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Show that the operation $*$ is commutative is a structural property.

Give a careful proof that the indicated property of a binary structure $\langle S,* \rangle$ is indeed a structural property. I've started this problem as: Let $\langle S,* \rangle$ be isomorphic to $\langle T,\Box\rangle$. Also let $f:S \to T$. This means that for $a,b \in S$, then $f(a*b)=f(a)\Box f(b)$ and $f(a*b)=f(b*a)=f(b)\Box f(a)$ and therefore $f(a)\Box f(b)=f(b) \Box f(a)$. This means that an operation $*$ is commutative is a structural property. Does this work?

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    @MathGems: And all it takes to remove the possible issue is to type out the subject line as the first line of the post. After all, it's easier if one sees the problem to be solved before the proposed solution, rather than after it...2012-02-13

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Start like this: Let $\langle S,* \rangle$ be isomorphic to $\langle T, \Box\rangle$. Assume $\langle S,* \rangle$ is commutative. I claim that $\langle T, \Box\rangle$ is commutative. To prove this, let $a,b \in T$. ...continue computation to get... $a\Box b = b \Box a$. Therefore ....

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    @user23793: You use the fact that the *function* you have that maps $S$ bijectively to $T$ (one-to-one and onto) is operation-preserving. The operation itself does not "preserve" anything. I suspect you merely misspoke/mistyped, but better to be sure.2012-02-14