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In a set $X$ we define a binary operation $*$ such that

$\forall x, y \in X,\ (x*y)*y=y*(y*x)=x.$

Is $*$ commutative and why?

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    Tried it, doesn't work. What would be $aa$? Probably with $ab=c$ and $ba=d$ for 2 new elements...2012-10-21

3 Answers 3

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Yes, it is commutative. Here is why : $ x * y = y*(y*(x*y)) = y * ((x * (x*y))*(x*y)) = y *x. $ I first use the identity to multiply by $y$ on the left twice, and then I replace the second $y$ by $x*(x*y)$. Then we use the fact that $ (x * (x * y))*(x*y) = x $ to remove the bigger chunk.

Hope that helps,

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    @ratchet freak : Read it again. Are you absolutely sure that I am wrong? Note that I used that for $X = x * y$ the fact that $y * (y * X) = X$, i.e. that $y * (y * (x*y)) = x*y$. The extreme left of my equality is $x*y$, the extreme right is $y*x$ and in the middle I just keep using the identity with appropriate choices of $X$ and $Y$ to plug in. Just read it carefully, there's only three steps to find. =)2012-10-21
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Ah, I think I got a counterexample:

Consider the free monoid $F_2$ on 2 letters $a,b$ (with empty word '$1$'), and consider its quotient $X:=F_2/(w^2\sim 1) $ for all words $w\in F_2$. Then, basically $X$ contains those words which doesn't have any substring of the form $ww$, and if such appears at a concatenation, they get just cancelled. For example $aba\cdot ab= a$.

Update: not good, as it is going to be a group, $xy=yx$ will follow from all $x^2=1$...

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    @tomasz : Perhaps it does, but I am a little tired and I got an answer, so I won't bother going through the details of the non-counter-example =) it's cool we discussed though, I had doubts for a second.2012-10-21
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Here's a proof that uses the automated theorem prover Prover9.

As a human, I find the output of these things difficult to read, but I find they can help by (a) suggesting an important in-between step, and (b) give a proof (even if it is unreadable), which means the result we're trying to prove is actually true, and it is possible there is a proof with a number of steps that could be understood by humans. (Also, there is a "coolness factor" to automated proofs.)

Here's the input:

formulas(assumptions).  (x * y) * y = x. y * (y * x) = x.  end_of_list.  formulas(goals).  x * y = y * x.  end_of_list. 

and a trimmed version of the output:

============================== PROOF =================================  % Proof 1 at 0.01 (+ 0.00) seconds. % Length of proof is 7. % Level of proof is 3. % Maximum clause weight is 7. % Given clauses 4.  1 x * y = y * x # label(non_clause) # label(goal).  [goal]. 2 (x * y) * y = x.  [assumption]. 3 x * (x * y) = y.  [assumption]. 4 c2 * c1 != c1 * c2.  [deny(1)]. 5 x * (y * x) = y.  [para(3(a,1),2(a,1,1))]. 7 x * y = y * x.  [para(2(a,1),5(a,1,2))]. 8 $F.  [resolve(7,a,4,a)].  ============================== end of proof ==========================