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In essence, I am looking for an example of a semigroup or a semicategory (closure is not that important, but it is useful) that is NOT a monoid or category.

Hopefully, there is a neat and simple-to-understand example of an only-associative operation on an infinite set.

Edit: Also, non-commutative!

Edit #2: From all the answers, it seems there are no associative, non-identity, non-inverse operations on the reals such that for $x,y \in \mathbb{R}$ the operation is $x \cdot y = f(x,y)$ where $f(x,y)$ is a some analytically expressible function, such that the reals form a semi-group under that operation.

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  • $\begingroup$ The real numbers with the operator being taking maximum? It is commutative, but withough inverses or identity. $\endgroup$ – Raclette Dec 5 '14 at 19:28
  • $\begingroup$ Sorry, I should've mentioned - non-commutative too! $\endgroup$ – XYZT Dec 5 '14 at 19:31
  • $\begingroup$ Do you really want to ensure nothing has an inverse? $\endgroup$ – rschwieb Dec 5 '14 at 20:00
  • $\begingroup$ From all the answers, it seems there are no operations on the reals such that for x,y∈R the operation is x⋅y=f(x,y) where f(x,y) is a some analytically expressible function. $x\cdot y=x$ is not "analytically expressible"? $\endgroup$ – rschwieb Dec 5 '14 at 20:02
  • $\begingroup$ I don't understand the question you are asking. $\endgroup$ – XYZT Dec 5 '14 at 20:06
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How about strictly upper triangular matrices with matrix multiplication? Matrix multiplication is clearly associative and the product of two strictly upper triangular matrices is again strictly upper triangular. There are no inverses since these matrices have nontrivial kernel and there is no identity since the identity matrix is not strictly upper triangular.

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  • $\begingroup$ You know, I have no idea what that is. It's not a simple enough example. $\endgroup$ – XYZT Dec 5 '14 at 19:35
  • $\begingroup$ Check out my edit, the second example should be clearer. $\endgroup$ – Cameron Williams Dec 5 '14 at 19:39
  • $\begingroup$ Don't upper triangular matrices have the identity matrix as an identity? By definition given here: mathworld.wolfram.com/UpperTriangularMatrix.html $\endgroup$ – XYZT Dec 5 '14 at 19:41
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    $\begingroup$ strictly upper triangular $\endgroup$ – rschwieb Dec 5 '14 at 19:42
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    $\begingroup$ the identity matrix is not strictly upper triangular $\endgroup$ – Jorge Fernández Hidalgo Dec 5 '14 at 19:42
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Consider the left-identity semigroup. It can be defined on a set $S$ as follows:

for $a,b\in S$ let $a\star b=b$. this algebraic structure is a semigroup but it has no identity (although every element is a left-identity).

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  • $\begingroup$ It also does not commute. $\endgroup$ – Jorge Fernández Hidalgo Dec 5 '14 at 19:36
  • $\begingroup$ That's not an example. It's just a definition. $\endgroup$ – XYZT Dec 5 '14 at 19:36
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    $\begingroup$ Yes, it is a definition of an example. Did you expect something tangible? $\endgroup$ – Jorge Fernández Hidalgo Dec 5 '14 at 19:39
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    $\begingroup$ Consider the aforementioned example with $S=\{1,2,3\}$ you now have a non-commutative semigroup that is not a monoid and is concrete. $\endgroup$ – Jorge Fernández Hidalgo Dec 5 '14 at 19:41
  • $\begingroup$ This is one of my favorite examples of a simple weird monoid. $\endgroup$ – rschwieb Dec 5 '14 at 19:57
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Continuous functions with convolution product form an algebraic structure where the product is associative but has no identity.

Given any set $X$ you can give it the following operation $$* \colon X \times X \to X$$ where $x*y=y$ for every pair $x,y \in X$. It's easy to see that this operation is associative but it clearly doesn't have any identity.

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    $\begingroup$ Convolution is commutative. $\endgroup$ – Bartek Dec 5 '14 at 22:07
  • $\begingroup$ @Bartek Yeah, but at the time I wrote the answer the OP didn't say anything about commutativity. $\endgroup$ – Giorgio Mossa Dec 6 '14 at 8:28
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The set of all square matrices of order $n$ over $2\Bbb Z$ under matrix multiplication is an infinite non commutative semigroup without left or right identities. The same if you consider, instead of $2\Bbb Z$, any other infinite ring without identity, commutative or not. Also, in $2\Bbb Z$ no element is invertible.

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