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I found elementary algebra exercise which I can't resolve. Let algebraic structure $(X,\cdot)$ where $\cdot$ has properties:

$$ x\cdot(x\cdot y)=y \\ (y\cdot x)\cdot x=y $$

How to proof that $\cdot$ is commutative? If $X=\mathbb{Z}$ and $m\cdot n=-m-n$ then it is true and $\cdot$ has expected poroperties.

$$ m\cdot(m\cdot n)=-m-(-m-n)=n\\ (n\cdot m)\cdot m=-(-n-m)-m=n $$


$$ m\cdot n=-m-n=-n-m=n\cdot m $$

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up vote 3 down vote accepted

Note first that that the right cancellation law holds:

If $y \cdot x = y' \cdot x$, then $y = (y \cdot x) \cdot x = (y' \cdot x) \cdot x = y'$.

Now given any elements $a$ and $b$, write

$$(a \cdot (b \cdot a))\cdot (b \cdot a) = a = b \cdot (b \cdot a)$$

Right-cancellation yields

$$a \cdot (b \cdot a) = b.$$

Now we have

$$ b \cdot a = a \cdot (a \cdot (b \cdot a)) = a \cdot b.$$

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