Last week I made a list of conjectures about Lemma 4.1 in this article before I went ahead and dug into the proof.
Recall: Let 
be a group, and 
subgroups of , and 
a selection of elements of . Suppose that $$G = \bigcup_{k \in \{1, \ldots, n\}} g_k H_k.$$ BH Neumann actually proves a result much stronger than what I discussed in last week’s post.
Theorem 1: Reorder the indices so that 
have finite index and 
do not. Then $$G = \bigcup_{k \in \{1, \ldots, m\}} g_k H_k.$$
That is, infinite index subgroup cosets are rendered redundant in the union. Thus one can take $$D = \bigcap_{k \in \{1, \ldots, m\}} H_k,$$ a finite index subgroup, and observe that all 
for 
are unions of cosets of 
. Since there are only finitely many cosets of , the situation is reduced to a finite problem, rendering Conjectures 1, 3, and 4 nearly trivially resolved in the affirmative; I won’t go into details here.
As for conjecture 5, I came up with the following counterexample: Consider to be the free group on generators 
and 
. Let 
and 
. Let 
and 
. Then $$g_1 H_1 = \{x^k \mid k \in \mathbb Z \} $$ and $$g_2 H_2 = \{yxy^k \mid k \in \mathbb Z\}$$ are disjoint, but the only group containing both 
and 
is all of .
Leave a Reply