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An anticommutative operation on a set X to be a function $\cdot:X\times X\rightarrow X$ satisfying two properties:

(i) Existence of right identity: $\exists r\in X:x\cdot r=x$ for all $x\in X$

(ii) $x\cdot y=r\iff(x\cdot y)\cdot(y\cdot x)=r\iff x=y$ for all $x,y\in X$

Then, why is it true that an anticommutative operation on a set $X$ of more than one element is not commutative and has no identity?

An obvious counterexample that I found is the set $X=\{r,a\}$ where $r\cdot r=r, a\cdot r=a, r\cdot a=a, a\cdot a=a$. If I am right then this is probably a typo since other sources seem to define an anticommutative operation to be one where $x\cdot y=y\cdot x \iff x=y$ for all $x,y\in X$.

Is there anything obvious that I am missing?

Timothy
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    What does the first equivalence in (ii) mean? – Davide Giraudo Jul 24 '12 at 15:19
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    Where did this come from? – Jonas Meyer Jul 24 '12 at 15:27
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    (ii) doesn't make sense to me; there's something missing or a wrong symbol being used... – Arturo Magidin Jul 24 '12 at 16:15
  • @ArturoMagidin and Davide: sorry i have a typo, the middle term should equal r. Jonas: this is exercise 4 page 28 of the book "Analysis I" by Amann and Escher. – Timothy Jul 24 '12 at 16:42
  • Your "counterexample" does not satisfy condition (ii). Note that by condition (ii), you must have $aa=r$ for all $a$. In particular, your $X$ does not satisfy condition (ii), since $aa\neq r$. – Arturo Magidin Jul 24 '12 at 16:54
  • Your counterexample isn't, because in it you cannot conclude from $x=y$ to $x\cdot y=r$ (equivalences go both ways!), as $x=y=a$ is a counterexample ($x=y$, yet $x\cdot y\neq r$). Therefore it violates property (ii). – celtschk Jul 24 '12 at 16:54

2 Answers2

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If we prove that there is no identity, then it will follow that the operation is not commutative, since we already have a right identity.

Note that by (ii), $xx=r$ for all $x$. Now suppose that $rx=x$. Then: $$(xr)(rx) = xx = r$$ hence by (ii) we conclude that $x=r$.

Therefore, if $X$ has more than one element, $x\neq r$, then $rx\neq x$, hence $r$ is not a two-sided identity, and moreover, $rx\neq x=xr$, so the operation is not commutative.

Arturo Magidin
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  • Are you using the uniqueness of the identity here tacitly? That if $e$ is an identity and if $r$ is (a priori) the right identity in this set, then $e = e \cdot r = r$ where the LHS follows from $r$ being a right identity and the RHS follows from $e$ being an identity? – EE18 Sep 14 '23 at 01:34
  • I suppose I should say a right inverse. Anyway, thank you (very belatedly!) for this very helpful solution. It is much slicker than mine was. – EE18 Sep 14 '23 at 01:40
  • @EE18 I do not understand your question. I showed that if $r$ is an element that makes both (i) and (ii) true (and in particular, $r$ is a right identity) and the set has at least two elements, then $r$ cannot be a left identity.. And we know such an $r$ exists. There is no uniqueness being invoked, tacitly or otherwise. – Arturo Magidin Sep 14 '23 at 01:46
  • Also when you conclude "then $rx\neq x$", is that because you've derived a contradiction (from $x \neq r$ and $r = x$)? How come you can't just conclude from the contradiction $x \neq r$ and $r = x$ that your original supposition that $r$ is a two-sided identity is wrong? – EE18 Sep 14 '23 at 01:46
  • Ah I see I think. I suppose I was arguing by contradiction in my head so that one would have to invoke uniqueness (again, I think), whereas you are simply giving a nice direct proof that any right identity cannot be a left identity. – EE18 Sep 14 '23 at 01:48
  • @EE18 I do not argue by contradiction. There is no argument by contradiction stated, in any way. It is shown that if $rx=x$, then $r=x$. The contrapositive is that if $r\neq x$, then $rx\neq x$. That is not an argument by contradiction. I also did not, in any way, shape, or form, assume that $r$ was a two sided identity. I know it's been over ten years, but I do not like people putting words in my mouth, either immediately or a decade later. – Arturo Magidin Sep 14 '23 at 01:50
  • Certainly I was not intending to put words in your mouth, and any hint of that was simply my inability to grasp your proof. Reading it again, I think I grasp now that you showed the contrapositive $r\neq x \implies rx\neq x$ in general so that, given that $r\neq x$ is true (which it is for some $x$), we conclude for that $x$ at least that $ rx\neq x$ (which is to say, any right identity $r$ cannot be a left identity, and therefore no identity can exist since an identity is a right identity in particular). Again, my apologies for not understanding the structure of your proof. – EE18 Sep 14 '23 at 01:59
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From property (ii) it is easy to derive that $x\cdot x=r$ for all $x\in X$. Now if the multiplication is commutative, you have $x\cdot y=y\cdot x$ for all $x,y\in X$, and thus $(x\cdot y)\cdot(y\cdot x)=(x\cdot y)\cdot(x\cdot y)=r$. Thus according to property (ii), $x\cdot y=r$ for all $x,y\in X$, especially if $y=r$. However (i) says $x\cdot r=x$. Combining both gives $x=r$ for all $x\in X$, thus $X$ has only one element.

In addition, it is also easy to show that any $X$ fulfilling the definition also fulfils the other definition you've seen:

As noted above. from property (ii) follows that $x\cdot x=r$ for all $x\in X$, and therefore with the same calculation as above (but now no longer for arbitrary $x,y\in X$) $(x\cdot y)\cdot(y\cdot x)=r$, and therefore according to (ii) $x=y$. That $x=y$ implies $x\cdot y=y\cdot x$ is trivial. Thus we have that from the definition above the other definition follows.

However the reverse is not true, as the example $a\cdot b=a$ for all $a,b\in X$ shows: It has a right inverse (indeed, every element is a right inverse) and from $a\cdot b=b\cdot a$ it follows trivially that $a=b$ (the left hand side evaluates to $a$ and the right hand side to $b$). However from $a=b$ it does not follow that $a\cdot b=r$ no matter which element you choose to be $r$ (as I said, any qualifies). However, it might be that those other definitions further demand that the right identity is unique, in which case my counterexample would no longer qualify.

celtschk
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