Suppose, $G$ is a finite group. Let’s call a series of subgroups of $G$ $\{H_k\}_{k=1}^n$ a verbal series of length $n$, iff $H_1 = E$, $H_n = G$ and $\forall k < n$ $H_k$ is a verbal subgroup of $H_{k+1}$. It is rather obvious, that in this case all $H_k$ are verbal subgroups of $G$.
Let’s define $V_s(G)$ as the set of all verbal series of $G$. Now let’s define partial order on $V_s(G)$ in a following way:
$$\{H_k\}_{k=1}^n > \{K_l\}_{l=1}^m \text{ iff } n > m \text{ and } \exists \{k_j\}_{j = 1}^{m} \forall j \leq m K_j = H_{k_j}$$
As such $V_s(G)$ is finite, it possesses minimal and maximal elements. Its minimal element is always unique and has length two (provided that $G$ is non-trivial) - it is $\{E, G\}$. However, the structure of maximal element of $V_s(G)$ is usually more complicated.
The question is:
Do all maximal verbal series (in respect to the order defined on $V_s(G)$) have the same length (which in this case will also be the largest possible length of verbal series)?
All I managed to prove about maximal verbal series was, that if $\{H_k\}_{k=1}^n$ is one, then $\forall k < n$ $\frac{H_{k+1}}{H_k}$ is verbally simple (and the structure of all verbally simple groups is described here: https://math.stackexchange.com/q/3112452/407165). However, that is clearly not enough.
This conjecture was inspired by the famous Jordan-Hoelder theorem about subnormal series, however, the methods used in its proof can not be applied there, as, unlike normality, verbality is not finite intersection closed
