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Let $G$ be a group such that for any $a,b,c\ne1$: $$abc=cba$$ Is $G$ abelian?

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  • $\begingroup$ Seems like a perfectly good question to me. I vote to leave open. $\endgroup$ – Brett Frankel Feb 11 '13 at 1:11
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The proof is easy to find by writing equations $\,x = y\,$ in the "relator" form $\,xy' = 1,\, $ for $\,y' = y^{-1}.\:$ Doing so below, we seek to transform the $\rm\color{#0A0}{green}$ term into the $\rm\color{#C00}{red}$ term, and the value of $\rm\:c\:$ that produces such a transformation is clear when written this way.

$$\begin{eqnarray} abc = cba &\iff&\color{#0A0} a&\color{#0A0} b&\color{#0A0} c&\color{#0A0}{ \!a'b'} &\color{#0A0}{\!\!c'} &=& \,1\\ &\iff& a& b& c& \!\!(ba)'&\!\!c' &=& \,1\\ &\ \ \Rightarrow&\color{#C00} a&\color{#C00} b& & &\!\!\!\!\!\!\!\color{#C00}{(ba)'} &=& \,1\quad\text{for}\ \ c\, =\, ba\\ &\iff& a&b&=& b\ a \end{eqnarray}$$

Remark $\ $ This method of matching-up equations normalized into relator form $\, r = 1\,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.

Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of

$$r= x^2yxy^3 = 1,\ s= y^2xyx^3 = 1\ \Rightarrow\ x^7 = 1$$

The puzzle pieces are the relators $\,r,s.\,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $\, x^7$ (invert the label if you traverse the edge opposite its arrow direction).

$\qquad$ Van Kampen diagram

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  • $\begingroup$ you have to assume $ab\ne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer. $\endgroup$ – user59671 Feb 10 '13 at 0:34
  • $\begingroup$ Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part. $\endgroup$ – Math Gems Feb 10 '13 at 0:37
  • $\begingroup$ I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it. $\endgroup$ – freehumorist Jan 3 at 16:38
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Suppose $a,b\in G$, $ab\ne 1$ and $a,b\ne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.

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  • $\begingroup$ Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!. $\endgroup$ – Matemáticos Chibchas Feb 10 '13 at 0:10
  • $\begingroup$ @MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions! $\endgroup$ – user59671 Feb 10 '13 at 0:20
  • $\begingroup$ @CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable. $\endgroup$ – Matemáticos Chibchas Feb 10 '13 at 0:59
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    $\begingroup$ Congrats on getting the Populist badge off of Math Gems' answer! $\endgroup$ – Zev Chonoles Feb 10 '13 at 22:29
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If we reformulate the question as follows:

Let $G$ be a group, and $a,b,c \in G$, Let's define two 3-cycles: $\tau_1 = (abc)$ and $\tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.

Then isn't it obvious that, as $\tau_{1,2} \in \mathscr{S}_3 $ and $\mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?

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  • $\begingroup$ What does this have to do with commutativity of $G$? $\endgroup$ – Tobias Kildetoft Jan 3 at 18:04
  • $\begingroup$ I should have detailed: Let's introduce $\sigma_1 = \left ( \begin{matrix} a & b & c\\ a & b & c \end{matrix} \right ) = Id_{\mathscr{S}_3}$ to represent left side of our equality; and $\sigma_2 = \left ( \begin{matrix} a & b & c\\ a & c & b \end{matrix} \right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$? $\endgroup$ – freehumorist Jan 3 at 18:41

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