# Show that an operation being commutative is a structural property

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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(1) Please include all relevant information in the body of the post; the subject is not part of the message, any more than the title of a book written on its spine is part of the narrative. (2) What is your definition of "structural property"? – Arturo Magidin Feb 13 '12 at 17:14
I'm not sure what you mean in the first part of your comment. Could you try explaining that again? By structural property I mean by definition; a structural property of a binary structure is one that must be shared by an isomorphic structure. – user23793 Feb 13 '12 at 18:15
In your original post, you had the task you needed to do, "Show that the operation * is commutative is a structural property" in the subject, but not in the post. Instead, the post started by saying "I've started this problem by:". But the post never said what "this problem" was, the only way to figure that out was to start reading at the "Title" line, instead of the first line of the post. So I added that title to be the first line of your post. The title is not part of the post, just like the title of a book, written on the spine, is not where we start reading a book. – Arturo Magidin Feb 13 '12 at 19:48
Also, in case you are not familiar, you might want to look at the FAQ for the process of accepting answers to your questions (very last end of that entry in the FAQ). – Arturo Magidin Feb 13 '12 at 20:16
@Arturo Perhaps you were reading a draft of the initial post, since the initial post does in fact include the titled question in the body - see the last two lines, which presumably mean "Does this suffice to prove that the operation....". – Math Gems Feb 13 '12 at 22:12

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 ....
Okay, let me see if I'm following you. For two binary operations to be isomorphic the have to have a function f that maps S onto T and that function is one-to-one, onto and is operation preserving. So I assume that $\langle S,* \rangle$ is commutative and I want to show that this leads to $\langle T, \Box\rangle$ being commutative, correct? Do I then use the fact that the operation is preserving to arrive at the answer? – user23793 Feb 13 '12 at 20:46
@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. – Arturo Magidin Feb 14 '12 at 5:09