Generalised triangle groups of type (3, q, 2)
If G is a group with a presentation of the form ⟨x, y|x³ = yq = W(x, y)² = 1⟩, then either G is virtually soluble or G contains a free subgroup of rank 2. This provides additional evidence in favour of a conjecture of Rosenberger.
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irk-123456789-1522592019-06-10T01:25:54Z Generalised triangle groups of type (3, q, 2) Howie, J. If G is a group with a presentation of the form ⟨x, y|x³ = yq = W(x, y)² = 1⟩, then either G is virtually soluble or G contains a free subgroup of rank 2. This provides additional evidence in favour of a conjecture of Rosenberger. 2013 Article Generalised triangle groups of type (3, q, 2) / J. Howie // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 1. — С. 1–18. — Бібліогр.: 22 назв. — англ. 1726-3255 2010 MSC:20F05, 20F06, 20E05. http://dspace.nbuv.gov.ua/handle/123456789/152259 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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If G is a group with a presentation of the form ⟨x, y|x³ = yq = W(x, y)² = 1⟩, then either G is virtually soluble or G contains a free subgroup of rank 2. This provides additional evidence in favour of a conjecture of Rosenberger. |
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Howie, J. |
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Howie, J. Generalised triangle groups of type (3, q, 2) Algebra and Discrete Mathematics |
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Howie, J. |
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Howie, J. |
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Generalised triangle groups of type (3, q, 2) |
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Generalised triangle groups of type (3, q, 2) |
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Generalised triangle groups of type (3, q, 2) |
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Generalised triangle groups of type (3, q, 2) |
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Generalised triangle groups of type (3, q, 2) |
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generalised triangle groups of type (3, q, 2) |
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Інститут прикладної математики і механіки НАН України |
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2013 |
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Generalised triangle groups of type (3, q, 2) / J. Howie // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 1. — С. 1–18. — Бібліогр.: 22 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT howiej generalisedtrianglegroupsoftype3q2 |
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2025-07-13T02:40:42Z |
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2025-07-13T02:40:42Z |
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1837497788800172032 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 15 (2013). Number 1. pp. 1 – 18
c© Journal “Algebra and Discrete Mathematics”
Generalised triangle groups of type (3, q, 2)
James Howie
Communicated by V. I. Sushchansky
Abstract. If G is a group with a presentation of the form
〈x, y|x3 = yq = W (x, y)2 = 1〉, then either G is virtually soluble
or G contains a free subgroup of rank 2. This provides additional
evidence in favour of a conjecture of Rosenberger.
1. Introduction
A generalised triangle group is a group G with a presentation of the
form
〈x, y|xp = yq = W (x, y)r = 1〉
where p, q, r ≥ 2 are integers and W (x, y) is a word of the form
xα(1)yβ(1) · · · xα(k)yβ(k)
(0 < α(i) < p, 0 < β(i) < q). We say that G is of type (p, q, r). The
parameter k is called the length-parameter. (The syllable-length, or free-
product length, of W regarded as a word in Zp ∗ Zq is 2k.) Without loss
of generality, we assume that p ≤ q.
A conjecture of Rosenberger [20] asserts that a Tits alternative holds
for generalised triangle groups:
Conjecture A (Rosenberger). Let G be a generalised triangle group.
Then either G is soluble-by-finite or G contains a non-abelian free sub-
group.
2010 MSC: 20F05, 20F06, 20E05.
Key words and phrases: Generalized triangle groups, Tits alternative.
2 Generalised triangle groups
This conjecture has been verified in a large number of special cases.
(See for example the survey in [10].) In particular it is now known:
• when r ≥ 3 [9];
• when 1
p + 1
q ≥ 1
2 [3, 14];
• when q ≥ 6 [19, 22, 4, 5, 1, 7, 17];
• when k ≤ 6 [20, 19, 21];
• for (p, q, r) = (3, 4, 2) [2, 18];
• for (p, q, r) = (2, 4, 2) and k odd [6].
In the present article we describe a proof of the Rosenberger Conjecture
for the cases (p, q, r) = (3, 3, 2) and (p, q, r) = (3, 5, 2), hence completing
the proof of the following
Theorem B. Let G be a generalised triangle group of type (p, q, r) with
p, q ≥ 3 and r ≥ 2. Then either G is soluble-by-finite or G contains a
non-abelian free subgroup.
Thus the Rosenberger Conjecture is now reduced to three cases, where
p = r = 2 and q ∈ {3, 4, 5}.
The results presented here have been posted online at [15, 16], where
the arguments are given in more detail. In particular, the proof in the
case q = 3 requires a certain amount of computer calculation using GAP
[11]: [15] provides full details of the computations involved, including code
and output. Also included in [15] are some partial results on the case
(p, q, r) = (2, 3, 2) of the Rosenberger Conjecture.
Our strategy of proof is essentially the same for the two cases q = 3
and q = 5, but the details differ substantially. A theoretical analysis of the
trace polynomial (see § 2.2 for details) reduces the problem to a finite set
of candidate words W by finding an upper bound for the length-parameter
k. In the case q = 5 the analysis is more detailed and yields the bound
k ≤ 4; the conjecture has already been proved when k ≤ 4 by Levin and
Rosenberger [19].
In the case q = 3 the analysis yields only the bound k ≤ 20. This
however is sufficient for a computer-based attack on the problem: a
computer search using GAP [11] refines the set of candidates to a list of
19 words. The conjecture is known for the 8 shortest words in the list,
by work of Levin and Rosenberger [19] and Williams [21]. The remainder
of the words satisfy a small cancellation condition, which ensures the
existence of nonabelian free subgroups.
J. Howie 3
Section 2 below contains some preliminary results on trace polynomials
and equivalence of words. The proof of the main result for the case q = 3
is contained in Section 3, and for the case q = 5 in Section 4.
Acknowledgement
I am grateful to Gerald Williams for helpful discussions on some of
the work presented in this paper.
2. Preliminaries
2.1. Equivalence of words
Our object of study is a group
G = 〈x, y|xp = yq = W (x, y)r = 1〉
where
W (x, y) = xα(1)yβ(1) · · · xα(k)yβ(k),
and 0 < α(i) < p, 0 < β(i) < q for each i.
We think of the word W as a cyclically reduced word in the free
product
Zp ∗ Zq = 〈x, y|xp = yq = 1〉.
We regard two such words W, W ′ as equivalent if one can be trans-
formed to the other by moves of the following types:
• cyclic permutation of W ,
• inversion of W ,
• automorphism of Zp or of Zq, and
• (if p = q) interchange of x, y.
It is clear that, if W, W ′ are equivalent words, then the resulting groups
G = 〈x, y|xp = yq = W (x, y)r = 1〉
and
G′ = 〈x, y|xp = yq = W ′(x, y)r = 1〉
are isomorphic. Hence for the purposes of studying the Rosenberger Con-
jecture (Conjecture A) it is enough to consider words up to equivalence.
4 Generalised triangle groups
2.2. Trace Polynomials
Suppose that X, Y ∈ SL(2,C) are matrices, and W = W (X, Y ) is a
word in X, Y . Then the trace of W can be calculated as the value of a
3-variable polynomial, where the variables are the traces of X, Y and
XY [12]. We can use this to find and analyse essential representations
from G to PSL(2,C). (A representation of G is essential if the images of
x, y, W (x, y) have orders p, q, r respectively.)
We can force the images x, y to have orders p, q in PSL(2,C) by
mapping them to matrices X, Y ∈ SL(2,C) of trace 2 cos(π/p) and
2 cos(π/q) respectively. Then the trace of W (X, Y ) ∈ SL(2,C) is given
by a one-variable polynomial τW (λ), where λ denotes the trace of XY .
We will refer to τW as the trace polynomial of W . Since we are in practice
interested in the case where r = 2, we obtain an essential representation
by choosing λ to be a root of τW .
We recall here some properties of τW . Details can be found, for example,
in [10]. (Complete formulae for the coefficients of τW are given in [17,
Appendix].)
Lemma 2.1. • τW has degree k;
• when p, q ≤ 3, τW (λ) is monic and has integer coefficients;
• in general, the coefficients of τW are real algebraic integers.
We also note a few more elementary properties.
Lemma 2.2. 1) Let J denote the interval
J = [2 cos(π/p + π/q), 2 cos(π/p − π/q)] ⊂ R.
Then τW (J) ⊂ [−2, 2].
2) If p = q = 3 and G does not contain a non-abelian free subgroup,
then the roots of τW belong to {0, 1, 1+
√
5
2 , 1−
√
5
2 }.
3) If p = 3, q = 5 and G does not contain a non-abelian free subgroup,
then the roots of τW belong to {0, 1, 1+
√
5
2 , −1+
√
5
2 }.
Proof. The space Mq of matrices in SU(2) ⊂ SL2(C) with trace 2 cos(π/q)
is path-connected. (Indeed, it is homeomorphic to the 2-sphere S2.) Writ-
ing Dp for the diagonal matrix with diagonal entries eiπ/p, e−iπ/p, fix
X = Dp ∈ Mp and let Y vary continuously in Mq from Dq to D−1
q . Then
λ = tr(XY ) will vary continuously from tr(DpDq) = 2 cos(π/p + π/q)
to tr(DpD−1
q ) = 2 cos(π/p − π/q). By the Intermediate Value Theorem
J. Howie 5
every j ∈ J arises as tr(XY ) for some Y ∈ Mq and X = Dp ∈ Mp, so
τW (j) = tr(W (X, Y )) ∈ [−2, 2] since W (X, Y ) ∈ SU(2).
Any root λ of τW corresponds to an essential representation ρ : G →
PSL(2,C). When p = q = 3, the image of ρ is a subgroup of PSL(2,C)
generated by two elements of order 3 and containing an element of order
2. Any such subgroup contains a free non-abelian subgroup unless it is
isomorphic to A4 or A5, in which case each of ρ(xy), ρ(xy−1) has order
2, 3 or 5. so λ = tr(ρ(xy)) and 1 − λ = tr(ρ(x))tr(ρ(y)) − tr(ρ(xy)) =
tr(ρ(xy−1)) both belong to {0, ±1, ±1±
√
5
2 }. This is possible only for
λ ∈ {0, 1, 1±
√
5
2 }, as claimed.
A similar argument applies when p = 3 and q = 5. Here the image of
ρ is generated by an element of order 3 and an element of order 5, and it
contains an element of order 2. Such a subgroup of PSL(2,C) contains
a non-abelian free subgroup unless it is isomorphic to A5, in which case
each of ρ(xy), ρ(xy±1) has order 2, 3 or 5. In this case tr(ρ(x))tr(ρ(y)) =
2 cos(π/5) = 1+
√
5
2 , so λ, 1+
√
5
2 − λ ∈ {0, ±1, ±1±
√
5
2 }, which is possible
only if λ ∈ {0, 1, ±1+
√
5
2 }, as claimed.
Lemma 2.3. If p = q = 3 and W, W ′ are equivalent with length-parameter
k, then either τW (λ) = τW ′(λ) or τW (λ) = (−1)kτW ′(1 − λ).
Proof. Since the trace of a matrix is a conjugacy invariant, it follows that
the trace polynomial is unchanged by cyclically permuting W . Moreover, if
X ∈ SL(2,C) then the traces of X, X−1 are equal, so the trace polynomial
is unchanged by inverting W .
If tr(X) = 1 = tr(Y ), then tr(Y −1) = 1 also. Interchanging x, y in
W has the effect on τW (λ) = tr(W (X, Y )) of replacing λ = tr(XY ) by
tr(Y X) = λ – in other words, no change.
Finally,
tr(XY −1) + tr(XY ) = tr(X)tr(Y ) = 1.
Hence replacing y by y2 has the effect of replacing τW (λ) = tr(W (X, Y )
by
tr(W (X, Y 2)) = tr(W (X, −Y −1)) =
= (−1)ktr(W (X, Y −1)) = (−1)kτW (1 − λ),
as claimed.
6 Generalised triangle groups
Theorem 2.4. Let G = 〈x, y|x3 = y3 = W (x, y)2 = 1〉 where W =
xα(1)yβ(1) · · · xα(k)yβ(k) with α(i), β(i) ∈ {1, 2} for each i. If G does not
contain a free subgroup of rank 2, then τW (λ) has the form
τW (λ) = λa(λ − 1)b(λ2 − λ − 1)c
with a, b ≤ 1 and c ≤ 3(a + b + 1). In particular k = a + b + 2c ≤ 20.
Proof. By Lemma 2.2 we may assume that the roots of τW all lie in
{0, 1, 1+
√
5
2 , 1−
√
5
2 }. Moreover, τW is monic with integer coefficients by
Lemma 2.1, so the the two potential roots 1±
√
5
2 occur with equal multi-
plicity c say, and τW has the form
τW (λ) = λa(λ − 1)b(λ2 − λ − 1)c
for some non-negative integers a, b, c.
We deduce the desired bounds on a, b, c from Lemma 2.2 as follows.
In this case the interval J in Lemma 2.2 is J = [−1, 2]. Hence 2a =
|τW (2)| ≤ 2 and 2b = |τW (−1)| ≤ 2, so a ≤ 1 and b ≤ 1. In addition,
(
5
4
)c (1
2
)a+b
=
∣∣∣∣τW
(
1
2
)∣∣∣∣ ≤ 2.
It follows that
c ln(5) ≤ (a + b + 2c + 1) ln(2),
which implies the desired conclusion
c ≤ 3(a + b + 1)
given that a + b ∈ {0, 1, 2}.
3. The case q = 3
3.1. Small Cancellation
In this section we prove a result on one-relator products of groups
where the relator satisfies a certain small cancellation condition. We will
apply this specifically to generalised triangle groups of types (3, 3, 2), but
as the result seems of independent interest, we prove it in the widest
generality available.
Suppose that Γ1, Γ2 are groups, and U ∈ Γ1 ∗Γ2 is a cyclically reduced
word of length at least 2. (Here and throughout this section, length means
J. Howie 7
length in the free product sense.) A word V ∈ Γ1 ∗ Γ2 is called a piece if
there are words V ′, V ′′ with V ′ 6= V ′′, such that each of V · V ′, V · V ′′ is
cyclically reduced as written, and each is equal to a cyclic conjugate of U
or of U−1. A cyclic subword of U is a non-piece if it is not a piece.
By a one-relator product (Γ1 ∗ Γ2)/U of groups Γ1, Γ2 we mean the
quotient of their free product Γ1 ∗ Γ2 by the normal closure of a cyclically
reduced word U of positive length. Recall [13] that a picture over the
one-relator product G = (Γ1 ∗ Γ2)/U is a graph P on a surface Σ (which
for our purposes will always be a disc) whose corners are labelled by
elements of Γ1 ∪ Γ2, such that
1) the labels around any vertex, read in clockwise order, spell out a
cyclic permutation of U or U−1;
2) the labels in any region of ΣrP either all belong to Γ1 or all belong
to Γ2;
3) if a region has k boundary components labelled by words W1, . . . , Wk ∈
Γi (read in anti-clockwise order; with i = 1, 2), then the quadratic
equation
k∏
j=1
XjWjX−1
j = 1
is solvable for X1, . . . , Xk in Γi. (In particular, if k = 1 then W1 = 1
in Γi).
Note that edges of P may join vertices to vertices, or vertices to the
boundary ∂Σ, or ∂Σ to itself, or may be simple closed curves disjoint
from the rest of P and from ∂Σ.
The boundary label of P is the product of the labels around ∂Σ. By a
version of van Kampen’s Lemma, there is a picture with boundary label
W ∈ Γ1 ∗ Γ2 if and only if W belongs to the normal closure of U .
A picture is minimal if it has the fewest possible vertices among all
pictures with the same (or conjugate) boundary labels. In particular every
minimal picture is reduced: no edge e joins two distinct vertices in such
a way that the labels of these two vertices that start and finish at the
endpoints of e are mutually inverse.
In a reduced picture, any collection of parallel edges between two
vertices (or from one vertex to itself) corresponds to a collection of
consecutive 2-gonal regions, and the labels within these 2-gonal regions
spell out a piece.
Since U is cyclically reduced, no corner of an interior vertex is contained
in a 1-gonal region.
8 Generalised triangle groups
Theorem 3.1. Let ℓ be an even positive integer. Suppose that U ≡
U1 · U2 · U3 · U4 · U5 · U6 ∈ Γ1 ∗ Γ2 with each Ui a non-piece of length at
least ℓ. Suppose also that A, B ∈ Γ1 ∗ Γ2 are reduced words of length ℓ
such that A is not equal to any cyclic conjugate of B±1 and such that
no Ui is equal to a subword of a power of A. Then G := (Γ1 ∗ Γ2)/〈〈U〉〉
contains a non-abelian free subgroup.
Proof. Since ℓ is even and positive, any reduced word of length ℓ in Γ1 ∗Γ2
is cyclically reduced. Replacing A by A−1 and/or B by B−1 if necessary,
we may assume that each of A, B begins with a letter from Γ1 and ends
with a letter from Γ2. Choose a large positive integer N > 20Kℓ, where
K is the length of U , and define X := AN BN , Y := BN AN . We claim
that X, Y freely generate a free subgroup of G.
We prove this claim by contradiction. Suppose that Z(X, Y ) is a
non-trivial reduced word in X, Y such that Z(X, Y ) = 1 in G. Then there
exists a picture P on the disc D2 over the one-relator product G with
boundary label Z(X, Y ). Without loss of generality, we may assume that
P is minimal, hence reduced.
Suppose that v is an interior vertex of P. The vertex label of v is
U or U−1 – by symmetry we can assume it is U . The subword U1 of U
corresponds to a sequence of consecutive corners of v; at least one of these
corners does not belong to a 2-gonal region of P, since U1 is a non-piece.
It follows that at least one of the corners of v within the subword U1 of
the vertex label does not belong to a 2-gonal region. The same follows for
the subwords U2, . . . , U6, so v has at least 6 non-2-gonal corners.
Now consider the (cyclic) sequence of boundary (that is, non-interior)
vertices of P, v1, . . . , vn say. This is intended to mean that the closed
path ∂D2, with an appropriate choice of starting point, meets a sequence
of arcs that go to v1, separated by 2-gons, then a sequence of arcs that go
to v2, separated by 2-gons, and so on, finishing with a sequence of arcs
that go to vn, separated by 2-gons, before returning to its starting point.
Note that it is possible that an arc of P joins two points on ∂D2; any
such arc is ignored here. Note also that we do not insist that vi 6= vj for
i 6= j in general. It is possible for the sequence of boundary vertices to
visit a vertex v several times. Nevertheless it is important to regard such
visits as pairwise distinct, so the notation v1, v2, . . . is convenient. We say
that a boundary vertex is simple if it appears only once in this sequence.
If vj is connected to ∂D2 by k arcs separated by k − 1 2-gons, then
this corresponds to a common (cyclic) subword Wj of Z(X, Y ) and U ,
of length k − 1. Let κ(j) ≤ 6 be the maximum integer t such that, for
J. Howie 9
some s ∈ {1, . . . , 6}, Wj contains a subword equal to (Us ·Us+1 · · · Us+t)
±1
(indices modulo 6). If no such t exsits, we define κ(j) = −1.
If vj is a simple boundary vertex with only r ≤ 4 corners not belonging
to 2-gons, then it is easy to see that κ(j) ≥ 5 − r:
There are more complex rules for non-simple boundary vertices. Nev-
ertheless, it is an easy consequence of Euler’s formula, together with the
fact that interior vertices have 6 or more non-2-gonal corners, that
n∑
j=1
κ(j) ≥ 6.
Now consider the word Z(X, Y ) as a cyclic word in Γ1 ∗ Γ2. Where a
letter X = AN BN or Y = BN AN is followed by another letter X or Y ,
then there is no cancellation in Γ1 ∗ Γ2. Similarly there is no cancellation
where X−1 or Y −1 is followed by X−1 or Y −1. Where X is followed by
Y −1 or vice versa, or where Y is followed by X−1 or vice versa, then there
is possible cancellation, but since A 6= B the amount of cancellation is
limited to at most ℓ letters from either side.
If Z has length L as a word in {X±1, Y ±1}, then after cyclic reduction
in Γ1 ∗ Γ2 it consists of L subwords of the form A±(N−1), L subwords of
the form B±(N−1), and L subwords V1, . . . , VL, each of length at most 2ℓ.
Now suppose that vj is a boundary vertex of P with κ(j) ≥ 0. Then
U±1
i is equal to a subword of Wj for some i. Since Ui cannot be a subword
of a power of A, Wj is not entirely contained within one of the segments
labelled A±(N−1).
If, in addition, κ(j) > 0, then Wj has a subword of the form (UiUi+1)±1
(subscripts modulo 6) As above, Wj cannot be contained in one of the
subwords A±(N−1). If it is contained in a subword of B±(N−1), then it is
a periodic word of period ℓ (that is, its i-th letter is equal to its (i + ℓ)-
th letter for all i for which this makes sense). Since Ui+1 has length at
least ℓ, there are at least two distinct subwords of UiUi+1 equal to Ui,
contradicting the fact that Ui is a non-piece in U .
Thus we see that the subwords Wj of Z(X, Y ) corresponding to
boundary vertices vj with κ(j) > 0 can occur only at certain points of
Z(X, Y ): where an A±(N−1)-segment meets a B±(N−1)-segment; or at
part of one of the words Vi.
In particular, the number of boundary vertices vj with κ(j) > 0 is
bounded above by L(2ℓ + 1). It follows that
κ :=
∑
j
κ(j) ≤ 5L(2ℓ + 1),
10 Generalised triangle groups
where the sum is taken over those boundary vertices vj with κ(j) ≥ 0.
The goal is to show that the total positive contribution to the sum κ
from those vj with κ(j) > 0 is cancelled out by negative contributions to
κ from other boundary vertices. This will show that κ ≤ 0, contradicting
the assertion above that κ ≥ 6.
Recall that K is the length of U . Thus each A±(N−1)-segment of ∂P is
joined to at least (N −1)ℓ/K boundary vertices, at most 2 of which (those
at the ends of the segment) can make non-negative contributions to κ. The
remaining vertices each contribute at most −1 to κ. Since N > 20Kℓ, it
follows that the negative contributions outweigh the positive contributions,
as required.
This gives the desired contradiction, which proves the theorem.
Corollary 3.2. Let Γ1 and Γ2 be groups, and suppose x ∈ Γ1 and y ∈ Γ2
are elements of order greater than 2. Suppose that W ≡ U1 ·U2 ·U3 ∈ Γ1∗Γ2
with each Ui a non-piece of length at least 4. Then G = (Γ1 ∗ Γ2)/〈〈W 2〉〉
contains a non-abelian free subgroup.
Proof. Let A1 = xyxy, A2 = xy−1xy−1, A3 = xyxy−1 and A4 = xyx−1y−1.
Then for i 6= j, Ai is not equal to a cyclic conjugate of A±1
j . Hence if
(say) U1 is equal to a subword of a power of Ai, it cannot be equal to a
subword of a power of Aj . Hence there is at least one A ∈ {Ai, 1 ≤ i ≤ 4}
with the property that no Ui is equal to a subword of a power of A. Now
choose B ∈ {Ai, 1 ≤ i ≤ 4} r {A} and apply the theorem, with U4 = U1,
U5 = U2 and U6 = U3.
3.2. Conclusion
Theorem 3.3. Let G = 〈x, y|x3 = y3 = W (x, y)2 = 1〉 be a generalised
triangle group of type (3, 3, 2). Then the Rosenberger Conjecture holds for
G: either G is soluble-by-finite, or G contains a non-abelian free subgroup.
Proof. Write
W = xα(1)yβ(1) · · · xα(k)yβ(k).
A computer search using GAP [11] (see [15] for details) produces a list of
all words W , up to equivalence, for which the trace polynomial τW has
the form indicated in Theorem 2.4: see Table 1. If W is not equivalent to
a word in the list, then G has a nonabelian free subgroup by Theorem 2.4,
so we may restrict our attention to the words W in Table 1.
J. Howie 11
W (x, y) SCC
1 xy NO
2 xyxy2 NO
3 xyx2y2 NO
4 xyxyx2y2 NO
5 xyxyx2yx2y2 NO
6 xyxy2x2yx2y2 NO
7 xyxyx2y2x2yxy2 NO
8 xyxyx2y2x2yx2yxy2 NO
9 (xyxyx)(y2x2y2x)(yx2yx2y2) YES
10 (xyxy)(x2y2x2yx)(y2x2yx2y2xy2) YES
11 (xyxy)(x2y2x2yx2)(y2xy2xyx2y2) YES
12 (xyxy)(x2y2xy2x2y2)(xyx2yx2y2) YES
13 (xyxy)(x2y2x2y2)(xy2x2y2xyx2yx2y2) YES
14 (xyxy)(x2y2xy2x2yxy)(x2y2x2yx2y2xy2) YES
15 (xyxy)(x2y2x2y2)(xy2x2yx2y2x2yxyx2y2xy2) YES
16 (xyxyx2y2)(x2yxy2xy2x2y2x2)(yxy2xyx2yx2y2x2yxy2) YES
17 (xyxyx2y2x2)(yxy2xyx2yx2y2x2yxy2x)(y2x2y2x2yxy2) YES
18 (xyxyx2y2x2yx2)(yxy2xyx2)(y2xyxy2x2y2x2yx2y2xy2) YES
19 (xyx2y2x2yx2)(y2xy2xyxy2x2)(y2x2yxy2x2yx2yxy2xy) YES
Table 1. Words in Z3 ∗ Z3 with trace polynomial as in Theorem 2.4.
The final column indicates whether or not W satisfies the small-cancellation
hypotheses of Corollary 3.2. In those cases where it does, the bracketing indicates
a subdivision of W into three non-pieces of length ≥ 4: W ≡ U1 · U2 · U3.
For those W in Table 1 for which k ≥ 7 (namely, numbers 9-19) the
small cancellation hypotheses of Corollary 3.2 are satisfied, and so G
contains a nonabelian free subgroup.
For k ≤ 6 (words 1-8) in the table, the result is known. Specifically,
groups 1-3 are well-known to be finite of orders 12, 180 and 288 respectively;
groups 4-6 were proved to have nonabelian free subgroups in [19]; and
finally groups 7 and 8 were shown in [21] (see also [15]) to be large. (That
is, each contains a subgroup of finite index which admits an epimorphism
onto a non-abelian free group.)
This completes the proof.
12 Generalised triangle groups
4. The case q = 5
To prove the result in the case q = 5, we first prove a number of
preliminary results.
Lemma 4.1. Let p : K → K be a regular covering of connected 2-
complexes with K finite, with covering transformation group abelian of
torsion-free rank at least 2. Let F be a field. If
H2(K, F ) = 0 6= H1(K, F ),
then
dimF H1(K, F ) = ∞.
Proof. Let {a, b} be a basis for a free abelian subgroup A of the group
of covering transformations of p : K → K, and let α be a cellular 1-
cycle of K over F that represents a non-zero element of H1(K, F ). If the
F [a]-submodule of H1(K, F ) generated by α is free, then H1(K, F ) is
infinite-dimensional over F , as claimed. So we may assume that there is a
cellular 2-chain β of K with d(β) = f(a)α for some non-zero polynomial
f(a) ∈ F [a].
For similar reasons, we may also assume that d(γ) = g(b)α for some
cellular 2-chain γ of K and some non-zero polynomial g(b) ∈ F [b].
Now f(a)γ − g(b)β ∈ H2(K, F ) = 0. In other words f(a)γ = g(b)β in
the group C2(K, F ) of cellular 2-chains of K, which is a free module over
the unique factorisation domain FA ∼= F [a±1, b±1]. Since f(a), g(b) are
coprime in F [a±1, b±1], it follows that there is a 2-chain δ with f(a)δ = β
and g(b)δ = γ. Hence f(a)(d(δ) − α) = d(β) − f(a)α = 0, in the group
C1(K, F ) of cellular 1-chains of K. But C1(K, F ) is also a free module
over the domain F [a±1, b±1], and f(a) 6= 0, so d(δ) = α, contradicting the
hypothesis that α represents a non-zero element of H1(K, F ).
This contradiction completes the proof.
Lemma 4.2. Let E be the set of midpoints of edges of a regular icosa-
hedron I ⊂ R3 centred at the origin, and let M = ZE its Z-span in R3.
Let V = {1, a, b, c} ⊂ Isom+(I) ⊂ SO(3) be the Klein 4-group, and let
C = {1, c} ⊂ V . Then, regarding M as a ZV -module via the action of V
by isometries of I, we have the following.
1) M ∼= Z6 as an abelian group.
2) H0(C, M) = Z ⊗ZC M ∼= Z4
2 ⊕ Z2.
J. Howie 13
3) The induced action of V/C on H0(C, M)/(torsion) is mutliplication
by −1.
Proof. If e is the midpoint of the edge joining two vertices u, v of I, then
e = (u + v)/2. Thus E is contained in the Q-span W of the set of vertices
of I. Since the vertices occur in 6 antipodal pairs, the Q-span QM of E
has dimension at most 6 over Q.
On the other hand, for any vertex v,
√
5 · v is the sum of the 5 vertices
adjacent to v in I. Thus
√
5 · v ∈ W . It also follows that
√
5 · e ∈ M for
any e ∈ E: specifically, (
√
5+3) ·e is the sum of the midpoints of the eight
edges of I that share a vertex with the edge containing e. If e1, e2, e3 ∈ E
are chosen to be linearly independent over R – and hence over Q[
√
5] –
then e1, e2, e3,
√
5 · e1,
√
5 · e2,
√
5 · e3 ∈ M are linearly independent over
Q. Thus QM = Q⊗Z M has dimension exactly 6 over Q. Since M ⊂ QM
is torsion-free and finitely generated, it follows that M ∼= Z6, as claimed.
If, in the above, we choose e1, e2, e3 to lie on the axes of the rotations
a, b, c ∈ V respectively, then we obtain a decomposition
QM = Q[
√
5]e1 ⊕ Q[
√
5]e2 ⊕ Q[
√
5]e3
of QM as a Q[
√
5]-vector space, with respect to which a, b, c act as the
diagonal matrices diag(1, −1, −1), diag(−1, 1, −1) and diag(−1, −1, 1)
respectively. Let
M+ := M ∩ Q[
√
5]e3 and M− := M ∩ (Q[
√
5]e1 ⊕ Q[
√
5]e2).
Then M− ∩M+ = {0}, while e1, e2,
√
5e1,
√
5e2 ∈ M− and e3,
√
5e3 ∈ M+,
so M−, M+ are free abelian of ranks 4 and 2 respectively.
Moreover, M/M− is naturally embedded in the vector space QM/QM−,
so is also free abelian – necessarily of rank 2. Note that M− is closed
under the action of V on M . Under the induced action on M/M−, each
of a, b acts as the antipodal map, multiplication by −1, and c acts as the
identity. Clearly also c acts on M− as the antipodal map.
Hence (1 − c)M = 2M−, so
H0(C, M) = M/(1 − c)M = M/2M− ∼= Z4
2 ⊕ Z2,
as claimed.
Finally, the quotient of H0(C, M) by its torsion subgroup is naturally
isomorphic to M/M−, and the induced action of V/C on this quotient is
via the antipodal map.
14 Generalised triangle groups
Lemma 4.3. Let G = 〈x, y|x3 = y5 = W (x, y)2 = 1〉 and suppose
that (λ − α)2 divides the trace polynomial τW (λ) of W , for some α ∈
{0, 1, (1+
√
5)/2, (−1+
√
5)/2}. Let ρ : G → A5 be the natural epimorphism
corresponding to the root α of τW (λ). Let C ⊂ A5 be a subgroup of
order 2 and V ⊂ A5 its centraliser of order 4. Then G has subgroups
N1 ⊳ N2 ⊳ ρ−1(V ) such that
1) ρ(N2) = {1};
2) ρ−1(C)/N2
∼= Z2;
3) ρ−1(V )/N2
∼= Z2 ⋊(−1) Z2;
4) N2/N1 is a non-zero vector space over Z2.
Proof. Let Λ = C[λ]/〈(λ − α)2〉, and choose matrices
X =
(
eiπ/3 0
1 e−iπ/3
)
, Y =
(
eiπ/5 λ − α − 2 cos(8π/15)
0 e−iπ/5
)
∈ SL2(Λ)
so that
tr(X) = 1, tr(Y ) =
1 +
√
5
2
and tr(XY ) = λ − α.
Then X, Y determine a representation ρ̂ : G → PSL2(Λ), since
tr(W (X, Y )) = τW (λ) = 0 in Λ. If φ : PSL2(Λ) → PSL2(C) is the
natural epimorphism obtained by setting λ = α, then the image of
ρ = φ ◦ ρ̂ is isomorphic to A5. Let K denote the kernel of ρ and let L
denote the kernel of ρ̂.
Clearly G/K ∼= A5. Now K/L ∼= ρ̂(K) is the normal closure of (xy)2.L,
so it is isomorphic to the subgroup of PSL(2, Λ) generated by
(XY )2 = −I + (λ − α)(XY )
together with its conjugates by elements of ρ̂(G). Let Z = φ(XY ) ∈ A5 ⊂
SU(2) denote the matrix obtained from XY by subsituting λ = α. Note
that tr(Z) = 0, in other words, Z ∈ sl2(C). Since (λ − α)2 = 0 in Λ, we
also have
(XY )2 = −I + (λ − α)Z.
For similar reasons, for any M ∈ ρ̂(G) we have
M(XY )2M−1 = −I + φ(M)Zφ(M)−1.
J. Howie 15
Moreover, since (λ − α)2 = 0 in Λ we have, for any A, B ∈ sl2(C),
(I − (λ − α)A)(I − (λ − α)B) = I − (λ − α)(A + B).
Thus K/L ∼= ρ(K) is isomorphic to the additive subgroup of sl2(C)
generated by MZM−1 for all M ∈ Â5 ⊂ SU(2). There are precisely 30
such conjugates of Z; geometrically they correspond to the midpoints of
the edges of a regular icosahedron centred at the origin in R3, where we
identify SU(2) with the 3-sphere of unit-norm quaternions, and R3 with
the space of purely imaginary quaternions. As an abelian group, therefore,
K/L ∼= ρ(K) ∼= Z6 by Lemma 4.2.
Now K/L is also an A5-module. Its structure as an A5-module does
not need to concern us, but Lemma 4.2 gives us some information about
its structure as a C-module and as a V -module. This in turn gives
information on the structure of ∆ := (ρ)−1(C).
Specifically, H0(C, K/L) = H0(∆/K, K/L) ∼= Z4
2 ⊕Z2. It follows from
the 5-term exact sequence
H2(∆/L) → H2(∆/K) → H0(∆/K, K/L) → H1(∆/L) → H1(∆/K) → 0
and the fact that ∆/K ∼= Z2 that H1(∆/L) has torsion-free rank 2, and
that the torsion subgroup of H1(∆/L) is a non-zero finite abelian 2-group.
Now let N0 = [∆, ∆].L and define N2 ⊃ N1 ⊃ N0 such that N2/N0 is
the torsion-subgroup of ∆/N0 = H1(∆/L) and N1/N0 = 2(N2/N0). Then
N0 ⊳ ρ−1(V ) since [∆, ∆] and L are both normal in ρ−1(V ). Hence also
N1, N2 ⊳ ρ−1(V ) since N1 and N2 are characteristic in ∆ which in turn
is normal in ρ−1(V ).
By construction, ∆/N2
∼= Z2, while N2/N1 is a non-zero Z2-vector
space.
Finally, since V/C acts on Z2 ∼= ∆/N2 by the antipodal map, it follows
that ρ−1(V )/N2
∼= Z2 ⋊(−1) Z2, as required.
5. Conclusion
Theorem 5.1. Let G = 〈x, y|x3 = y5 = W (x, y)2 = 1〉. If the trace
polynomial τW (λ) of W has a multiple root, then G contains a nonabelian
free subgroup.
Proof. We may assume that the root α is one of 0, 1, (±1 +
√
5)/2, for
otherwise the result is immediate from Lemma 2.2. Let ρ : G → A5
be the essential representation corresponding to α, let c = ρ(W ) ∈ A5,
16 Generalised triangle groups
C = {1, c} ⊂ A5 the subgroup generated by c, and V = {1, a, b, c} ⊂ A5
its centraliser in A5.
Let N1 ⊳ N2 ⊳ ρ−1(V ) < G be the subgroups promised by Lemma 4.3.
Let Γ = ρ−1(C) < ρ−1(V ) be the unique index 2 subgroup such that
N2 ⊂ Γ and Γ/N2
∼= Z2. Then Γ has index 30 in G and contains no
conjugate of x or of y.
Applying the Reidemeister-Scheier process to the presentation of G in
the statement of the Theorem, we obtain a presentation of Γ of the form
Γ = 〈k1, · · · , k31|r1, . . . , r30, s2
1, s2
2〉,
where r1, . . . , r10 are rewrites of conjugates of x3; r11, . . . , r16 are rewrites
of conjugates of y5; and r17, . . . , r30 and s2
1 = W 2, s2
2 = âW 2â−1 are
rewrites of conjugates of W 2, with ρ(â) = a and so s1 = W, s2 = âW â−1 ∈
Γ.
Let K be the 2-complex model of this presentation, F = Z2, and
p : K → K the regular cover correspdonding to the normal subgroup
N2 ⊳ Γ. Let L ⊂ K be the subcomplex obtained by omitting the 2-cells
corresponding to the relators s2
1, s2
2, and let L := p−1(L) ⊂ K.
Now, since Γ/N2 is torsion-free, and since s2
1 = 1 = s2
2 in Γ, s1, s2 ∈ N2.
Hence each lift of each 2-cell s2
i (i = 1, 2) to K is bounded by the square
of some path in K
(1)
. As a consequence, the 2-cells in K \ L represent
elements of H2(K, F ), and it follows that the inclusion-induced map
H1(L, F ) → H1(K, F ) is an isomorphism.
Since N2/N1 is a nonzero F -vector space, we have
H1(L, F ) ∼= H1(K, F ) = H1(N2, F ) 6= 0.
If H2(L, F ) = 0, then by Lemma 4.1 it follows that dimF H1(N2, F ) = ∞.
On the other hand, if H2(L, F ) 6= 0 then H2(L, F ) contains a free F (Γ/N2)-
module of rank > 0 = χ(L), since F (Γ/N2) is an integral domain. In this
case H1(L, F ) contains a non-zero free F (Γ/N2)-submodule, by [14, Propo-
sition 2.1 and Theorem 2.2]. Again we deduce that dimF H1(N2, F ) = ∞.
Thus the Bieri-Strebel invariant Σ of the F (Γ/N2)-module N2/N1 is a
proper subset of S1 [8, Theorem 2.4]. But by Lemma 4.2 (3) it follows that
Σ is invariant under the antipodal map: Σ = −Σ. Hence Σ ∪ −Σ 6= S1,
and it follows [8, Theorem 4.1] that Γ contains a nonabelian free subgroup,
as claimed.
Corollary 5.2 (Main Theorem). Let G be a generalised triangle group of
type (3, 5, 2). Then either G is virtually soluble or G contains a nonabelian
free subgroup.
J. Howie 17
Proof. By Theorem 5.1 and Lemma 2.2 the result follows unless τW (λ)
has only simple roots in the set {0, 1, (1 +
√
5)/2, (−1 +
√
5)/2}, in which
case the degree k of τW (λ) is at most equal to 4.
But the Rosenberger Conjecture is known for k ≤ 4 [19].
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Contact information
J. Howie Department of Mathematics and
Maxwell Institute for Mathematical Sciences
Heriot-Watt University
Edinburgh EH14 4AS
UK
E-Mail: J.Howie@hw.ac.uk
URL: www.ma.hw.ac.uk/∼jim
Received by the editors: 03.05.2012
and in final form 20.06.2012.
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