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Let $f\colon \mathbb{R}^2 \to\mathbb{R} $ be a smooth function.

Can there exist an algebraic structure $(\mathbb{R}, \cdot)$ such that for $x,y \in \mathbb{R}$, $x \cdot y = f(x,y)$ that is a non-commutative semigroup that is strictly not a monoid or a group?

I can't think of an example, but it seems so unlikely that you can't have such an object.

If not, how does one prove so?

XYZT
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1 Answers1

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Sure you can, just take $f(x,y)=y$

Asinomás
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  • I don't think you read the question correctly. :-) – Asaf Karagila Dec 06 '14 at 00:06
  • oh, sorry I haven't taken calculus yet, is that function not smooth? – Asinomás Dec 06 '14 at 00:08
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    @Asaf Karagila: What's wrong with it? $f(f(x, y), z) = f(y, z) = z = f(x, f(y, z)) = f(x, z)$, so it forms a semigroup. It is not a monoid or group since it has no right identity. – The_Sympathizer Dec 06 '14 at 00:10
  • How can I phrase my question while ruling out left identities? No offence, but your answer is the kind of uninteresting answer I have been trying to avoid. I am looking for something that does not reduce all operations to finding the right-most argument. – XYZT Dec 06 '14 at 00:11
  • Or more equivalently, what should I call an algebraic structure that strictly has neither left nor right identities. I supposed using the word "semigroup" is misleading to what I am looking for. – XYZT Dec 06 '14 at 00:13
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    You should have said that earlier, there is also right identities. – Asinomás Dec 06 '14 at 00:14
  • It there a specific name for a semigroup with strictly no identities? I will have to accept your answer however, for it is correct and answers the question I inadvertently posed. – XYZT Dec 06 '14 at 00:16
  • you don't have to accept my answer if you don't want to. Let me see if I can find a different example. – Asinomás Dec 06 '14 at 00:19