Grushko decomposition and rank additivity

Fix $j$. Apply [L1] to the subgroup $A_j\le G$ inside the decomposition $G\cong B_1\ast \cdots \ast B_n\ast F_s$. Because $A_j$ is freely indecomposable and not infinite cyclic, its Kurosh decomposition can contain neither a positive-rank free part nor two distinct nontrivial factors. By [L2], any subgroup of a conjugate of $F_s$ is free, so a nontrivial such subgroup would be either infinite cyclic or freely decomposable. Therefore the unique nontrivial Kurosh factor is $A_j$ itself, and it is contained in a conjugate of some $B_k$.

Let $B_k^{\mathrm{\prime }}=g{B}_k{g}^{-1}$ be a conjugate containing $A_j$. Now view $B_k^{\mathrm{\prime }}$ as a subgroup of the first decomposition $G\cong A_1\ast \cdots \ast A_m\ast F_r$ and apply [L1] again. The identity double coset for the factor $A_j$ contributes the nontrivial subgroup $A_j$ to the Kurosh decomposition of $B_k^{\mathrm{\prime }}$. Since $B_k^{\mathrm{\prime }}$ is also freely indecomposable and not infinite cyclic, its Kurosh decomposition has no second nontrivial factor and no free part. Hence $B_k^{\mathrm{\prime }}=A_j$. So every $A_j$ is conjugate to some $B_k$.

Apply the same argument with the roles of the two decompositions reversed: every $B_k$ is conjugate to some $A_j$. Also, applying [L1] to the subgroup $A_j$ inside its own decomposition shows that $A_j$ meets every conjugate of $A_{\mathrm{\ell }}$ for $\mathrm{\ell }\mathrm{\ne }j$ and every conjugate of $F_r$ trivially, because the identity double coset for $A_j$ already supplies the only possible nontrivial Kurosh factor. Therefore two distinct $A_j$ cannot both be conjugate to the same $B_k$. By symmetry the correspondence is bijective, so after permuting indices we get $m=n$ and each $A_j$ is conjugate to $B_j$.

After that permutation, quotient $G$ by the normal closure of the factors $A_1,\dots ,A_m$. In the first decomposition this kills the nonfree factors and leaves $F_r$; in the second decomposition it kills the conjugate factors $B_1,\dots ,B_m$ and leaves $F_s$. Thus $F_r\cong F_s$. Since both free groups have finite rank, [L3] gives $r=s$. This proves the uniqueness and rank statement.