Let $M$ be a Monoid and, for any subsets $A, B ⊆ M$ define the left-quotient
$$A \backslash B = \{ m ∈ M: (∃a ∈ A) am ∈ B\}.$$
For $m ∈ M$ and $B ⊆ M$, define $m \backslash B = \{m\} \backslash B$. The following identities hold:
$$
(A ∪ B) \backslash C = (A \backslash C) ∪ (B \backslash C),\quad
(A B) \backslash C = B \backslash (A \backslash C),\quad
1 \backslash A = A,
$$
where $A, B, C ⊆ M$ and $1$ denotes the identity of $M$.
For the free monoid $X^*$ generated by a set $X$, and for any $A ⊆ M$, the set
$$Q(A) = \{ w \backslash A: w ∈ X^* \}$$
is the state space for the minimal deterministic automaton corresponding to $A$, whose start state is $A$, whose state transitions on $x ∈ X$ from any $B ∈ Q(A)$ are given by $B \overset{x}→ x \backslash B$ and whose final states are those $B ∈ Q(A)$ for which $1 ∈ B$.
It resides in a single state diagram that contains the state diagram of every single automaton over $X$ - both finite and infinite. That, by the way, includes all Turing machines and push-down automata, as well as all oracles for subsets $A ⊆ X^*$ that lie beyond the Chomsky hierarchy. It is the Univeral State Diagram for $X^*$.
The set $Q(A)$ is finite if and only if the minimal deterministic automaton is finite if and only if $A$ is a regular subset of $X^*$.
Example:
Consider the subset of $X^*$, given by the grammar $S → u S v$, $S → w$, where $X = \{u, v, w\}$ and denote the subset by $S$. Let $S(n) = S \{v^n\}$ and $T(n) = \{v^n\}$, for $n = 0, 1, 2, ⋯$. Then,
$$Q(S) = \{ S(n): n=0,1,2,⋯ \} ∪ \{ T(n): n=0,1,2,⋯ \} ∪ \{∅\},$$
and the following are its state transitions in the Universal State Diagram for $X^*$:
$$
S(n) \overset{u}→ S(n+1),\quad S(n) \overset{v}→ ∅,\quad S(n) \overset{w}→ T(n),\\
T(n) \overset{u}→ ∅,\quad T(n+1) \overset{v}→ T(n),\quad T(0) \overset{v}→ ∅,\quad T(n) \overset{w}→ ∅,\\
∅ \overset{u}→ ∅,\quad ∅ \overset{v}→ ∅,\quad ∅ \overset{w}→ ∅.
$$
for $n = 0, 1, 2, ⋯$. The start state is $S = S(0)$ and the sole final state is $T(0)$
Therefore, $S$ is not regular.
The state diagram for the minimal deterministic automaton for $S$ is an infinite ladder with $S$ and $T$ rungs, where $u$ raises the level by one, $w$ crosses over from the $S$ rungs to the $T$ rungs, and $v$ lowers the level by one, and by convention, the $∅$ state is not displayed.
Example: Consider the subset of $X^*$ given by the grammar $S → u S$, $S → S v S$, $S → S w$, $S → x$, where $X = \{u, v, w, x\}$. The grammar specifies the subset that is given as the least fixed-point solution to the following
$$S ⊇ \{u\} S ∪ S \{v\} S ∪ S \{w\} ∪ \{x\},$$
or algebraically $S ≥ u S + S v S + S w + x$, where $≥$ denotes the subset ordering relation and $+$ the union operation. The least fixed point solution to $y ≥ f(y)$ is written as $μy·f(y)$.
The one in the previous example could have been written $μs·(u s v + w)$.
In Kleene algebra, the least fixed point solution to $x ≥ a + x b$ is also the least fixed point solution to $x ≥ a b^*$, i.e. $μx·(a + x b) = μx·(a b^*)$. As such, the grammar for this example could just as well be written in EBNF form as $S → (u S | x) (v S | w)^*$, with the corresponding fixed point equation given by $S = μs·(u s + x) (v s + w)^*$.
With that having been said, let $S$ also denote the subset of $X^*$ recognized by the grammar in this example. The least fixed point solution is also a solution to the corresponding equation, i.e. $μx·f(x) = f(μx·f(x))$. Here, that means
$$S = (\{u\} S ∪ \{x\}) T,$$
where
$$T = (\{v\} S ∪ \{w\})^*,$$
and - since $a b^* b^* = a b^*$ is a Kleene-algebraic identity - it also means that
$$S = S T.$$
This is enough to determine that
$$Q(S) = \{ S, T, ∅ \},$$
that the state transitions are
$$
S \overset{u}→ S,\quad S \overset{v}→ ∅,\quad S \overset{w}→ ∅,\quad S \overset{x}→ T,\\
T \overset{u}→ ∅,\quad T \overset{v}→ S,\quad T \overset{w}→ T,\quad S \overset{x}→ ∅,\\
∅ \overset{u}→ ∅,\quad ∅ \overset{v}→ ∅,\quad ∅ \overset{w}→ ∅,\quad ∅ \overset{x}→ ∅,
$$
that the start state is $S$, that the sole final state is $T$, that $S$ is regular and that it is given by the regular expression $S = u^* x w^* (v u^* x w^*)^*$. Thus,
$$μs·(u s + s v s + s w + x) = u^* x w^* (v u^* x w^*)^*.$$
On the other hand, to write the previous example as a regular expression, you would have to add in the following extra symbols $Z = \{b,d,p,q\}$, subjecting them to the identities
$$
b d = 1 = p q,\quad b q = 0 = p d,\quad z x = x z\quad (x ∈ X,\quad z ∈ Z),
$$
(and, optionally, also to the identity $d b + q p = 1$), in which case you may write $S = b (u p)^* w (q v)^* d$ - i.e. what shall come to be known as a Chomsky-Schützenberger context-free expression, once the new algebraic framework becomes more widely-known.
