$n \in N$, $n | m$ implies

We can now identify in this $M,N,G_M$ to subsets of $G_N$. We consider the following extra assumption. We assume that, for all $m \in M, n \in N$, if $n$ divides $m$ in $N$ then $n \in M$.

Under this assumption, $U,V$ in $M$ have the same lcm in $M$ and in $N$. Moreover, $M = N \cap G_M$.

Since $\lcm_M(U,V)$ divides $U,V$ in $N$, we have that $\lcm_N(U,V))$ divides $\lcm_M(U,V)$ in $N$. Conversely, since $\lcm_N(U,V)$ divides $U$ in $N$ and $U \in M$, by the assumption we get $\lcm_N(U,V) \in M$. From the lemma we thus get that $\lcm_N(U,V)$ divides $U$ and $V$ in $M$ hence $\lcm_M(U,V)$ divides $\lcm_N(U,V)$ in $N$. It follows that $\lcm_M(U,V) = \lcm_N(U,V)$.

We have $M \subset N \cap G_M$. Let $n \in N \cap G_M$. Since $n \in G_M$ there exists $a,b \in M$ with $n = ab^{-1}$, hence $nb = a \in M$. Hence $n \in N$ divides $a \in M$ in $M$. By the assumption we get $n \in M$ and the conclusion.

The groups ${{\mathrm{B}}}(e,e,r)$ {#s:beer}
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The Corran-Picantin monoid
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We denote ${{\mathrm{B}}}(e,e,r)$ for $e \geq 1$ and $r \geq 2$ the braid 