denoted by $N(\psi) = N(w)$.

The next lemma is a direct consequence of Theorem \[thm:relations\].

\[lem:normalized\_words\] Let $w  \in \{a, b, \alpha, \beta\}^*$. We have $w = N(w)$ if and only if $w$ does not contain $\alpha a^k\beta$ or $a\alpha^kb$ as a factor for any $k \in \mathbb{N}$. In particular, if $w \in \{ a, b, \alpha, \beta\}^* \setminus \{a, \alpha\}^*$, the normalized name $N(w)$ has prefix either $a^i\beta$ or $\alpha^ib$ for some $i \in \mathbb{N}$.

Since $\psi =  \varphi_a\varphi_b\varphi_\alpha\varphi_b =   \varphi_b\varphi_a\varphi_\alpha\varphi_b =   \varphi_b\varphi_b\varphi_\beta\varphi_a$, the normalized name of $\psi$ is $N(\psi) = bb\beta a$.

The morphism $E: 0\to 1, 1\to 0$ which exchanges letters in words over $\{0,1\}$ cannot change the factor complexity of an infinite word. Thus, $E$ is clearly a Sturmian morphism. But $E$ does not belong to the monoid $\mathcal{M} = \langle  \varphi_a,\varphi_b,\varphi_\alpha,\varphi_\beta\rangle  $. In fact, $E$ is the only missing morphism. More precisely, any Sturmian morphism $\psi$ either belongs to $\mathcal{M}$ or $\psi = \eta\circ E$, where $\eta \in \mathcal{M}$ (see [@MiSe93]). To generate the whole monoid of Sturmian morphisms $St$, one needs only three morphisms, say $E$, $\varphi_a$ and $\varphi_b$ (see [@Lo83]). We have $$\label{zamena}
\varphi_\alpha = E\varphi_aE  \quad \text{ and } \quad \varphi_\beta = 