If $u\cdot v \in (z \cdot z')^*$ and $v \cdot u \in (z' \cdot z)^*$ then $u \in (z \cdot z')^* \cdot z$ I'm searching for a reference for the following property:

Fact.  Let $u, v \in A^*$.  Write $w$ for the primitive root of $u\cdot v$, i.e., $u\cdot v =  w^c$.  There are unique words $z, z'$ such that:


$w = z \cdot z'$
$v \cdot u = (z' \cdot z)^c$
$u \in (z \cdot z')^* \cdot z$ and $v \in z' \cdot (z \cdot z')^*$


Proof. Most of the statement is given by, e.g., Lothaire 83:

What is left to show is the last point.  From $u \cdot v = (z\cdot z')^c$, we obtain that $u = (z\cdot z')^k\cdot z_{\text{p}}$ and $v = z_{\text{s}}\cdot z' \cdot (z\cdot z')^{k'}$ where $z = z_{\text{p}}\cdot z_{\text{s}}$ (the case where $u$ ends with a prefix of $z'$ is similar).  Since $v \cdot u = (z' \cdot z)^c$, it holds that $$z' \cdot z = z_{\text{s}}\cdot z' \cdot z_{\text{p}}\enspace.$$
  As $z' \cdot z$ is primitive, all of its permutations are distinct, hence the right-hand side of the previous equation is not a permutation, that is, $z_\text{s} = \varepsilon$. $\square$
The full statement (including the last point) seems to follow directly from the proof of Proposition 1.3.3 in (Lothaire '83).  The last point is part of the proof: using the notation in the original post, the proof states that $u = (z \cdot z')^{k_1} \cdot z$ and $v = z' \cdot (z \cdot z')^{k_2}$ for some $k_1$ and $k_2$ such that $k_1+k_2+1 = c$.  Perhaps the proof of this proposition could be cited...?