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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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Seems like a perfectly good question to me. I vote to leave open. –  Brett Frankel Feb 11 '13 at 1:11
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3 Answers

up vote 19 down vote accepted

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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you have to assume $ab\ne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer. –  user59671 Feb 10 '13 at 0:34
    
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. –  Math Gems Feb 10 '13 at 0:37
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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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+1. Beautiful solution. –  Sigur Feb 9 '13 at 23:49
    
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!. –  Matemáticos Chibchas Feb 10 '13 at 0:10
    
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions! –  user59671 Feb 10 '13 at 0:20
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+1 Very nice answer! –  Isaac Solomon Feb 10 '13 at 2:59
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Congrats on getting the Populist badge off of Math Gems' answer! –  Zev Chonoles Feb 10 '13 at 22:29
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I'm assuming that $abc = (ab)c$ and $cba=(cb)a$. To show G is abelian, we need for all $x,y \in G$, $xy=yx$.

$G$ is abelian iff $abc = (ab)c = c(ab) = c(ba) = (cb)a = cba$. The last equality is due to associativity, which we can assume holds if $G$ is indeed a group.

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you don't need to worry about associativity in a group. –  user58512 Feb 9 '13 at 23:35
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This argument isn't correct; you've shown that a sufficient criterion for the property to hold is that $G$ is abelian, but you have not shown that it is necessary. –  Zev Chonoles Feb 9 '13 at 23:36
    
You use associativity by assumption in your first statement. You don't need to. It's a group, and that's an axiom. –  Sam DeHority Feb 9 '13 at 23:37
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