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I was wondering if there exists a semi-group in which some elements have a two-sided identity, while some others do not (they either have one-sided identities or no identities at all).

I came up with the 2-membered set $\{a,b\}$ with the binary operation defined as below, which seems to be an associative binary operator: $$a^2 = a, \quad b^2 = b, \quad ab = a, \quad ba = b$$ I do not know how to make sure this operator is associative (therefore making the structure a semi-group); however, assuming that this is indeed a semi-group fulfilling the condition I mentioned at the beginning of the post, are there any other semi-groups with a similar property in which the "local identities" are not just the elements being their own identities?

Mehdi Rejali
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    Note that this always gives $pqr=p$ however it is bracketed - the operation maps the product to the first factor. – Mark Bennet Aug 17 '24 at 11:59
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    Note the semigroup of square matrices of order $2$ whose second row is zero. – kabenyuk Aug 17 '24 at 13:05
  • Idempotents are local identities to themselves. – Shaun Aug 17 '24 at 13:05
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    Your example is a left-zero semigroup, i.e., a semigroup defined by the identity $xy=x$. This is trivially associative (see the first comment), and each element is itself a semigroup (indeed, a group!) having that element as the identity. It's rather trivial, but it might work for your proposes. – amrsa Aug 17 '24 at 18:29

2 Answers2

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A subgroup of a semigroup is a subsemigroup that is a group. Its identity is a local one in general.

An example can be found here.

Any semigroup with at least two subgroups that do not share an identity should suffice.

Shaun
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  • Maybe more can be added as to how these properties help solve the question? – Тyma Gaidash Aug 17 '24 at 15:27
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    Done, @TymaGaidash. – Shaun Aug 17 '24 at 15:29
  • If I have understood what you said correctly, then any proper semigroup which has a non-trivial subgroup should satisfy the condition. Is this correct? – Mehdi Rejali Aug 17 '24 at 15:30
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    Any semigroup with two subgroups that do not share an identity should suffice, @MehdiRejali. – Shaun Aug 17 '24 at 15:31
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    Given any idempotent $e \in S$, one could consider the subset of those $x \in S$ for which there exists $y \in S$ for which $xy=yx=e$ and $xyx=x$ and $yxy=y$. This is a subgroup with identity $e$, and there is one such subgroup for each idempotent. – Geoffrey Trang Aug 17 '24 at 15:58
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Here's another approach than the one with subgroups of the semigroup.

A very easy way to come up with semigroups (which might not have a global neutral element) having many different sub-semigroups with different neutral elements is start with a finite semilattice $(S,\wedge)$ (here, I'm considering $a\wedge b$ to be the greatest lower bound of $a$ and $b$).
Each chain contained in $S$ is finite, and as such it has a greatest element, which is the neutral element of the sub-semigroup which is that chain.

amrsa
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