Subgroups form a sublattice

Every group $G$ is associated with a lattice of subgroups $Sub(G,*,1,/)$ but it also associated with a lattice of submonoids $Sub(G,*,1)$. By this, we will see that the subgroups lattice of $G$ is a sublattice of the submonoid lattice.

Lemma 1. let $G$ be a group and $S$ an inverse-closed subset then composition closure $cl(S)$ is inverse closed.

Proof. let $a,b \in S$ then by composition closure $ab \in cl(S)$. By inverse closure $a^{-1} \in S$ and $b^{-1} \in S$. So by composition closure, $b^{-1}a^{-1} \in cl(S)$, but this is equal to $(ab)^{-1}$ so $(ab)^{-1} \in cl(S)$. It follows that $cl(S)$ is inverse closed. $\square$

Theorem. let $G$ be a group then the lattice of subgroups $Sub(G,*,1,/)$ is a sublattice of the lattice of submonoids $Sub(G,*,1)$.

Proof. let $A,B$ be subgroups of $G$. Then $A,B$ are inverse closed, so $A \cup B$ is inverse closed. By lemma 1, $cl(A \cup B)$ is a subgroup. With respect to the lattice of submonoids $A \vee B$ coincides with $cl(A \cup B)$ so the monoid join of subgroups is a group. It follows that subgroups are a sublattice of submonoids. $\square$

By extending previous results we get the following chain of sublattices: normal subgroups $\subseteq$ groups $\subseteq$ monoids $\subseteq$ subsemigroups. Previously we also saw that the lattice of partitions $Part(A)$ is a sublattice of the lattice of preorders. With similar results from both order theory and semigroup theory, it is reasonable to assume that this can be extended to categories.

This is indeed possible using lattice of subcategories, but for now I am interested in applications of this approad to semigroup theory. A significant result to that effect is that if $S$ is a finitely generated cancellative submonoid of a group $G$ then its subgroup closure is finitely generated as well.

Lemma 2. let $G$ be a group and $S$ a submonoid then if $S$ is finitely generated its group closure $H$ is a finitely generated as well.

Proof. let $X$ be the finite generating set of $S$. Then since $X$ is finite $X \cup -X$ is finite as well. By lemma 1, $X \cup -X$ is a finite generating set for $H$, so $H$ is finitely generated as well. $\square$

Theorem. finitely generated commutative cancellative torsion-free J-trivial monoids are affine

Proof. let $S$ be a finitely generated commutative cancellative J-trivial monoid. Then let $G$ be its Grothendeick group, then since $S$ is J-trivial $G$ is torsion-free. Since $S$ is finitely generated, $G$ is as well. By the classification of torsion-free finitely generated commutative groups, $G$ is isomorphic to $\mathbb{Z}^d$ for some $d$. $\square$

It is very easy to construct affine semigroups, by taking subsets of $\mathbb{Z}^d$. However, it is harder to determine when a given commutative cancellative semigroup is affine. This is a partial result to that effect.