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Ok so a semigroup only has associativity.

Suppose i have a set $S = \{a,b,c,d\}$

Someone give me an example for the function $+:S \times S \rightarrow S$ (in table form) so that is only associative so that $(S, +)$ is just a semigroup and nothing more (meaning not commutative or anything else). Neutral element does not matter.

userSd93r0294029480294850934t0
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  • Great question --- it's always good to look for the "why is this even part of the definition?" kind of things as you learn a new subject. – John Hughes Nov 25 '19 at 02:47
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    I don't think it's what you really want, but "or anything else" is too broad - you might be able to come up with semigroups lacking common properties (e.g. commutativity, identity, and inverses to name some common ones) - but there are certainly less common named ones that might accidentally happen (e.g. medial semigroups, idempotent semigroups, semilattices) and there will always, for a finite set, be some universal equation that holds beyond the semigroup axioms (e.g. there's always one of the form $x^{a} = x^b$ for some $a,b$ and all $x$ just because of finiteness) – Milo Brandt Nov 25 '19 at 02:54

2 Answers2

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Two simple addition operators which would make the set $S$ a semigroup are $+_L:x+_Ly\mapsto x$ and $+_R:x+_R y\mapsto y$.

It's obvious that both of these operators are associative$$(a+_L b)+_Lc=a+_Lc=a$$$$a+_L(b+_Lc)=a+_Lb=a$$ $$(a+_Rb)+_Rc=b+_Rc=c$$$$a+_R(b+_Rc)=a+_Rc=c$$

But, neither of these operators have an identity, nor are commutative.

Rushabh Mehta
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Actually 3 elements suffice to obtain a semigroup that is neither a monoid, nor a group, and is neither commutative nor idempotent. Take $S = \{a,b,0\}$ with $aa = a$, $ab=b$, $ba =0$ and $0a=a0=0b=b0$.

If you insist to have 4 elements, take $S =\{a,b,c,0\}$ with $aa = a$, $c = ab$, $ac = c$ and all other products equal to $0$. Technically speaking, this is the syntactic semigroup of the language $a^+b$.

J.-E. Pin
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