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Let our signature be that of a single binary operation $+$. Suppose I have an equational identity $E$ such that $E$ is equivalent to the commutative identity $x+y=y+x$. In other words, $E$ implies and is implied by the commutative identity. Must $E$ simply be an alphabetic variant of the commutative identity? That is, must $E$ be of the form $v_1+v_2=v_2+v_1$ for distinct variables $v_1$ and $v_2$?

user107952
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    If the structure is a group, then there are other equivalent conditions that it is abelian other than that expression. An easy example using the structure theorem, is that all finite abelian groups are direct products of cyclic groups, and vice versa. However, I’m unsure about structures other than groups. – Malady Aug 05 '23 at 01:50
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    @D.W. I think the "general definition" was given. It was not an "example". – Anne Bauval Sep 23 '23 at 14:07

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Yes. Note that the two sides of $E$ must be equal when interpreted in the free commutative monoid on the variables, so in particular each variable must occur the same number of times on each side. If there are three or more occurrences of variables on each side of $E$, then $E$ does not imply commutativity, since it does not imply any equation about sums of only two elements. (Explicitly, consider a structure $\{a,b,a+b,b+a,\infty\}$ where any sum besides $a+b$ and $b+a$ is defined to be $\infty$; this will satisfy $E$ but not commutativity.) So there are at most two occurrences of variables on each side of $E$, and then it is clear that $E$ must have the form $v_1+v_2=v_2+v_1$ in order to be equivalent to commutativity.

Eric Wofsey
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    When you say "it is clear that E...", I don't understand why it is clear. Perhaps you can justify that little bit, because I don't quite see why it follows. – user107952 Aug 05 '23 at 15:24
  • There are very few possibilities for what $E$ could be; just list them all. – Eric Wofsey Aug 05 '23 at 15:34
  • @D.W.: I don't know what you think is missing. Nowhere do I assume that $E$ contains only 2 variables. – Eric Wofsey Sep 24 '23 at 02:06
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    @D.W. Eric has established that each side of the equality has at most two occurrences of variables. So each side is either $x$ or $x+y$, where $x$ and $y$ are variables (which could be the same). Up to renaming variables, there are only a small finite number of identities of this form. I enumerated them, and I agree with Eric that "it is clear". Since the OP is also satisfied (having accepted the answer), I don't think anything else needs to be done here. – Alex Kruckman Sep 24 '23 at 17:15
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    Moreover, the number of possibilities to list is even smaller since each variable has to appear the same number of times on each side of the equation. Explicitly, up to renaming variables, the only ones are $x=x$, $x+x=x+x$, $x+y=x+y$, and $x+y=y+x$. – Eric Wofsey Sep 24 '23 at 17:37