Normalizing group intersection problem

Description

Given data

Our universe is some group $U$ (such as a linear group or a permutation group) in which products and inverses can be readily computed.

Two groups $G_{1}$ and $G_{2}$ in $U$ are specified by means of their generating sets $A_{1}$ and $A_{2}$. We are also given that $G_{1}$ is a normalizing subgroup for $G_{2}$, viz every element of $G_{1}$ normalizes the subgroup $G_{2}$ of $U$.

Goal

We are required to determine a generating set for $G_{1}$ ∩ $G_{2}$.

Solution

The underlying descending chain

We have the goal of computing $G_{1}\cap G_{2}$. First, consider the descending chain of subgroups:

$G_{2}\geq G_{2}^{(1)}\geq G_{2}^{(2)}\geq \ldots \geq G_{2}^{(n-1)}$

Here, $G_{2}^{(i)}$ denotes the subgroup of $G_{2}$ comprising permutations that fix the first $i$ elements in the set. In other words $G_{2}^{(i)}=G_{2}\cap Stab_{1,2,...,i}(Sym(S))$.

Multiplying all terms by $G_{1}$:

$G_{1}G_{2}\geq G_{1}G_{2}^{(1)}\geq G_{1}G_{2}^{(2)}\ldots G_{1}G_{2}^{(n=1)}=G_{1}$

Each of these is a subgroup because $G_{1}$ normalizes every element of $G_{2}$.

Now, intersect each member of this descending chain with $G_{2}$, to get:

$G_{2}\geq G_{2}\cap G_{1}G_{2}^{(1)}\geq \ldots G_{1}\cap G_{2}$

Now notice that at each stage (intersection the pointwise stabilizer with <math.G_2</math>, multiplying with $G_{1}$, and again intersecting with $G_{2}$, the index does not increse. Since we know that $[G_{2}^{(i)}:G_{2}^{(i+1)}]\leq n-i$, we conclude that even in the final descending chain, the indices are bounded by $n-i$.

The idea is now to start from a generating set of $G_{1}G_{2}^{(i)}$ and use that to manufacture a generating set of $G_{1}G_{2}^{(i+1)}$.