On the linearity of HNN-extensions with abelian base group

0.12 MB·English

by  V. Metaftsis

On the linearity of HNN-extensions with abelian base
groups
8
0
0 V. Metaftsis, E. Raptis and D. Varsos
2

t
c
O
Abstract

4
We show that an HNN-extension with finitely generated abelian base
1
group is Z-linear if and only if it is residually finite.

]
R
1 Introduction
G
.
h
AgroupGis linearif itadmits afaithfulrepresentation intosome matrixgroup
t
a GL (k) for some commutative ring k. The linearity of groups seems to be a
m n
difficult property to recognize and in most cases there is no certain method to
[
do so. The first that systematically studied linearity of groups was Mal’cev [8].

1 Since then many authors have shown the linearity of certain families of groups
v
but the list of linear groups remains short. We must mention here that in 1988,
9
8 Lubotzky (see [7]) gave necessary and sufficient conditions for a group to be
3 linearover C. Unfortunately,provingthatcertaingroupssatisfytheLubotzky’s
2
criterion, appears to be not an easy job.
.
0
In case we choose k to be the ring of integers Z, the range of examples
1
8 of linear groups shortens even further. The purpose of the present note is to
0 investigate the linearity of HNN-extensions of the form G = ht,K | tAt−1 = Bi
:
v with K a finitely generated abelian group. Our main result shows that those
Xi groups are Z-linear if and only if are residually finite.
Moreover, an interesting side result shows that when a certain isolated sub-
r
a group of K is trivial and A∩B 6= 1 then there is a finite index subgroup K′ of
K such that the isomorphism between A and B is induced by an isomorphism
of K′ of finite order. This allows us to embed a certain finite index subgroup
of the HNN-extension into a larger group which in turn has a finite index sub-
groupwhichisaright-angled Artingroup. Thatisenoughtoprovethelinearity
of the original HNN-extension. At the end of the paper we also give various
consequences of our main result concerning HNN-extensions.
We conjecture that fundamental groups of trees with finitely generated
abelian vertex groups are always linear and even that fundamental groups of
graphs with finitely generated abelian vertex groups are linear if and only if
are residually finite, but we have not been able to prove such result with the
techniques at hand.
1
2 On the isomorphisms of subgroups of finitely gen-
erated free abelian groups
Let G be a group and H a subgroup of G. The subgroup H is isolated in G, if
whenever gn ∈ H for g ∈ G and n > 0, then g ∈ H. By i (H) we denote the
G
isolated closure of H in G, that is, the intersection of all isolated subgroups of
G that contain H. For more on isolated subgroups and the isolated closure of
a group, the author should consult [3]
For the sequel, G is always the HNN-extension G = ht,K | tAt−1 = B,φi
where K is a finitely generated free abelian group and A,B isomorphic sub-
groups of K with ϕ : A −→ B the isomorphism induced by φ. Let D be the
subgroup of G with
D = {x ∈ K | for each ν ∈ Z there exists λ = λ(ν)∈ N such that t−νxλtν ∈ K}.
Then D is an isolated subgroup of K (see Proposition 2 in [1]) and therefore a
direct factor of K. Moreover D ≤ i (A∩B) and i (A∩B)/D is a free abelian
K K
group. If fact D plays a central roˆle in the proofs of the main results of [1, 2]
and apparently in the present work as well.
We can also describe D as follows: let M = A∩B, M = ϕ−1(M )∩M ∩
0 1 0 0
ϕ(M ) and inductively M = ϕ−1(M )∩M ∩ϕ(M ). Then M ≤ M and
0 i+1 i i i i+1 i
since K is finitely generated, there is k ∈ N such that the rank rank(M ) =
k+1
rank(M ). Consequently, D = i (M ) (see again [1, 2]). Notice also that if
k K k
H = T∞ M , then H contains every subgroup L of K with ϕ(L) = L.
i=0 i
Finally, we can give an alternative description of D using standard Bass-
Serre theory (see [13]). Let T be the standard tree on which G acts. Then D is
the subgroup of K such that for every finite subtree T′ of T, there is a positive
integer n ∈ Nsuch thatDn stabilizes T′ pointwise. Notice that thesubgroupH
defined above is the subgroup of K that stabilizes the entire tree T pointwise.
In this section we show that if D = 1 and A ∩ B 6= 1, then there is an
algorithm that allows us to consider a certain finite index subgroup of K, such
that the isomorphism induced to it by ϕ, has finite order.
Step I. Take a non-trivial c ∈ A∩B and the powers
1
...,ϕ−2(c ),ϕ−1(c ),c ,ϕ(c ),ϕ2(c ),....
1 1 1 1 1
Since D = 1 there are λ ,µ ∈ N with the property
1 1
S = {ϕ−λ1+1(c ),...,ϕ−1(c ),c ,ϕ(c ),...,ϕµ1−1(c )} ⊆ A∩B,
1 1 1 1 1 1
but ϕ−λ1(c ) ∈ A\(A∩B) and ϕµ1(c ) ∈ B\(A∩B).
1 1
Herewecansupposethathϕ−λ1(c )i∩(A∩B) = 1andhϕµ1(c )i∩(A∩B) =
1 1
1. For if (ϕ−λ1(c ))κ ∈A∩B for some κ∈ N and (ϕ−µ1(c ))ν ∈A∩B for some
1 1
ν ∈ N, then we can replace c by c′ = (c )κ·ν and obtain (possibly some others)
1 1 1
λ ,µ ∈ N with the property that they are the greatest positive integers such
1 1
that
{ϕ−λ1+1(c′),...ϕ−1(c′),c′,ϕ(c′),...,ϕµ1−1(c′)} ⊆ A∩B,
1 1 1 1 1
2
hϕ−λ1(c′)i∩(A∩B)= 1 and hϕµ1(c′)i∩(A∩B)= 1.
1 1
The elements of the set
S¯ = {ϕ−λ1(c ),...ϕ−1(c ),c ,ϕ(c ),...,ϕµ1(c )} =
1 1 1 1 1 1
{ϕ−λ1(c )}∪S ∪{ϕµ1(c )}
1 1 1
are(Z−)linearlyindependentinK. Indeed,letξ ϕ−λ1(c )+···+ξ ϕ−1(c )+
−λ1 1 −1 1
ξ c + ξ ϕ(c ) + ··· + ξ ϕµ1(c ) = 0 with ξ ∈ Z. If j with −λ ≤ j ≤
0 1 1 1 µ1 1 j 0 1 0
µ is the smallest subscript such that ξ 6= 0 we have that ξ ϕj0(c ) =
1 j0 j0 1
ξ ϕj0+1(c )···+ξ ϕ−1(c )+ξ c +ξ ϕ(c )+···+ξ ϕµ1(c ) and from
j0+1 1 −1 1 0 1 1 1 µ1−1 1
this (by applying ϕ−λ1−j0) we have that ξ ϕ−λ1(c ) = [ξ ϕj0+1(c )··· +
j0 1 j0+1 1
ξ ϕ−1(c )+ξ c +ξ ϕ(c )+···+ξ ϕµ1(c )]ϕ−λ1−j0 ∈ B, absurd by the
−1 1 0 1 1 1 µ1−1 1
previous observation.
So the set S¯, as linearly independent, generates a free abelian group S =
1 1
hS¯iofrankλ +1+µ andthe(cyclic)permutationj −→ j+1for−λ ≤ j ≤ µ
1 1 1 1 1
taken mod(λ +1+µ ) on the exponents of ϕj(c) defines an automorphism,
1 1
say ϕ , of S , of order λ +1+µ ).
1 1 1 1
Notice that, by the definition of ϕ , we have that ϕ | = ϕ. Moreover,
1 1 A∩S1
we have that ϕ (A∩S ) = ϕ(A∩S )= B∩S . Indeed, if a∈ A∩S then a is
1 1 1 1 1
a linear combination of
ϕ−λ1(c ),...,ϕ−1(c ),c ,ϕ(c ),...,ϕµ1−1(c ).
1 1 1 1 1
Moreover, ϕ(a) is a linear combination of
ϕ−λ1+1(c ),...,ϕ−1(c ),c ,ϕ(c ),...,ϕµ1(c ).
1 1 1 1 1
Step II. Let c ∈ A∩B such that hc i∩S = 1. We repeat the above process
2 2 1
in order to construct the set S = { ϕ−λ2+1(c ), ... ϕ−1(c ), c , ϕ(c ), ...,
2 2 2 2 2
ϕµ2−1(c )} ⊆ A∩B and the set S¯. Again, S¯ has similar properties, namely
2 2 2
S¯ is ( Z−) linearly independent in K and there are λ ,µ ∈ N being the
2 2 2
greatest positive integers such that
{ϕ−λ2+1(c ),...ϕ−1(c ),c ,ϕ(c ),...,ϕµ2−1(c )} ⊆ A∩B,
2 2 2 2 2
hϕ−λ2(c )i∩(A∩B)= 1 and hϕµ2(c )i∩(A∩B)= 1.
2 2
Also, the set S¯, as linearly independent in K, generates the free abelian
2
group S = hS¯i of rank λ +1+µ and the (cyclic) permutation j −→ j +1
2 2 2 2
for −λ ≤ j ≤ µ taken mod(λ +1+µ ) on the exponents of ϕj(c) defines the
2 2 2 2
automorphism ϕ , of S of order λ +1+µ .
2 2 2 2
Notice that S¯ ∩S¯ = ∅. For if ϕi(c ) = ϕj(c ), then c = ϕj−i(c ) with
1 2 2 1 2 1
−λ ≤ j −i ≤ µ (the only powers of ϕ that can be applied to c are integers
1 1 1
between −λ and µ ), and this contradicts the choice of c .
1 1 2
StepIII.Repeattheabove procedure,forc ∈ A∩B suchthathc i∩(S +S )=
3 3 1 2
1, in order to construct the sets S and S¯ with
3 3
S = {ϕ−λ3+1(c ),...ϕ−1(c ),c ,ϕ(c ),...,ϕµ3−1(c )} ⊆ A∩B
3 3 3 3 3 3
3
and
S¯ = {ϕ−λ3(c ),...ϕ−1(c ),c ,ϕ(c ),...,ϕµ3(c )} = {ϕ−λ1(c )}∪S ∪{ϕµ1(c )}.
3 3 3 3 3 3 3 3 3
Since the free rank of A∩B is finite, there will be an n ∈ N such that the
sets S ,S ,...,S , constructed in finitely many steps, by the above procedure,
1 2 n
almost generate thesubgroupA∩B in thesensethat hS ,S ,...,S i is of finite
1 2 n
index in A∩B.
For each S and S¯, let S = hS¯i. If we take the equivalence relation up
i i i i
to commensurability on the subgroups S ,S ,...,S , we can suppose that for
1 2 n
i 6= j, S ∩ S is of infinite index in both S and S . This means that the
i j i j
intersection S ∩ S belongs to A∩B for all i 6= j. By their definitions, the
i j
automorphisms ϕ act on the elements of A ∩ B as the isomorphism ϕ. So,
i
for all i = 1,...,n all ϕ coincide on the elements of the intersections S ∩S .
i i j
Thereforeall ϕ can beextended toacommon automorphismϑ of thesubgroup
i
S = hS ,S ,...,S i, which is of finite order, since it acts as a permutation on
1 2 n
the generators of S .
i
So we can now show the following.
Proposition 2.1. Let G = ht,K | tAt−1 = B,φi where ϕ is the isomorphism
induced by t. Suppose that D is defined as above and that D = 1. Then there
exists a finite index subgroup K¯ of K in which ϕ induces an automorphism ϕ¯,
i.e. ϕ¯| =ϕ. Moreover , ϕ¯ has finite order.
K¯∩A
Proof. Since D = 1, the subgroups A and B must be of infinite index in K.
Suppose that A∩B = 1. Then for the isolator closures i (A) and i (B) of
K K
A and B we have i (A)∩i (B)= 1 and therefore, there is a subgroup N ≤ K
K K
such that K = N⊕i (A)⊕i (B). Thegroup K¯ = N⊕A⊕B has the required
K K
property. Indeed, we can define the automorphism ϕ¯ = 1 ⊕ϕ⊕ϕ−1 which
N
has order two and ϕ¯| = ϕ.
A
Suppose now that A∩ B 6= 1. Let S = hS ,S ,...,S i be the subgroup
1 2 n
constructed in the above procedure. Then S is a subgroup of hA,Bi. If S is
of finite index in hA,Bi, then we take a (direct) complement of K in the sense
K = N ⊕i (hA,Bi). Then the group K¯ = N ⊕S has the required property
K
since we can define the automorphism ϕ¯= 1 ⊕ϑ which has finite order, where
N
ϑ is the automorphism defined above.
IfS isofinfiniteindexinhA,Bi,thenthereare(free)generators{a ,...,a }
1 m
of A such that A = ha ,...,a i ⊕ i (A ∩ hS ,S ,...,S i). If we take b =
1 m A 1 2 n 1
a ϕ,...,b = a ϕ, we obtain the group C = ha ,...,a i⊕hS ,S ,...,S i⊕
1 m m 1 m 1 2 n
hb = ϕ(a ),...,b = ϕ(a )i, which by its construction is of finite index in
1 1 m m
hA,Bi. Evidently, the automorphism ϑ can be extended to an automorphism
of C, and then we proceed as above taking C in place of S.
3 Linearity of HNN-extensions
Theorem 3.1. Let K be a finitely generated free abelian group, ϕ an au-
tomorphism of K of finite order and A a subgroup of K. Then, the multiple
4
HNN-extension
G = ht ,...,t ,K |t at−1 = ϕ(a),a ∈ A,i =1,...,ni
1 n i i
is Z-linear.
Proof. WetaketheHNN-extensionG˜ generatedbytheelementsξ ,...,ξ ,ζ,K,
1 n
satisfying the following relations ξ kξ−1 = ϕ(k),k ∈ K,i = 1,...,n, [ζ,a] = 1
i i
for all a∈ A. Notice that each ξ acts on K as an automorphism of finite order,
i
the same order for all i= 1,...,n.
The map f :G −→ G˜ with f(k) = k for every element of K and f(t ) = ξ ζ
i i
defines a monomorphism. Indeed, the relations in G are preserved by f, so it
is a homomorphism, and if 16= g ∈ G, then by a t-length argument we can see
that f(g) 6= 1, namely f is 1-1.
Let ν be the order of the automorphism ϕ. Since hξ ,i = 1,...,ni generate
i
a free group, we can consider the epimorphism ψ : hξ ,...,ξ i −→ Z with
1 n ν
ψ(ξ ) = 1 for all i = 1,...,n. This epimorphism, extends to an epimorphism
i
of G¯, which for simplicity we also denote ψ, ψ : G˜ −→ Z by sending all
ν
elements of K and ζ to zero. Let H be the kernel of ψ. Obviously, H is a
subgroup of finite index in G˜. In order to find a presentation of H we choose
a Schreier transversal U for H to be the set U = {1,ξ ,ξ2,...,ξν−1}. Then
1 1 1
H is generated by {ξ ξ−1,ξ ξ ξ−2,ξ2ξ ξ−3,...,ξν−2ξ ξ−(ν−1),ξν−1ξ ξ−ν} for all
i 1 1 i 1 1 i 1 1 i 1 1 i 1
j
i = 1,...,n along with the generators of K, ζ and all ξ conjugates of the
1
j −j
generators of K and ξ ζξ for all j = 1,...,ν −1.
1 1
Now,fromtheSchreierrewritingprocess,therelationsofH aretherelations
uru−1 where u ∈ U and r ∈ R, where R is the set of relations of G˜. One can
easily see that the relations that do not involve any ξ are commutators and
i
therefore all its conjugates are also commutators in the generating set of H.
On the other hand, any relation of the form
ξkξ aξ−1ξ−k = ξkϕ(a)ξ−k,
1 i i 1 1 1
transformed into the generating set of H, becomes
ξkξ ξk−1·ξ−(k−1)aξk−1·ξ−(k−1)ξ−1ξ−k = ξkϕ(a)ξ−k
1 i 1 1 1 1 i 1 1 1
or equivalently
ξkξ ξk−1·ξ−(k−1)aξk−1·ξ−(k−1)ξ−1ξ−k = ξk(ξ−1aξ )ξ−k = ξ−(k−1)aξk−1,
1 i 1 1 1 1 i 1 1 1 1 1 1 1
which is again a commutator in the generators of H. Therefore, all relations
that involve ξ , rewritten in the generators of H become commutators.
i
But the above implies that H has a presentation where all relations are
commutators of the generators and therefore is a right-angled Artin group.
Consequently H is Z-linear. (For more on right-angled Artin groups and its
linearity the reader can see [4, 6].) But it is known that the linearity is closed
under taking finite extensions or subgroups. Therefore the group G is Z-linear.
5
The above technique can actually show that if K is a right-angled Artin
group with standard generating set S, and S ,S are subsets of S then any
1 2
HNN-extension ht,S | tS t−1 = S i is Z-linear. This is only a special case of a
1 2
more general result by Hsu and Leary [5].
Proposition 3.2. Let K be a finitely generated free abelian group and A,B
isomorphic subgroups of K with ϕ : A −→ B an isomorphism. Suppose that
D = 1, then the HNN-extension
G = ht,K | tat−1 = ϕ(a),a ∈ A,i
is Z-linear.
Proof. Fromtheproposition2.1thereexistsafiniteindexsubgroupK¯ inK and
anautomorphismϕ¯ofK¯ of finiteordersuchthatϕ¯| = ϕ. LetG = ht,K¯iG
K¯∩A 1
be the normal closure of ht,K¯i in K. Evidently G is of finite index in G. Let
1
k ,k ,...,k be representatives of K¯ in K. We take t = k tk−1,i = 1,2,...,n,
1 2 n i i i
then using the Schreier rewriting process we obtain for G the presentation
1
G = ht ,...,t ,K¯ | t at−1 = ϕ¯(a),a ∈ K¯ ∩A,i = 1,...,ni. The group G is
1 1 n i i 1
Z-linear by the previous theorem. So G is, as finite extension of G .
1
Proposition 3.3. Let K be a finitely generated abelian group and A,B iso-
morphic subgroups of K with ϕ : A −→ B an isomorphism. Suppose that D is
finite, then the HNN-extension
G = ht,K |tat−1 = ϕ(a),a ∈ Ai
is Z-linear.
Proof. ThegroupGisresiduallyfinite,sinceDisfinite(see[1]). Thereforethere
exists a finiteindex normal subgroupN such that N∩D = 1. Let K = K∩N,
1
A = A∩N,B = B ∩N and ϕ : A −→ B the induced isomorphism (the
1 1 1 1 1
group N is normal in G). It is clear that the corresponding subgroup D =
1
{x ∈ K | for each ν ∈ Z there exists λ = λ(ν) ∈ N such that t−νxλtν ∈ K }
1 1
is trivial. So the hypotheses of Proposition 2.1 are satisfied, consequently there
is a finite index subgroup K of K and an automorphism ϕ¯ of K of finite
2 1 1 2
orderwhich extendsϕ . Thenormalclosureht,K iG is of finiteindex inGand,
1 2
as in the previous proposition, we get that ht,K iG (and therefore the group
2
G) is Z-linear.
Let K be a finitely generated abelian group and A,B isomorphic subgroups
of K with ϕ: A−→ B an isomorphism and
G = ht,K |tat−1 = ϕ(a),a ∈ Ai
be the corresponding HNN-extension. Let also H be the largest subgroup of
K such that ϕ(H) = H. Then, since K is abelian, it is easy to see that H is
the larger normal subgroup of G contained in K, in other words H = K , the
G
core of K in G.
6
Theorem 3.4. Let K be a f.g. abelian group and A,B proper, isomorphic
subgroups of K with ϕ: A−→ B an isomorphism and
G = ht,K |tat−1 = ϕ(a),a ∈ Ai
be the corresponding HNN-extension. The group G is Z-linear if and only if
the group G¯ = G/H is Z-linear.
Proof. Assume G to be linear. Then G is residually finite, therefore, by the
main Theorem in [1], we have that H is of finite index in D. Thequotient G/H
has the HNN-presentation G/H = ht,K | tA/Ht−1 = B/H,ϕ i, where ϕ is
H H
the induced isomorphism (since ϕ(H) = H). The corresponding subgroup D
H
is equal to D¯ = D/H which is finite. Therefore by be previous proposition
G/H is linear.
For the converse, since the group G¯ = G/H is linear, there is an homomor-
phism ϑ :G −→ R to a linear group R with Kerϑ≤ H. On the other hand the
linearity of G¯ = G/H implies the residually finiteness of it, butthe correspond-
ing H¯ = (K/H) is trivial. Therefore, by the main Theorem in [1] the group
G¯
D¯ = D/H must be finite, which, by the Theorem 2 in [2], implies that there
exists a finitely generated abelian group X such that K ≤ X and an automor-
phismϕ¯ofX withϕ¯| = ϕ. Nowtheobvioushomomorphism̺:G −→ X⋊hϕ¯i
A
is an embedding on K, so Ker̺∩K = 1. The linearity of G follows from the
linearity of the groups R, X ⋊hϕ¯i and the fact that Kerϑ∩Ker̺ = 1.
Corollary 3.5. Let K be a f.g. abelian group and A,B proper isomorphic
subgroups of K with ϕ: A−→ B an isomorphism and
G = ht,K |tat−1 = ϕ(a),a ∈ Ai
be the corresponding HNN-extension. The group G is Z-linear if and only if it
is residually finite.
Proof. Supposethat G is residually finite. By the main Theorem in [1] we have
that H is of finite index in D, where H and D are the subgroups of K defined
above. Therefore the group G/H is linear by proposition 3.3. The result now
follows from the previous theorem.
For the converse, it is well known that a finitely generated linear group is
residually finite.
Remark 1. In the proof of Theorem 3.4, in order to prove that the linearity
of G implies the linearity of G/H it is worth remarking that we used only that
G is residually finite and proposition 3.3, which in turn depends on proposition
2.1. We can give a direct proof using heavily that G is linear as follows.
The quotient G/H has the HNN-presentation G/H = ht,K | tA/Ht−1 =
B/H,ϕ i, where ϕ is the induced isomorphism (since ϕ(H) = H). The
H H
corresponding subgroup D is equal to D¯ = D/H which is finite, so G/H is
H
residually finite. Therefore there exists a normal subgroup N of finite index in
G such that H ≤ N and N ∩D = H. By the structure theorem of Bass-Serre
7
theory (see [13]), the structure of the subgroups of an HNN-extension N is the
fundamental group of a finite graph of groups with vertex groups of the form
N ∩ gKg−1, g ∈ G and edge groups of the form N ∩ gAg−1, g ∈ G. Since
H = K and N is normal in G, H is contained in every edge (and vertex)
G
group. Now the group D is isolated in K, which implies that D∩N is isolated
in K∩N. So H = gHg−1 is isolated (therefore a direct factor) in every vertex
group. This means, by the normal form of elements of N, that N = H ⋉M
for some subgroup M of the linear group G. Consequently N/H ≃ M is linear.
But N/H is of finite index in G/H, so G/H is linear.
Remark 2. The value of the propositions 2.1 and 3.3 consists in the fact that
exhibiting an internal property of finitely generated abelian groups we obtain a
tangible criterion for the linearity of HNN-extensions with base group a finitely
generated abelian group.
Corollary 3.6. Let K be a finitely generated abelian group, A,B proper sub-
groups of K and ϕ: A−→ B an isomorphism.
1. The subgroup D is finite.
2. There exists a finite index subgroup K of K and an automorphism ϕ of
1 1
K of finite order which extends ϕ in the sense that ϕ =ϕ.
1 1|A∩K1
3. There exists an abelian group X which contains the group K as a subgroup
of finite index and an automorphism ϑ of X of finite order such that
ϑ = ϕ.
|A
For the above statements we have the following:
1) implies 2).
2) is equivalent to 3).
1) implies 3).
Proof. 1) implies 2). This is the proposition 3.3.
2) is equivalent to 3). Assuming 2), it is easy to construct the group X
adding roots to the group K .
1
Assuming 3), we take K = Xn, where n is the index of K in X.
1
1) implies 3) is a consequence of the above two.
Remark 3. In account of Theorem 2 in [2], we see that there is a (weak)
equivalence where there is no needed for the automorphism ϑ to have finite
order. On the other hand, the statements 2) or 3) do not imply 1). For example
the free abelian group K = ha,b | [a,b]i with A = hai = B and ϕ the identity
satisfies (trivially) both 2) and 3), but not 1).
In Corollary 3.5 it is assumed that the associated subgroups A and B are
proper subgroups in the base group K. In the case where one of them is all the
base group (e.g. K = A), then the HNN-extension G = ht,K | tKt−1 = Bi is
a residually finite group, as a solvable constructible group, it is Q-linear. (This
result and a good account of basic properties of solvable constructible groups
can be found in [14]). Then G is not a Z-linear group. This is concluded form
8
the fact that the subgroup B is not closed in the profinite topology of G (see
e.g. in [10]), on the other hand if G was Z-linear, then by Theorem 5 p. 61 in
[11], all subgroups of K must be closed in the profinite topology of G.
In the case where the base group of an HNN-extension is not a f.g. abelian
group only miscellaneous cases are known for the linearity of these groups.
Proposition 3.7. Let K be any finitely generated linear group and ϕ an iso-
morphism between finite subgroups A and B of K. The HNN-extension G =
ht,K |tat−1 = ϕ(a),a ∈ Ai is linear.
Proof. At first, K is residually finite and since A and B are finite groups, the
group G is residually finite. Therefore there exists a normal subgroup N of
finite index in G such that N ∩A= N ∩B = 1. From the structure theorem of
Bass-Serre theory, the subgroup N is a free product of a free group and a finite
family of subgroups of kind N ∩gKg−1, g ∈ G. Then the result of Nisneviˇc
([9], see also [15]) implies that if the group K is linear of degree d≥ 2, then N
is linear of degree at most d+1. So G is linear.
Proposition 3.8. (cf. Theorem 1.3. in [11]) The HNN-extension G = ht,K |
tat−1 = ϕ(a),a ∈ Ai with base group K a polycyclic-by-finite group and proper
associated subgroups A and B = ϕ(A) of finite index in K is Z-linear if and
only if it is subgroup separable.
In the case where the associated subgroups are not of finite index we have
a simple (very) special result.
Proposition 3.9. Let K be the split extension of a polycyclic-by-finite group
A by a polycyclic-by-finite group C (K = A⋊ C). The HNN-extension G =
ht,K |tAt−1 = A,ϕi is linear.
Proof. Evidently G = A⋊ht,Ci. Therefore G/A ≃ hti∗C, so it is linear. On
the other hand the map ϑ : G −→ A⋊hϕi which sends every element a ∈ A
to itself, every element of C to the trivial element and t to ϕ is a well defined
homomorphism with kerϑ∩A =1. Therefore, G is linear.
References
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finite HNN-extensions of finitely generated abelian groups, Archiv Math. 50
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don Math. Soc. (3) 27 (1973), 425-439.
Department of Mathematics, University of the Aegean, Karlovasi 832 00,
Samos, Greece. Email: [email protected].
Department of Mathematics, University of Athens, Panepistimiopolis 157
84, Athens, Greece.
Email of E. Raptis: [email protected].
Email of D. Varsos: [email protected].
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