Given two groups \(A\) and \(B\), consider the following question: what is the minimal commutator length of the \(n\)-th power of an element \(g \in A * B\) not conjugate to elements of the free factors?
Recall that the commutator length of a group element is the minimal integer \(k\) such that it decomposes into a product of \(k\) commutators.
A simplified version of the main theorem of this paper goes as follows: Suppose that, in a free product of groups \(G = \star_{j\in J}\) without non-identity elements of order less than \(N\), the following equality holds: \[ c_1\cdots c_k d_1\cdots d_l = u_1^{n_1}\cdots u_m^{u_m} \] where

1.

the \(c_i\) are commutators;

2.

the \(d_i\) are conjugate to elements of the union of the groups \(A_j\);

3.

the \(u_i\) are conjugate to each other and not conjugate to elements of the union of the groups \(A_j\); and

4.

the \(n_i\) are positive integers.

Then \[ 2k+l \geq \sum_{i=1}^m (n_i-1) - 2\left[ \frac{1}{N}\sum_{i=1}^m n_i\right] +2 \]
As a corollary the authors show that: if \(c_1 \cdots c_k = u^n\) holds in the free product of groups \(\star_{j\in J} A_j\), where the \(c_i\) are commutators and \(u\) is not conjugate to elements of the free factors, then \[ 2k \geq n - 2\left[ \frac{n}{N}\right] +1. \]
Here, \([x]\) is the floor of \(x\). The results of of this paper recover and improve results from [L. P. Comerford jun. et al., Proc. Am. Math. Soc. 122, No. 1, 47–52 (1994; Zbl 0821.20008); L. Chen, Proc. Am. Math. Soc. 146, No. 7, 3143–3151 (2018; Zbl 1387.57001); S. V. Ivanov and A. A. Klyachko, Bull. Lond. Math. Soc. 50, No. 5, 832–844 (2018; Zbl 1436.20042)].