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Assume that $*$ is an associative operation on $S$ and that $a \in S$.

Let $C(a) = \{x: x \in S\text{ and } a*x = x*a\}$.

Prove that $C(a)$ is closed with respect to $*$.

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As a general rule of thumb, you will usually get much better answers if you ask a question rather than give an assignment and state what you have tried. –  JavaMan Jan 28 '13 at 4:41
    
Ok, I will keep that in mind –  a koala bear Jan 28 '13 at 4:47

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up vote 2 down vote accepted

Let $x_1,x_2\in C(a)$. Then $$x_1*a=a*x_1,x_2*a=a*x_2$$ We want to show that $x_1*x_2\in C(a).$ We have: $$(x_1*x_2)*a=x_1*(x_2*a)=x_1*(a*x_2)=(x_1*a)*x_2=(a*x_1)*x_2=a*(x_1*x_2)$$ This is what you need.

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short and sweet! –  amWhy Jan 28 '13 at 4:20
    
Thanks amWhy. ;-) –  Babak S. Jan 28 '13 at 4:22
    
Why do they tell me the a*x=x*a part? To tell me that it's commutative? –  a koala bear Jan 28 '13 at 4:27
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Bull, the operation may not be commutative --- but if $x$ is in $C(a)$, then $x$ and $a$ commute, and you can see where that was used, twice, in the proof. –  Gerry Myerson Jan 28 '13 at 4:44

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