Kurosh subgroup theorem

Let $X$ be the star-shaped graph with one central vertex of trivial group, one leaf vertex $v_i$ of group $G_i$ for each $i\in I$, and one trivial edge joining the center to $v_i$. The whole star is a maximal subtree, so [L2] and [L3] kill all edge symbols and leave only the leaf vertex groups with no cross-relations. By the universal property in [L1], this relative fundamental group is exactly ${\ast }_{i\in I}{G}_i$, so we may regard $G$ as the fundamental group of this graph of groups.

Let $T$ be the Bass-Serre tree of the star graph of step 1.1. The subgroup $H$ acts on $T$ without inversions, so [L5] identifies $H$ with the fundamental group of the quotient graph of groups $H\mathrm{\backslash }T$. By [L4], every edge stabilizer in $T$ is trivial, and every vertex stabilizer over the leaf orbit of $v_i$ has the form $H\cap g{G}_i{g}^{-1}$ for some $g\in G$.

Choose a maximal subtree $T_0$ of $H\mathrm{\backslash }T$. Because every edge group is trivial, the quotient graph-of-groups path group has no conjugacy relations, only the reversal relations from [L2]. Killing the edges of $T_0$ via [L3] therefore leaves a free product of the nontrivial vertex stabilizers together with a free group generated by the geometric edges outside $T_0$. Hence $H$ is a free product of the groups $H\cap g{G}_i{g}^{-1}$ that occur as nontrivial vertex stabilizers, together with some free group $F$.

A vertex of $T$ above the leaf vertex $v_i$ is a left coset $g{G}_i$. Two such vertices, $g{G}_i$ and $g^{\mathrm{\prime }}{G}_i$, lie in the same $H$-orbit exactly when $g^{\mathrm{\prime }}=hga$ for some $h\in H$ and $a\in G_i$, equivalently when $Hg{G}_i=H{g}^{\mathrm{\prime }}{G}_i$. Thus the $H$-orbits of vertices above $v_i$ are indexed by the double cosets $H\mathrm{\backslash }G\mathrm{/}{G}_i$, and choosing one representative from each double coset with nontrivial stabilizer gives exactly the index set in the statement. Combining this with step 3.1 proves the theorem.