Combinatorial Mathematics V Proceedings of the Fifth Australian Conf
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E
is residually finite.
The proof of this theorem, although on a different level of abstraction, is essentially based on the same idea as the proof of the theorem in this section.
9.
THE KUROSH SUBGROUP THEOREM In this section we will prove the Kurosh subgroup theorem and related results.
The construction used in the proof of the Kurosh theorem is the same as the one in Higgins [8], the selection of spanning trees and maximal forests corresponding to the use of transfinite induction in the original proof [15]. Kurash Subgroup Theorem:
Let G be the free product of the groups A , a a
TI*
G
and let H be a subgroup of G.
aEI
E
I,
A a
Then H
= D
* TI* v ga
(H n g v A (gv ) - 1 ) ,
a a
a
where D is a free group and the gV are a set of representatives of the double cosets a HgAa' Proof: Y.
We set
S
= aEI U Aa , form = cr(rr(Y, H))
We know that H
X
= rIG,
s) , Y
= X/H,
and select a spanning tree T of
and that rr(Y, H) is generated by the walks w(y) for
all chords y of Y with respect to T;
w(y) being defined as in
§ 3.
However, we want
to find a generating set of H using only a subset of the w(y) . To this end we color every edge y of X or Y with the color a if cr(y) E A and a consider, for all a, the subgraphs Y of Y consisting of the edges of color a and their a endpoints. Further, we extend the Y n T to maximal spanning forests Fa of Y The a. a components of the Fa are spanning trees of the components of Ya. In every component
we choose a vertex
We can assume that is the unique reduced walk from H to HgV in T.
=
where
a
Recalling that w(y) = r(o(y))y r(t(y))l we define the following sets of walks
15
in 1T(Y, H):
v) We note that o(Lv) = H n gVA (gV)-l and denote the union of the sets o(C by B. a a a a a Later we will show that B is a basis for the factor D in our decomposition of H as a free product. We want to show that the gv are a system of representatives for the double cosets a
HgAa'
We do this by showing that two elements b, c in G are in the same coset HgAa V if and only if Hb and Hc are in the same component y of Y . a
a
Suppose b , c E HgA Then there are elements aI' a 2 E A with Hba 1 = Hca 2. a. -1 a -1 But then ( Hb, a 1a2 ) is an edgE in Y with the endpoints Hb and Hba1a2
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