Example of an associative binary operation, without identities or inverses.

Question

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.

The real numbers with the operator being taking maximum? It is commutative, but withough inverses or identity. — Raclette, Dec 05 '14 at 19:28
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"? — rschwieb, Dec 05 '14 at 20:02
@Nikhil I think rschwieb doesn't understand what you're asking either. The sentence he's quoted seems strange. You did not ask a question about operations on reals that are "analytically expressible", whatever that means exactly. — Bartek, Dec 05 '14 at 21:06

Cameron L. Williams · Accepted Answer · 2014-12-05T19:53:03.583

7

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.

edited Dec 05 '14 at 19:53

answered Dec 05 '14 at 19:32

Cameron L. Williams

30,485

You know, I have no idea what that is. It's not a simple enough example. – XYZT Dec 05 '14 at 19:35
Check out my edit, the second example should be clearer. – Cameron L. Williams Dec 05 '14 at 19:39
Don't upper triangular matrices have the identity matrix as an identity? By definition given here: http://mathworld.wolfram.com/UpperTriangularMatrix.html – XYZT Dec 05 '14 at 19:41
2

strictly upper triangular – rschwieb Dec 05 '14 at 19:42
2

the identity matrix is not strictly upper triangular – Asinomás Dec 05 '14 at 19:42
Notice I said strictly upper triangular meaning that all entries on the diagonal are zero. – Cameron L. Williams Dec 05 '14 at 19:42
Hmm, that is very interesting. With enough multiplications, you eventually always reach the zero matrix, that is very interesting. Thanks, that is a fun example. – XYZT Dec 05 '14 at 19:49

score 6 · Answer 2 · answered Dec 05 '14 at 19:35

6

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).

answered Dec 05 '14 at 19:35

Asinomás

107,565

It also does not commute. – Asinomás Dec 05 '14 at 19:36
That's not an example. It's just a definition. – XYZT Dec 05 '14 at 19:36
3

Yes, it is a definition of an example. Did you expect something tangible? – Asinomás Dec 05 '14 at 19:39
1

Consider the aforementioned example with $S={1,2,3}$ you now have a non-commutative semigroup that is not a monoid and is concrete. – Asinomás Dec 05 '14 at 19:41
This is one of my favorite examples of a simple weird monoid. – rschwieb Dec 05 '14 at 19:57

score 1 · Answer 3 · answered Dec 05 '14 at 19:43

1

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.

answered Dec 05 '14 at 19:43

Giorgio Mossa

18,698

1

Convolution is commutative. – Bartek Dec 05 '14 at 22:07
@Bartek Yeah, but at the time I wrote the answer the OP didn't say anything about commutativity. – Giorgio Mossa Dec 06 '14 at 08:28

score 0 · Answer 4 · answered May 01 '15 at 18:38

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.

Example of an associative binary operation, without identities or inverses.

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