Relationship Between Semigroups and Monoids

In the paper, product intersection sets are defined using subsets of an arbitrary algebraic structure that is a semigroup , and the authors show that you can replace some semigroups by monoids without changing the relevant sets . The paper defines the relevant global construction using a semigroup \(S\): \(H^ \mathbb{N} := \{ H S^\mathbb{N}(A q) : S \text{ is a semigroup and } (A q) {q\ge 1}\text{ is asymptotically strictly decreasing}\}\).[‌:cite[1]{ln=3}‌] More generally, it defines \(H {\!S}^{Q}(A q)\) for a \(Q\) indexed family of subsets of a semigroup \(S\), i.e. “for a family (A q) {q\in Q} of subsets of a semigroup, the product intersection set records those exponents \(h\in\mathbb{N}\) for which …”.[‌:cite[2]{ln=1}‌] The authors remark: “the same results hold true if we restrict to monoids instead of semigroups” (for non empty \(Q\)).[‌:cite[3]{ln=2}‌] They justify this by observing that if \(S\) is embedded as a subsemigroup of a larger semigroup/monoid \(T\), then the products \(A^h\) and \(A^{hq}\) (and hence the product intersection set) are unchanged whether you compute them inside \(S\) or inside \(T\).[‌:cite[3]{ln=4}‌], [‌:cite[3]{ln=5}‌] In particular, if \(T\) is the monoid obtained from \(S\) by adjoining an identity element , then any product intersection set realized by a semigroup is also realized by a monoid .[‌:cite[3]{ln=6}‌] So: monoids are semigroup like structures with an identity element , and in this setting (non empty \(Q\)) adding an identity does not change which exponents \(h\) occur in the product intersection set , meaning semigroup results transfer to monoids .[‌:cite[3]{ln=2}‌], [‌:cite[3]{ln=4}‌], [‌:cite[3]{ln=6}‌]