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Operation $\circ$ in set $S$ satisfies the conditions:

(1) $\forall x \in S \hspace{0.5cm} x \circ x = x$,

(2) $\forall x,y,z \in S \hspace{0.5cm} (x \circ y) \circ z = (y \circ z) \circ x$

You need to prove that:

$y \circ (x \circ y) = (y \circ x) \circ y$

My friend showed me such proof, but I'm not sure if it's right. (even if it's right, I don't understand why)

$(y \circ y) \circ (x \circ y) = [(x \circ y) \circ y] \circ y = [(y \circ y) \circ x] \circ y = (y \circ x) \circ y$

Can you please explain me, why this "works"? Especially, I want to know, where from come out this part: $(y \circ y) \circ (x \circ y) = [(x \circ y) \circ y] \circ y$

It looks like 2 nd condition of given operation, but I'm not sure.

Also, I'm trying to prove associative property, but I completely have no idea where to start. Can you guve me a hint (I don't want straight answer)?

In general - can you give me some links and/or list books/articles which can be helpful and interesting about operations theory? I would be glad if I can use some kind of problem set with a few solved examples.

Thanks for help

PS: I hope you can understand me.

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As for associativity, it is fairly easy to see that given your two axioms, associativity implies commutativity and vice versa. It is probably a bit easier to make headway on commutativity, because then there are only two variables around -- not that I've been able to find a solution. How sure are you that these axioms imply associativity at all? –  Henning Makholm Oct 6 '11 at 18:36
    
@HenningMakholm I am not sure at all, to be honest. I mean - it's just a problem from some problem set, so it's probably associative. –  exTyn Oct 6 '11 at 19:41
    
Does the problem set instruct you specifically to prove associativity, or something more open-ended (such as "prove or disprove")? –  Henning Makholm Oct 6 '11 at 21:14

2 Answers 2

Yes, $(y \circ y) \circ (x \circ y) = [(x \circ y) \circ y] \circ y$ is your second axiom applied from left to right on the elements $x\circ y$, $y$, and $y$.

Start with the axiom: $$(x \circ y) \circ z = (y \circ z) \circ x$$ To prevent confusion, rename the variables to $a$, $b$, $c$: $$(a \circ b) \circ c = (b \circ c) \circ a$$ Flip left and right (since equality is symmetric): $$(b \circ c) \circ a = (a \circ b) \circ c$$ Substitute $y$ for both $b$ and $c$: $$(y \circ y) \circ a = (a \circ y) \circ y$$ Finally substitute $x\circ y$ for $a$, and voilà: $$(y \circ y) \circ (x\circ y) = ((x\circ y) \circ y) \circ y$$

In the next step of the proof a similar rotation happens to $(x\circ y)\circ y$.

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Let's rewrite property (2) using different letters, $$(a \circ b) \circ c = (b \circ c) \circ a \qquad \forall a,b,c \in S $$

Then to prove the identity $y \circ (x \circ y) = (y \circ x) \circ y$, start with lhs:

$$ (y \circ x) \circ y \stackrel{(1)}{=} (y \circ x) \circ (y \circ y) \stackrel{ (2) \text{, use } c=(y \circ y)}{=} ( ( y \circ x) \circ y) \circ y \stackrel{(2) \text{, with } a=(y\circ x)}{=} (y \circ y) \circ (y \circ x) \stackrel{(1)}{=} y \circ (x \circ y) $$

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@HenningMakholm You beat me to it, but we even used the same letters :) –  Sasha Oct 6 '11 at 18:20
    
Thank's for help (for Sasha & @Henning Makholm). I'm glad, that You and Henning Makholm used a bit different solutions, it really helped me understand this problem. One more thing: both answers are helpful, so I'm not going to mark one particular as accepted answer. –  exTyn Oct 6 '11 at 19:42

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