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(Warning: This might be a silly question.) Suppose we have a commutative and associative binary operator $\cdot$ over a set of elements $S$. If $a \cdot t = b\cdot t$ for all $t\in S$, does $a=b$ necessarily? Intuitively I would say "yes", because in the world of this binary operator/set, $a$ and $b$ have the exact same properties. But I'm not sure what being "equal" entails in this context.

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I wonder, is there a counterexample where $S^2=S$, i.e., $S=\{xy:x,y\in S\}$? – Jonas Meyer Dec 4 '12 at 6:30

No: if $S=\{0,1\}$ and $x\cdot y=0$ for all $x,y\in S$, you can easily check that $\cdot$ is commutative and associative.

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Definitely agree, thanks. I guess it still bothers me a little bit in the sense of what "equals" means. For example, what if we defined $S= \{\mathbb{Z} \cup v\}$ with the operator of addition $+$, where $v$ is essentially some symbolic element. We define the addition as usual over $\mathbb{Z}$, but when needing to add $v$, we treat it as $1$; e.g. $z+v = z+1$. Since $v$ does the exact same thing as $1$ to everything in this little world of the operator/set, it bothers me that it wouldn't be "equal" to $1$. Any way you could help formalize this? (Maybe the difference between $=$ vs $\sim$?) – E.R. Dec 4 '12 at 4:18
@E.R.: You could define relations $\sim_\ell,\sim_r$, and $\sim$ on $S$ by $a\sim_\ell b$ iff $as=bs$ for all $s\in S$, $a\sim_r b$ iff $sa=sb$ for all $s\in S$, and $a\sim b$ iff $a\sim_\ell b$ and $a\sim_r b$ and prove easily that all are equivalence relations. – Brian M. Scott Dec 4 '12 at 4:32

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