Proposition 12

Let H be a Hall set relative to X and let x, y be in X. (a) $H \cap M(\{x\}) = \{x\}$. (b) Suppose that $x < y$ and let $d_y$ be the homomorphism of $M(X)$ into $\mathbf{N}$ such that $d_y(y) = 1$ and $d_y(z) = 0$ for $z \in X, z \neq y$. The set of elements $w \in H \cap M(\{x, y\})$ such that $d_y(w) = 1$ consists of the elements $x^n y$ with n an integer $\geq 0$.

By Definition 2 (B), $x \in H$ and $H \cap M^2(\{x\}) = \varnothing$. If $w \in H \cap M(\{x\})$, where $n = l(w) \geq 3$, the elements $\alpha(w)$ and $\beta(w)$ also belong to $H \cap M(\{x\})$ by Definition 2 (C). It immediately follows by induction on $n$ that $H \cap M^n(\{x\}) = \varnothing$ for $n \geq 2$, whence (a).

We now prove (b). By Definition 2 (B), $y \in H$ and $xy \in H$. We show by induction on $n$ that $x^n y \in H$ for $n$ an integer $\geq 2$. Now $x^n y = x(x(x^{n-2}y))$ and the induction hypothesis implies that $x^{n-2}y \in H$. Now $l(x) < l(x^{n-2}y)$ for $n > 2$ and $x < y$, whence $x < x^{n-2}y$ in any case; condition (C) of Definition 2 shows that $x^n y \in H$. On the other hand, certainly $d_y(x^n y) = 1$. Conversely, let $w \in H \cap M(\{x, y\})$, with $d_y(w) = 1$. If $l(w) = 1$, then $w = y$; if $l(w) = 2$, then $w = xy$ by Definition 2 (B). If $l(w) \geq 3$, then $w = a(bc)$, with $a, b, c, bc$ in $H \cap M(\{x, y\})$ (Definition 2 (C)). $d_y(bc) = 0$ is impossible, since this would imply $bc \in M(\{x\})$, which is impossible by (a). Hence $d_y(bc) = 1$ and $d_y(a) = 0$, whence $a = x$ by (a). It follows immediately by induction on $n = l(w)$ that $w = x^{n-1}y$, which completes the proof of (b).