computes a large but finite fragment of Z * Z * Z, with generators a,b, and c. (Or even the free semi group of words representing positive exponents.) The desired condition about five distinct elements (thanks Ycor) implies two short words commute. Surely we can find in the large fragment three elements which generate exactly four more elements? Just trying all words with at most 7 letters should be a feasible computation. If it fails, then you still have a partial result. Gerhard "Not A Group Theory Programmer" Paseman, 2018.08.22. – Gerhard Paseman Aug 22 '18 at 19:00
• @GerhardPaseman: Sorry, but I don't understand. -- Could you perhaps use somewhat more standard terminology and notation? – Stefan Kohl Aug 22 '18 at 20:42
• @StefanKohl It is known that if $6\not\in {\rm P}_3(G)$ then $G$ is metabelian and if $|G'|\leq 5$ then $6\not\in {\rm P}_3(G)$. – Alireza Abdollahi Aug 29 '18 at 19:55

Suppose $$abc=cab$$ in the group G. Then the elements $$c$$ and $$ab$$ commute. Thus $$a$$, $$ab$$, and $$c$$ are three candidates for $$d$$, $$e$$, and $$f$$, and it can be shown that they produce at least three and at most four distinct products. If the group they are in satisfies $$abac$$ is distinct from $$caab$$, then we have the required $$d$$, $$e$$, and $$f$$. Otherwise we have the additional derived relation $$abaca=caaba$$, so we have two other words commuting, in addition to the words $$ab$$ and $$c$$ commuting. If $$G$$ is finite, we may look for a contradiction by examining consequences of these two relations and show that they imply a forbidden relation such as $$abc=cba$$, which would give the result when $$ab$$ and $$c$$ commute (or $$a$$ and $$bc$$ commute). Using Ycor's presentation of the group with relation $$abc=cba$$, we might be able to do something similar to the above for this case. If $$G$$ is not finite, we might push on and consider $$b$$, $$ab$$, $$c$$ as candidates or $$ca$$, $$aba$$, and $$b$$ as candidates for $$d$$, $$e$$, and $$f$$. I suspect that we will either hit on a good triple for $$d$$, $$e$$, $$f$$, or else the assumption and failure of two or three candidate sets will lead to a contradiction.