Taken from An Introduction to Semigroup Theory by J.M. Howie, aLb if and only if a and b are generating the same principal ideals.
What does it really means to generate the same ideal?
If someone has asked the same matter and has been answered, please kindly send the link to the answer! Thank you in advance.
The notion of ''substructure generated by a certain subset'' is widely encountered all throughout mathematics and can be formalized in a very general setting, but let us refer specifically to your particular context.
First, given a semigroup $S$ one defines a left ideal on $S$ to be a subset $I \subseteq S$ such that $SI \subseteq I$; let us agree to denote the set of all left ideals of $S$ by $\mathscr{Id}_{\mathrm{s}}(S)$. Having introduced this notion, we notice two remarkable properties:
$$\bigcap \mathscr{M} \in \mathscr{Id}_{\mathrm{s}}(S)$$
This being so, given an arbitrary subset $X \subseteq S$ we can consider the collection of all left ideals including $X$, in other words the set
$$\mathscr{M}=\{I \in \mathscr{Id}_{\mathrm{s}}(S)|\ I \supseteq X\}$$
which will be nonempty by virtue of property 2 and thus contain a smallest element (with respect to inclusion of course) by virtue of 1; this minimum of $\mathscr{M}$ is given explicitly as the intersection of all the left ideals including $X$ and it is what we are going to call by definition the left ideal generated by $X$, denoted as (in my favourite notation):
$$(X)_{\mathrm{s}}=\bigcap_{I \in \mathscr{Id}_{\mathrm{s}}(S) \\ \ \ \ \ I \supseteq X} I$$
This applies in particular to the case when $X$ is a singleton (one-element set); in such an instance we simplify the notation by removing the braces, as follows:
$$(\{t\})_{\mathrm{s}}=(t)_{\mathrm{s}}$$
and we refer to this object as the left ideal generated by element $t$. A left ideal is called principal whenever it is generated by a certain element of the semigroup.
First of all, the $\cal L$-relation is related to left ideals, not to ideals. There are two equivalent ways to define this Green's relation. Let $S$ be a semigroup and let $S^1$ be the semigroup equal to $S$ if $S$ has an identity and to $S \cup \{1\}$, where $1$ is a new identity, otherwise.
Two elements $s$ and $t$ are $\cal L$-equivalent (notation $s \mathrel{\cal L} t$) if they generate the same left ideal, that is, if $S^1s = S^1t$. Equivalently, $s \mathrel{\cal L} t$ if one can pass from $s$ to $t$ and back by left multiplication, that is, if there exists $x, y \in S^1$ such that $s = xt$ and $t = ys$.
