d} \in \Sigma_{p,d,P}$. Hence $d|(m-n-d)$, that is, $m-n-d=td$ for some $t \in \mathbb{N}^0$ and so $m-n=(t+1)d$. Hence for any $a^ib^j \in S$ we have that $d|i-j$. By Lemma \[dimpact\], the first part of the following lemma is clear.

\[dequal1l\] If a two-sided subsemigroup $S=F_{D}\cup \widehat{F}\cup \widehat{\Lambda}_{I,p,d}\cup \Sigma_{p,d,P}$ of $\mathcal{B}$ is a left I-order in $\mathcal{B}$, then $d=1$ and $q=0$.

Suppose that $q\neq 0$, let $a^0b^k\in \mathcal{B}$ where $k\in \mathbb{N}$. Then $$a^0b^k=(a^ib^j)^{-1}(a^mb^n)=a^{j-i+t}b^{n-m+t}$$ where $t=\max\{m,i\}$, so that $0=j-i+t$. Hence we can deduce that $j=0$. If $i=0$, then $a^0b^k=a^mb^n$ so that $a^0b^k\in S$ a contradiction and so $i>0$. Hence $a^ib^0\in S$, but $a^ib^0 \in \widehat{\Lambda}_{0,p,1}\cup \widehat{F}$ as $\widehat{F}\subset T_{0,p}$ a contradiction again. Therefore $q=0$ as required.

\[L1\] In the case where $q=0$ it is easy to see that $a^pb^0\in S$. If $m\notin I$, then $a^ub^m\notin \widehat{F}$ for any $0\leq u<p$. For, if $a^ub^m \in \widehat{F}$, then $ a^pb^0a^ub^m =a^{p+u}b^m \in \widehat{\Lambda}_{I,p,d}$ a contradiction.

\[gcase\] The subsemigroup $S=F_{D}\cup \widehat{F}\cup \widehat{\Lambda}_{I,p,d}\cup \Sigma_{p,d,P}$ of $\mathcal{B}$ is a left I-order in $\mathcal{B}$ if and only if $d=1$ and $I=\{0,...,p-1\}$.

($\Longrightarrow$) Suppose that $S$ is a left I-order in $\mathcal{B}$. Then any element $q=a^mb^n \in \mathcal{B}$ can be written as $(a^ib^j)^{-1}(a^kb^l)$ for some $a^ib^j,a^kb^l \in  S$. By Lemma \[dequal1l\], $d=1$ and $0\in I$. It is remains to show that $I=\{0,...,p-1\}$.

Let $0<m<p$. Then $$\begin{array}{rcl}
a^mb^m&=&(a^ib^j)^{-1}(a^kb^l)\\ &=& a^{j-i+t}b^{l-k+t}\end{array}$$ where $t=$max$\{i,k\}$, for some $a^ib^j,a^kb^l\in S$. Then $m=j$ or $m=l$; so that $a^ub^m \in S$ for some $u$. If $m \notin I$, so $u< p$, then $a^pb^0a^ub^m=a^{p+u}b^m \in S$, in contradiction to Remark \[L1\].

($\Longleftarrow$) Suppose that $d=1$ and $I=\{0,...,p-1\}$. Then for any $a^mb^n\in \mathcal{B}$ we hav