Some doubts on the algorithm of graph isomorphism of bounded degree graph

Question

There is an well known algorithm given by E.M Luks for bounded graph isomorphism. There are three important steps of the algorithm two of them are given below:

1. If the group action on vertex set of input graph is transitive (single orbit) then solve the isomorphism problem in each coset:

$$ISO(G,X,Y) = ISO(K_i\sigma_1,X,Y) \cup ISO(K_i\sigma_2,X,Y)\cdots ISO(K_i\sigma_k,X,Y)$$

where $k$ is number of cosets. The $K_i$ is a kernel of group action $\pi : G \times B \mapsto B$.

2. If the group action on vertex set of input graph is not transitive and let $B_1,B_2,\cdots B_l$ are the orbits (see group action $\pi$). Then we solve problem in each orbit : $$ISO_B(K_i\sigma_1,X,Y)=ISO_{B_{1}} (ISO_{B_{2}}\cdots ISO_{B_l}(K_i\sigma_1))$$

Questions :

1. From step we will be getting the solutions , but how to patch these all solutions to get a solution to original problem.

2. In step set $B = \{B_1,B_2, \cdots, B_l\}$ is a collection of orbits , If $B$ is a minimal system of imprimitivity then (step 2) still works or not.

Answer 1

1. We know that each $ISO$ is either a coset or an empty set. Suppose that non-empty answers to children $ISO$'s are cosets $g_i H_i$. Take $g_1$ as a representative for the resulting coset, then the subgroup $H$ corresponding to the resulting coset is induced by $g_1^{-1}g_i h_i$, where $h_i$ ranges over generators of $H_i$.

2. No, why would it work? The transitive case is treated with enumerating elements of the primitive action on the minimal system of imprimitivity (and corresponding cosets of the action kernel) rather than with direct restriction to the blocks, so the question seems totally irrelevant to the understanding of the algorithm.