CONJUGACY GROWTH OF COMMUTATORS

arXiv:1802.09507v1 [] 26 Feb 2018

PETER S. PARK
Abstract. For the free group Fr on r > 1 generators (respectively, the free product G1 ∗ G2 of
two nontrivial finite groups G1 and G2 ), we obtain the asymptotic for the number of conjugacy
classes of commutators in Fr (respectively, G1 ∗ G2 ) with a given word length in a fixed set of free
generators (respectively, the set of generators given by the nontrivial elements of G1 and G2 ). A
geometric interpretation of our result is that for any connected CW-complex X with fundamental
group isomorphic to Fr or G1 ∗ G2 , we obtain the asymptotic number of free homotopy classes
of loops γ : S 1 → X with word length k such that there exists a torus Y with one boundary
component and a continuous map f : Y → X satisfying f (∂Y ) = Im γ. Our result is proven by
using the classification of commutators in free groups and in free products by Wicks, and builds
on the work of Rivin, who asymptotically counted the conjugacy classes of commutator-subgroup
elements in Fr with a given word length.

1. Introduction
Let G be a finitely generated group with a finite symmetric set of generators S. Any element
g ∈ G can then be written as a word in the letters of S, and one can define the length of g by
inf

k∈Z≥0 :∃c1 ,...,ck ∈S
g=c1 ···ck

k.

Consider the closed ball Bk (G, S) ⊂ G of radius k in the word metric, defined as the subset
consisting of elements with length ≤ k. One can then ask natural questions about the growth of
G: how large is |Bk (G, S)| as k → ∞, and more generally, what connections can be made between
the properties of G and this notion of its growth rate? For example, one of the pioneering results
on this question is that of Gromov [11], who classified the groups G with polynomial growth.
Since the middle of the 20th century, the growth of groups has been widely studied in various
contexts largely arising from geometric motivations, such as characterizing the volume growth of
Riemannian manifolds and Lie groups. In fact, in addition to applying knowledge on the growth
of groups to describe growth in such geometric settings, one can also pass information in the
other direction, i.e., use information on volume growth in geometric settings to yield consequences
regarding the growth of the relevant groups. For instance, Milnor [19] used inequalities relating
the volume and curvature of Riemannian manifolds to prove that the fundamental group of any
compact, negatively-curved Riemannian manifold has at least exponential growth.
A group that arises especially commonly in such geometric contexts is the free group Fr on r > 1
generators, which also has exponential growth; more precisely, after fixing a symmetric generating
−1
set S ··= {x1 , . . . , xr , x−1
1 , . . . , xr }, it is easy to see that

k
k
X
X
r (2r − 1)k − 1
i−1
2r(2r − 1)
=1+
|∂Bi (G, S)| = 1 +
,
|Bk (G, S)| = 1 +
r−1
i=1

i=1

where ∂Bk (G, S) denotes the subset of length-k elements.
Date: February 27, 2018.
1

2

PETER S. PARK

1.1. Motivation and main results. In certain contexts, it is more natural to consider the growth
rate of the conjugacy classes of G. For a given conjugacy class C of G, define the length of C by
inf length(g),

g∈C

and define ∂Bkconj (G, S) as the set of conjugacy classes of G with length k. In the case of Fr ,
the minimal-length elements of a conjugacy class are precisely its cyclically reduced elements, all
of which are cyclic conjugates of each other. The conjugacy growth of Fr can be described as
∂Bkconj (G, S) ∼ (2r − 1)k /k, which agrees with the intuition of identifying the cyclic conjugates
among the 2r(2r − 1)k−1 words of length k; for the full explicit formula, see [15, Proposition 17.8].
One context for which conjugacy growth may be a more natural quantity to study than the growth
rate in terms of elements is when characterizing the frequency with which a conjugacy-invariant
property occurs in G. An example of such a property is membership in the commutator subgroup
[G, G]. On this front, Rivin [21] computed the number ck of length-k cyclically reduced words in
Fr that are in the commutator subgroup (i.e., have trivial abelianization) to be the constant term
in the expression
!
r 
X
√
1
1
k
√
,
xi +
(2 2r − 1) Tk
xi
2 2r − 1 i=1
where Tk denotes the kth Chebyshev polynomial of the first kind. This quantity can asymptotically
be described as ck ∼ Cr (2r − 1)k /kr/2 for some positive constant Cr depending only on r. Furthermore, from the number of cyclically reduced words with trivial abelianization, one can derive
the growth of conjugacy classes with trivial abelianization by using Möbius inversion, due to the
following relationships:
X
ck =
pd ,
d|k

where pd denotes the number of primitive (i.e., not a proper power of any subword) length-d words
with trivial abelianization, and
X pd
,
|∂Bkconj (G, S) ∩ [G, G]| =
d
d|k

which together imply by Möbius inversion that




′
X1X
X ce X µ(d ) 
d
µ
|∂Bkconj (G, S) ∩ [G, G]| =
ce =


d
e
e
d′
′ k
d|k

(1.1)

=

e|d

e|k

d |e

X ce φ(k/e) X φ(k/e)
·
=
ce
.
e
k/e
k
e|k

e|k

In the above, φ denotes the Euler totient function. For details on this derivation, the reader is
directed to [15, Chapter 17] .
In this paper, we answer the analogous question for commutators rather than for commutatorsubgroup elements. This new inquiry is structurally different in that it aims to solve a Diophantine equation over a group G (whether, for a given W ∈ G, there exist X, Y ∈ G such that
XY X −1 Y −1 = W ), rather than a subgroup-membership problem (whether W is in [G, G]). In
particular, the set of commutators is not multiplicatively closed, so we cannot use primitive words
as a bridge between counting cyclically reduced words and counting conjugacy classes as above.
Instead, we use a theorem of Wicks [28], which states that an element of Fr is a commutator if and
only if it is a cyclically reduced conjugate of a commutator satisfying the following definition.

CONJUGACY GROWTH OF COMMUTATORS

3

Definition 1.1. A Wicks commutator of Fr is a cyclically reduced word W ∈ Fr of the form
ABCA−1 B −1 C −1 . Equivalently, it is a word of the form ABCA−1 B −1 C −1 such that the subwords
A, B, and C are reduced; there are no cancellations between the subwords A, B, C, A−1 , B −1 , and
C −1 ; and the first and last letters are not inverses.
Using this classification, we count the number of conjugacy classes of commutators in Fr with
length k by counting the number of Wicks commutators with length k.
Theorem 1.2. Let k ≥ 0 be even. The number of distinct conjugacy classes of commutators in Fr
with length k is given by
k

(2r − 2)2 (2r − 1) 2 −1 2
k + Or (k) ,
96r
where the implied constant depends only on r and is effectively computable.
Note that the number of conjugacy classes of commutators in Fr is roughly proportional to the
square root of the number (1.1) of all conjugacy classes with trivial abelianization.
We also employ a similar argument, using Wicks’ characterization of commutators in free products, to answer the analogous question for the free product G1 ∗ G2 of two nontrivial finite groups.
We consider the set of generators S ··= (G1 \ {1}) ∪ (G2 \ {1}) of G1 ∗ G2 . Then, a theorem of
Wicks [28] analogous to the previous one implies that an element of G1 ∗ G2 is a commutator if and
only if it is a cyclic conjugate of a commutator satisfying the following definition.
Definition 1.3. A Wicks commutator of G1 ∗ G2 is a word W ∈ G1 ∗ G2 that is fully cyclically
reduced, which is to say that the adjacent letters (i.e., nonidentity elements of G1 and G2 in the
word) are in different factors of the free product, as are the first and last letters; and in one of the
following forms:
(1) α ∈ G1 that is a commutator of G1 ,
(2) α ∈ G2 that is a commutator of G2 ,
(3) Aα1 Aα−1
2 for α1 , α2 ∈ G2 that are conjugates,
(4) α1 Aα−1
2 A for α1 , α2 ∈ G1 that are conjugates,
(5) ABA−1 B −1 ,
(6) Aα1 Bα2 A−1 α3 B −1 α4 for α1 , α2 , α3 , α4 ∈ G2 satisfying α4 α3 α2 α1 = 1,
(7) α1 Aα2 Bα3 A−1 α4 B −1 for α1 , α2 , α3 , α4 ∈ G1 satisfying α4 α3 α2 α1 = 1,
(8) Aα1 Bβ1 Cα2 A−1 β2 B −1 α3 C −1 β3 for α1 , α2 , α3 , β1 , β2 , β3 ∈ G2 satisfying α3 α2 α1 = 1 and
β3 β2 β1 = 1, with A, B, and C nontrivial,
(9) β1 Aα1 Bβ2 Cα2 A−1 β3 B −1 α3 C −1 for α1 , α2 , α3 , β1 , β2 , β3 ∈ G1 satisfying α3 α2 α1 = 1 and
β3 β2 β1 = 1, with A, B, and C nontrivial.
A fully cyclically reduced element W ∈ G1 ∗ G2 with length k > 1 alternates between k/2 letters
in G1 \ {1} and k/2 letters in G2 \ {1}, where k/2 is necessarily an integer. Thus, if k > 1 is odd,
then there are no fully cyclically reduced elements of G1 ∗ G2 with length k. Furthermore, for W
to be a Wicks commutator not of the form (1) or (2) in Definition 1.3, it is necessary not only that
k is even, but also that k/2 is an even integer. Thus, all but possibly finitely many length-1 Wicks
commutators of G1 ∗ G2 have length divisible by 4. Thus, for any k divisible by 4, we obtain the
number of length-k conjugacy classes in G1 ∗ G2 comprised of commutators.
Theorem 1.4. Let k ≥ 0 be a multiple of 4. The number of distinct conjugacy classes of commutators in G1 ∗ G2 with length k is given by

k
k
1 
(|G1 | − 1) (|G2 | − 2)2 + (|G1 | − 2)2 (|G2 | − 1) k2 (|G1 | − 1) 4 −1 (|G2 | − 1) 4 −1
192


k
k
+O|G1 |,|G2| k(|G1 | − 1) 4 (|G2 | − 1) 4 ,
where the implied constant only depends on |G1 | and |G2 |, and is effectively computable.

4

PETER S. PARK

Suppose that 4 | k > 0. Then, the set of cyclically reduced elements of G1 ∗ G2 (whose letters of
odd position are elements of G1 , without loss of generality) with length k and trivial abelianization
bijectively maps to the Cartesian product of two sets: the set of closed paths of length k/2 on the
complete graph K|G1 | with fixed basepoint P1 , and the set of closed paths of length k/2 on the
complete graph K|G2 | withP
fixed basepoint P2 . For a graph E, the number of closed paths on E
with length n is given by
λn , where λ ranges over the eigenvalues (counted with multiplicity)
of the adjacency matrix of E. For m ≥ 2, the adjacency matrix of Km is the m × m matrix with
diagonal entries 0 and all other entries 1; its eigenvalues are m − 1 (with multiplicity 1) and −1
(with multiplicity m − 1). Consequently, for even integers n > 0, the number of closed paths on
Km with length n is given by (m − 1)n + (m − 1)(−1)n = m(m − 1)n−1 . By symmetry between
the m vertices, the number of such closed paths with a given basepoint is (m − 1)n−1 . Thus, the
number of cyclically reduced words with length k and trivial abelianization in G1 ∗ G2 is given by
k

k

(|G1 | − 1) 2 −1 (|G2 | − 1) 2 −1 .
Applying Möbius inversion as done in (1.1), we see that the number of conjugacy classes of commutators in G1 ∗ G2 with length k is roughly comparable to the square root of the number of all
length-k conjugacy classes with trivial abelianization.
1.2. Geometric applications. Counting conjugacy classes of commutators has a topological application. Let X be a connected CW-complex with fundamental group π1 (X), and let C be a
conjugacy class of π1 (X) with trivial abelianization, corresponding to the free homotopy class of a
homologically trivial loop γ : S 1 → X. Then, the commutator length of C, defined as the minimum
number of commutators whose product is equal to an element of C, is also the minimum genus of
an orientable surface (with one boundary component) that continuously maps to X so that the
boundary of the surface maps to γ [3, Section 2.1]. Thus, a conjugacy class of π1 (X) is comprised of
commutators if and only if its corresponding free homotopy class γ : S 1 → X satisfies the following
topological property:
(⋆)

There exists a torus Y with one boundary component
and a continuous map f : Y → X satisfying f (∂Y ) = Im γ.
Consequently, we obtain the following.
Corollary 1.5. Let X be a connected CW-complex.
(1) Suppose X has fundamental group Fr with a symmetric set of free generators S. Then, the
number of free homotopy classes of loops γ : S 1 → X with length k (in the generators of S)
satisfying Property (⋆) is given by
k


(2r − 2)2 (2r − 1) 2 −1 2
k + Or (k) ,
96r
where the implied constant depends only on r and is effectively computable.
(2) Suppose X has fundamental group G1 ∗ G2 with the set of generators
S ··= (G1 \ {1}) ∪ (G2 \ {1}).

Then, the number of free homotopy classes of loops γ : S 1 → X with length k (in the
generators of S) satisfying Property (⋆) is given by

k
k
1 
(|G1 | − 1) (|G2 | − 2)2 + (|G1 | − 2)2 (|G2 | − 1) k2 (|G1 | − 1) 4 −1 (|G2 | − 1) 4 −1
192


k
k
+O|G1 |,|G2| k(|G1 | − 1) 4 (|G2 | − 1) 4 ,
where the implied constant depends only on |G1 | and |G2 |, and is effectively computable.

CONJUGACY GROWTH OF COMMUTATORS

5

For a connected CW-complex with a natural geometric definition of length for loops, one expects
the word length of a free homotopy class of loops to correlate with the geometric length of the free
homotopy class, taken to be the infimum of the geometric lengths
of its loops, in some way. A
Wr
trivial example of this phenomenon is in the case of a wedge j=1 S 1 of r unit circles, for which
there is perfect correlation; a free homotopy class in this space has word length k if and only if the
minimal geometric length of a loop in the class is k.
A more interesting example of the connection between word length and geometric length arises
from hyperbolic geometry. Let X ∼
= Γ\H be a hyperbolic orbifold, where Γ ⊂ PSL2 (R) denotes
a Fuchsian group. Then, every free homotopy class of loops in X that does not wrap around
a cusp has a unique geodesic representative, so the geometric length of the class is realized as
the geometric length of this unique closed geodesic [2, Theorem 1.6.6]. In particular, we have a
canonical correspondence between hyperbolic free homotopy classes of loops (i.e., ones that do not
wrap around a cusp) and closed geodesics. Let us specialize to the case that X is a pair of pants
(surface obtained by removing three disjoint open disks from a sphere). We have that X ∼
= Γ\H for
Γ∼
F
,
so
fix
a
symmetric
set
of
free
generators.
Then,
Chas,
Li,
and
Maskit
[4]
have
conjectured
= 2
from computational evidence that the geometric length and the word length of a free homotopy
class of loops are correlated in the following fundamental way.
Conjecture 1.6 (Chas–Li–Maskit). For an arbitrary hyperbolic metric ρ on the pair of pants that
makes its boundary geodesic, let Dk denote the set of free homotopy classes with word length k in
a fixed choice of free generators for the fundamental group. Then, there exist constants µ and σ
(depending on ρ) such that for any a < b, the proportion of C ∈ Dk such that the geometric length
h(C) of the unique geodesic representative of C satisfies
h(C) − µk
√
∈ [a, b]
k

converges to
Z b
x2
1
√
e− 2σ2 dx
σ 2π a
as k → ∞. In other words, as k → ∞, the distribution of geometric lengths g(C) for C ∈ Dk
approaches the Gaussian distribution with mean µk and standard deviation σk.
It is natural to expect that even when restricting to commutators, the resulting distribution would
still exhibit a fundamental correlation between geometric length and word length. If this were
proven true, then one could use Corollary 1.5 to indirectly count closed geodesics in X with geometric length ≤ L satisfying Property (⋆).
For a general hyperbolic orbifold Γ\H, we have the aforementioned natural correspondence between the closed geodesics of Γ\H and the hyperbolic conjugacy classes of Γ. In the case that Γ is
finitely generated, we claim that the number of non-hyperbolic conjugacy classes of commutators
with word length k is small. Indeed, Γ only has finitely many elliptic conjugacy classes. Moreover,
Γ only has finitely many primitive parabolic conjugacy classes, corresponding to equivalence classes
of cusps. Denoting the primitive parabolic conjugacy classes by {Ci }1≤i≤n , we can conjugate each
Ci by elements in Γ to assume without loss of generality that the representative gi is fully cyclically
reduced. Then, since every parabolic conjugacy class is a power of a primitive one, we see that the
number of parabolic classes with word length k is at most n.
Suppose further that Γ is isomorphic to Fr (respectively, to G1 ∗G2 for two nontrivial finite groups
G1 and G2 ) and fix a set of generators. We see that the number of non-hyperbolic conjugacy
classes of commutators with length k is bounded by a constant, and thus clearly dominated by
the error term of ≪ k(2r − 1)k/2 (respectively, ≪ k(|G1 | − 1)k/4 (|G2 | − 1)k/4 ). In fact, it is
known [24] that a commutator of a free group cannot be a proper power, which shows that the
number of non-hyperbolic conjugacy classes of commutators is finite in this case. For the latter case

6

PETER S. PARK

of Γ ∼
= G1 ∗ G2 , the work of Comerford, Edmunds, and Rosenberger [5] gives the possible forms
that a proper-power commutator can take in a general free product. After discarding the O(1)
number of non-hyperbolic conjugacy classes of commutators with word length k, one obtains from
Corollary 1.5 the asymptotic number of closed geodesics γ with word length k satisfying Property
(⋆), with the error term unchanged. It is natural to expect for Γ\H a correlation between the
word length and the geometric length similar to that conjectured for the pair of pants. If such a
correlation were proven true, then Corollary 1.5 would yield information on the number of closed
geodesics in Γ\H with length ≤ L satisfying Property (⋆).
For an example of such a hyperbolic orbifold Γ\H whose fundamental group Γ is isomorphic to
G1 ∗ G2 , we introduce the Hecke group H(λ) for a real number λ > 0, defined by the subgroup of
PSL2 (R) generated by




0
−1
1
λ
S ··=
and
Tλ ··=
.
1 0
0 1
Denote λq ··= 2 cos(π/q) for integers q ≥ 3. Hecke [12] proved that H(λ) is Fuchsian if and only if
λ = λq for some integer q ≥ 3 or λ ≥ 2. Specializing to the former case, H(λq ) has the presentation
hS, Rλ : S 2 = Rλq = Ii ∼
= Z/2Z ∗ Z/qZ,
where the generator Rλ can be taken to be STλ [13]. Thus, our desired example H(λq )\H has
fundamental group isomorphic to Z/2Z ∗ Z/qZ. In particular, note that H(λ3 ) = PSL2 (Z) (with
the standard generators S and T1 ), which has a presentation as the free product Z/2Z ∗ Z/3Z.
1.3. Connections to number theory. Counting closed geodesics on hyperbolic orbifolds Γ\H
by geometric length is a well-studied problem that often has analogies (in both the asymptotic
of the resulting counting function and the techniques used to prove it) to counting problems in
number theory. To illustrate, the primitive hyperbolic conjugacy classes of Γ correspond precisely
to the primitive closed geodesics of Γ\H. These are called prime geodesics because when ordered
by trace, they satisfy equidistribution theorems analogous to those of prime numbers, such as the
prime number theorem (generally credited to Selberg [25], while its analogue for surfaces of varying
negative curvature was proven by Margulis [16]) and Chebotarev’s density theorem (proven by
Sarnak [22]). Specifically, this analogue of the prime number theorem is called the prime geodesic
theorem, which states that the number of prime geodesics γ ∈ Γ\H with norm ≤ N is asymptotically
given by ∼ N/ log N , where the norm n(γ) denotes the real 
number ρ> 1 such that the Jordan
±ρ 0
normal form of the PSL2 (R)-matrix corresponding to ℓ is
0 ±ρ−1 , which is related to the
geometric length ℓ(γ) by the relation (see [14, p. 384]) given by
(1.2)



n(γ) = cosh

ℓ(γ)
+
2

s

cosh

ℓ(γ)
2

2

2

− 1 .

Furthermore, Sarnak’s analogue of Chebotarev’s density theorem implies that for every surjective
homomorphism ϕ : Γ → G onto a finite abelian group G, the number of prime geodesics of Γ\H
with norm ≤ N in the inverse image of a given g ∈ G is asymptotically given by ∼ N/(|G| log N ).
In particular, if [Γ, Γ] is of finite index in Γ, then letting φ be the quotient map Γ → Γ/[Γ.Γ],
one sees that the number of prime geodesics of Γ\H with norm ≤ N and trivial abelianization is
asymptotically given by
∼

N
.
|Γ : [Γ, Γ]| log N

CONJUGACY GROWTH OF COMMUTATORS

7

The result of Corollary 1.5 can be seen as the answer to an analogous problem of counting closed
geodesics of Γ\H that correspond to conjugacy classes of commutators rather than arbitrary conjugacy classes with trivial abelianization, the main difference being that our formula counts closed
geodesics ordered by word length.
We mention one more number-theoretic setting which involves counting conjugacy classes of
commutators of a Fuchsian group, specialized to the modular group Γ = PSL2 (Z) = H(λ3 ). A
commutator of Γ is precisely a coset {C, −C} for a commutator C = ABA−1 B −1 of SL2 (Z), where
A, B ∈ SL2 (Z); this choice is unique because −I has nontrivial abelianization, which implies that
−C does, as well. Thus, a conjugacy class of commutators in Γ with such a representative {C, −C}
is the union of the SL2 (Z)-conjugacy class of the SL2 (Z)-commutator C and the SL2 (Z)-conjugacy
class of −C.
For the free group F2 on generators X and Y , the map Hom(F2 , SL2 (C)) → C3 defined by
ρ 7→ (Tr ρ(X), Tr ρ(Y ), Tr ρ(XY )) induces an isomorphism between the moduli space of SL2 -valued
representations of F2 and the affine space C3 , by the work of Vogt [27] and Fricke [8]. Labeling
this parametrization of C3 as (x, y, z), we note that Tr ρ(XY X −1 Y −1 ) is given by the polynomial
M (x, y, z) ··= x2 + y 2 + z 2 − xyz − 2; in other words, we have the trace identity [9, p. 337]
(1.3)

Tr(A)2 + Tr(B)2 + Tr(AB)2 − Tr(A) Tr(B) Tr(AB) − 2 = Tr(ABA−1 B −1 )

for all A, B ∈ SL2 (C). In particular, note that every conjugacy class of commutators in SL2 (Z)
(say, with representative ABA−1 B −1 of SL2 (Z) such that A, B ∈ SL2 (Z)) with a given trace T
gives rise to an integral solution
(Tr(A), Tr(B), Tr(AB))
of the Diophantine equation M (x, y, z) = T . The integral solutions to the Markoff equation
M (x, y, z) = −2 play an important role in the theory of Diophantine approximation through their
relation to extremal numbers via the Lagrange–Markoff spectrum [17, 18], and have been studied
recently in [1] by applying strong approximation. Moreover, integral solutions to the Markoff-type
equation M (x, y, z) = T for varying T are also of independent number-theoretic interest and have
been studied recently in [10].
The conjugacy classes of PSL2 (Z) with trace T correspond to the SL2 (Z)-orbits of binary quadratic forms with discriminant T 2 − 4. This allows one to use Gauss’ reduction theory of indefinite
binary quadratic forms, which yields a full set of representatives for the SL2 (Z)-orbits of binary
quadratic forms of any positive discriminant, to obtain the conjugacy classes of PSL2 (Z) with trace
T . Furthermore, there is a straightforward condition determining whether each such conjugacy
class is in the commutator subgroup [PSL2 (Z), PSL2 (Z)], since [PSL2 (Z), PSL2 (Z)] is precisely
( 



a b
a b
,−
: (1 − c2 )(bd + 3(c − 1)d + c + 3) + c(a + d − 3) ≡ 0 (mod 12)
c d
c d
)
or (1 − c2 )(bd + 3(c + 1)d − c + 3) + c(a + d + 3) ≡ 0 (mod 12)

a congruence subgroup of index 6; this follows from the fact that [SL2 (Z), SL2 (Z)] = ker χ for the
surjective homomorphism χ : SL2 (Z) → Z/12Z defined by


a b
7→ (1 − c2 )(bd + 3(c − 1)d + c + 3) + c(a + d − 3),
c d
as shown in [6, Proof of Theorem 3.8]. However, there is no known number-theoretic condition determining whether a conjugacy class of PSL2 (Z) with trivial abelianization is in fact a commutator;
the only known method to do so is to decompose an element of the conjugacy class into a word in
the generators S and T1 , then combinatorially checking whether there exists a cyclic conjugate of

8

PETER S. PARK

this word in the form given by Definition 1.3. For an implementation of this algorithm in magma,
see [20, Appendix].
Since every commutator of SL2 (Z) with trace T gives rise to an integral point on MT , the
natural follow-up question is whether the converse is true. A local version of this holds, in that
for every power pn of a prime p ≥ 5, there exists a commutator ABA−1 B −1 satisfying (1.3)
mod pn [10, Lemma 6.2]. The global version of the converse, however, does not hold in general,
nor do we know of any number-theoretic condition determining whether an integral point on MT
comes from a commutator. Then, a question of Sarnak [23] asks:
What is the asymptotic number of integral points on MT that arise from SL2 (Z)-commutators by
the trace identity (1.3) as |T | → ∞?
On this front, one can consider SL2 (Z)-commutators as elements of PSL2 (Z) ∼
= Z/2Z ∗ Z/3Z, for
which Theorem 1.4 gives the asymptotic number of conjugacy classes of commutators with a given
word length in the generators. Any correlation between the word length of a hyperbolic conjugacy
class γ of PSL2 (Z) and the geometric length of the corresponding geodesic in Γ\H would also yield
correlation between the word length of γ and its trace t(γ), defined to be the absolute value of the
trace of any representative matrix. This is due to the relation between t(γ) and ℓ(γ) given by
t(γ) = 2 cosh

ℓ(γ)
,
2

which is equivalent to the relation (1.2). Thus, a proof of such a correlation would allow Theorem 1.4
to yield information on the number of integral points on MT and M−T for T > 2 (the condition
for hyperbolicity, where we have without loss of generality taken the representative trace T to be
positive) arising from commutators by (1.3).
1.4. Strategy of proof. The main idea behind the proof of Theorem 1.2 given in Section 2
(respectively, the proof of Theorem 1.4 given in Section 3) is to count the Wicks commutators of Fr
(respectively, those of G1 ∗ G2 ) and show that each such Wicks commutator is on average cyclically
conjugate to precisely six Wicks commutators (including itself) while keeping the overall error term
under control. Thus, we obtain the asymptotic number of conjugacy classes of commutators with
length k by dividing the asymptotic number of Wicks commutators of length k by 6. Finally, in
Section 4, we conclude the paper with a discussion of future directions for generalization.
2. Proof of Theorem 1.2
2.1. Counting the Wicks commutators of Fr . Since the cyclically reduced conjugacy representative of Fr is unique up to cyclic permutation, it suffices to count equivalence classes (with
respect to cyclic permutation) of Wicks commutators of length k = 2X. Let RX denote the set of
reduced words of length X, of which there are 2r(2r − 1)X−1 . For each such word W , the number
of ways to decompose W into A, B, and C (i.e., W = ABC without cancellation) is given by the
number of ordered partitions p of X into three (not necessarily nontrivial) parts.
Define (W, p) to be a viable pair of length n if W is a length-n reduced word, p = (n1 , n2 , n3 ) is
a partition of n into three parts n1 , n2 , n3 > 0, and the word W ′ ··= ABCA−1 B −1 C −1 (obtained
by decomposing W as specified by p, in the method described above) is a Wicks commutator. It is
natural to first count the viable pairs (W, p) of length X. To do so, we will need the following.
Lemma 2.1. In the set of all reduced words of length n beginning with a given letter, the proportion
of words ending in any one letter is given by


1
(2r − 1)n−1 + O(r)
1
1
·
=
+O
.
(2r − 1)n−1
2r
2r
(2r − 1)n−1

CONJUGACY GROWTH OF COMMUTATORS

9

Proof. The number of cyclically reduced words with length n is (2r − 1)n + (r − 1)(−1)n + r
(see [15, Proposition 17.2]). Another way of stating this is that the number an of reduced words of
length n whose initial and final letters are inverses (i.e., are not cyclically reduced) is
an = 2r(2r − 1)n−1 − (2r − 1)n − (r − 1)(−1)n − r = (2r − 1)n−1 − (r − 1)(−1)n − r
Similarly, let bn denote the number of reduced words whose initial and final letters are equal. Note
that {an }n∈Z>0 and {bn }n∈Z>0 are solutions to the recurrence
cn = (2r − 1)cn−2 + (2r − 2)(2r(2r − 1)n−3 − cn−2 ) = cn−2 + 2r(2r − 2)(2r − 1)n−3 .

Indeed, for n ≥ 3, a reduced word whose initial and final letters are inverses is precisely of the
form sW s−1 , and there are 2r − 1 possible choices for the letter s if W also has the aforementioned
property (initial and final letters are inverses) and 2r − 2 possible choices if W does not; an
analogous discussion holds for the property that the initial and final letters are equal. It follows
that the difference {bn −an }n∈Z>0 is a solution to the associated homogeneous recurrence dn = dn−2 ,
and thus a linear combination of the functions 1 and n. In light of the initial data a1 = a2 = 0 and
b1 = b2 = 2r, we see that bn − an = 2r, which yields
bn = (2r − 1)n−1 − (r − 1)(−1)n + r.

For s1 , s2 ∈ S, let Rns1 ,s2 ⊂ Rn denote the subset of words that begin
with s1 and end with s2 ,
S
which gives us a decomposition of Rn into the disjoint union Rn = s1 ,s2 ∈S Rns1 ,s2 . For any s ∈ S,
−1

|Rns,s | =

and
|Rns,s | =

(2r − 1)n−1 − (r − 1)(−1)n − r
an
=
2r
2r

(2r − 1)n−1 − (r − 1)(−1)n + r
bn
=
.
2r
2r

It follows that for t ∈ S \ {s, s−1 },

−1


(2r − 1)n−1 − |Rns,s | − |Rns,s |
1
(2r − 1)n−1 − (r − 1)(−1)n
n−1
=
=
(2r − 1)
−
2r − 2
2r − 2
r
n−1
(2r − 1)
+ O(1)
,
=
2r
which completes the proof of our claim.


|Rns,t |

Now, fix a partition p = (n1 , n2 , n3 ) of n into three parts n1 , n2 , n3 > 0, and consider RX with
the uniform probability measure placed on its elements. Within the decomposition W = ABC in
accordance with p, let the first letter and last letter of A respectively be a0 and a1 , and define b0 , b1 ,
c0 , and c1 similarly. We will compute the probability that (W, p) is viable for a random W ∈ RX ,
−1
i.e., the probability that b1 6= a−1
0 , c0 6= a0 , c1 6= a1 , and c1 6= b0 .
Suppose that a0 = s. Then, by our work above, the set of possible candidates for a1 b0 is the
2r(2r − 1)-element set S = {wz : w, z ∈ S, w 6= z −1 }, each of which has probability





1
1
1
1
1
+O
+O
.
=
2r − 1 2r
(2r − 1)n1 −1
2r(2r − 1)
(2r − 1)n1

We fix a choice of a1 b0 in S, and all probabilities from now on are conditional on this event. The set
of possible candidates for b1 c0 is also S, each of which has probability 1/2r(2r−1)+O(1/(2r−1)n2 ).
Let S ′ ⊂ S be the subset of possible candidates for b1 c0 that satisfy the conditions a0 6= b−1
1 and
a0 6= c0 . The cardinality of S ′ can be computed as follows: there are 2r − 1 choices for b1 satisfying
−1
s 6= b−1
1 , and conditional on this, there are 2r − 2 choices for c0 satisfying s 6= c0 and b0 6= c0 , for a
total of (2r − 1)(2r − 2) elements of S ′ . Fix a choice of b1 c0 in S ′ , and all probabilities from now on

10

PETER S. PARK

−1
are conditional on this event. Since a1 6= b−1
0 , the conditions c1 6= a1 and c1 6= b0 leave precisely
2r − 2 (out of 2r) possible values for c1 , so the probability that c1 satisfies these conditions is



1
1
+O
.
(2r − 2)
2r
(2r − 1)n3 −1

Overall, we have that the probability that (W, p) is viable is



1
1
2r(2r − 1)
+O
2r(2r − 1)
(2r − 1)n1



1
1
+O
· (2r − 1)(2r − 2)
2r(2r − 1)
(2r − 1)n2



1
1
· (2r − 2)
+O
2r
(2r − 1)n3 −1
 



2 

2r − 2
1
1
=
1+O
1+O
2r
(2r − 1)n1 −2
(2r − 1)n2 −2



1
· 1+O
(2r − 1)n3 −2


 

2r − 2 2
1
1
1
+
+
=
.
1+O
2r
(2r − 1)n1 −2 (2r − 1)n2 −2 (2r − 1)n3 −2
Thus, the number of viable pairs (W, p) of length X is given by
2

X
X−1 2r − 2
2r(2r − 1)
2r
0<n ,n ,n
1

2

3

n1 +n2 +n3 =X



· 1+O

1
1
1
+
+
(2r − 1)n1 −2 (2r − 1)n2 −2 (2r − 1)n3 −2

(2r − 2)2 (2r − 1)X−1
=
2r


 (X − 2)(X − 1)
+
·

2
=

0<n1 ,n2 ,n3
n1 +n2 +n3 =X

·
=



O







1
1
1

+
+

n
−2
n
−2
n
−2
(2r − 1) 1
(2r − 1) 2
(2r − 1) 3



(2r − 2)2 (2r − 1)X−1 
 (X − 2)(X − 1) + 3 · O 


2r
2

(2r − 2)2 (2r − 1)X−1
=
2r
=

X

(2r −

2)2 (2r

1)X−1

−
2r
(X − 2)(X − 1)
+O
2

(2r − 2)2 (2r − 1)X−1
4r



(X − 2)(X − 1)
+O
2

X

0<n1 ,n2 ,n3
n1 +n2 +n3 =X

X−2
X




1

(2r − 1)n1 −2 

1
(X − 1 − n1 )
(2r − 1)n1 −2

n1 =1

 !!
(2r − 1)2 (2r − 1)2−X + X(2r − 2) − 4r + 3
(2r − 2)2



r2
2
+ rX
.
X +O
(2r − 1)X

!!

CONJUGACY GROWTH OF COMMUTATORS

11

There could a priori be a Wicks commutator ABCA−1 B −1 C −1 arising from distinct viable pairs,
say (W, p1 ) and (W, p2 ) for p1 = (n1 , n2 , n3 ), and p2 = (m1 , m2 , m3 ). We show that the number of
such commutators is small.
Let W = ABC be its decomposition with respect to p1 , and W = A′ B ′ C ′ its decomposition
with respect to p2 . Consider the function f : {1, . . . , X} → {1, . . . , X}2 that maps i to (j, k) in
the following way: the ith letter of A−1 B −1 C −1 = A′−1 B ′−1 C ′−1 , when viewed in terms of the
decomposition given by p1 , is the inverse of the jth letter of W , and when viewed in terms of the
decomposition given by p2 , is the inverse of the kth letter of W . For example, the first letter of
A−1 B −1 C −1 is defined to be the inverse of the n1 th letter when the decomposition is in terms of
p1 , and is defined to be the m1 th letter of W when in terms of p2 , so f (1) = (n1 , m1 ). We consider
two cases: when the two entries of f (i) are distinct for all i, and otherwise. In the first case, the
following algorithm allows us to reduce the degrees of freedom for the letters of W by at least half.
(1) Let i = 1. For f (i) = (ji , ki ), do the following:
• If neither the ji th or the ki th position has an indeterminate variable assigned to it,
then assign a new indeterminate variable simultaneously to the ji th and ki th positions.
This increases the number of indeterminate variables by two.
• If just one of the ji th and the ki th positions has an indeterminate variable assigned to
it, but the other does not, then assign this indeterminate variable to the former.
• If both the ji th and the ki th positions have the same indeterminate variable assigned
to them, then make no changes.
• If the ji th and the ki th positions have distinct indeterminate variables assigned to
them, then set these indeterminate variables equal to each other. This decreases the
number of indeterminate variables by one.
(2) Incrementing i by one each time, repeat this procedure for all 1 ≤ i ≤ X.
By our hypothesis that the two entries of f (i) are distinct for all i, the number of indeterminate
variables, which precisely represents the number of degrees of freedom for the word W such that
(W, p1 ) and (W, p2 ) give rise to the same commutator, is ≤ X/2. It follows that there are only
O((2r − 1)X/2 ) of such words for each pair p1 , p2 .
Now, consider the next case that there exists an i such that the two entries of f (i) are equal.
Then, we consider the following three cases for W , p1 , and p2 .
Case 1. Suppose the smallest i such that the two entries of f (i) are equal satisfies that this
entry is a position in A. Then, n1 = m1 must be this entry and i must equal 1, since otherwise
we can continue to decrement i so that the two entries of f (i) are incremented and remain equal,
a contradiction. Next, the subwords B −1 C −1 and B ′−1 C ′−1 must be equal. Without loss of
generality, suppose that n2 > m2 . Decompose B = B ′ D so that our condition B −1 C −1 = B ′−1 C ′−1
is precisely CB ′ D = DCB ′ . Since CB ′ and D commute, we have that they are both powers of a
common subword V . We can bound the number of (W, p) satisfying this case by counting, for each
i = X − n1 and for each divisor d | i (denoting the length of V ), the number of ways to divide the
right subword of length i into two subwords B and C, and the number of degrees of freedom. Thus,
the number of double-countable Wicks commutators in this case can be bounded from above by
X−1
XX
i=1 d|i

(i − d + 1)(2r − 1)X−i+d

= (X − 1)(2r − 1)X +

X−1
XX
i=1 d|i
d6=i

(i − d + 1)(2r − 1)X−i+d

12

PETER S. PARK

≤ X(2r − 1) + (2r − 1)

X−1
X

= X(2r − 1)X + (2r − 1)X

X−1
X

≤ X(2r − 1)X + (2r − 1)X

X−1
X

X

≪ X(2r − 1)X ,

X

X

i=2 1≤d≤ i
2

i=2

i=2

(i − d + 1)(2r − 1)−i+d

(2r − 1)−i

X

1≤d≤ 2i

(2r − 1)−i+1

(i − d + 1)(2r − 1)d





i
i
i(2r − 2) (2r − 1) 2 − 2 + 2(2r − 1) (2r − 1) 2 − 1
2(2r − 2)2

which is dominated by our error term.
Case 2. Suppose the smallest i such that the two entries of f (i) are equal satisfies that this entry
is a position in C. An argument symmetric to the one above can be given to show that the above
expression is also an upper bound for the number of (W, p) satisfying this case.
Case 3. Suppose the smallest i such that the two entries of f (i) are equal satisfies that this
entry is a position in B. Without loss of generality, suppose n1 > m1 . Then, f (m1 + 1) =
(n1 − m1 , m1 + m2 ), and by an argument similar to that in Case (1), we have that the simultaneous
entry of the aforementioned f (i) must be n1 + n2 , with n1 − m1 = n3 − m3 so that f (m1 +
1) = (n1 − m1 , n1 + n2 + (n1 − m1 )). Thus, divide W into DEF GH so that |D| + |E| = n1 ,
|G| + |H| = n3 , and |E| = |G|. Then, (W, p1 ) gives rise to the commutator W E −1 D −1 F −1 H −1 G−1 ,
while (W, p2 ) gives rise to the commutator W D −1 G−1 F −1 E −1 H −1 . Since these are equal, it follows
that DE = GD and GH = HE. Note that if a word  satisfies the equality E = G without
cancellation, then  is uniquely determined, since one can inductively identify the letters of  from
left to right (or right to left). It follows that D = H. But this contradicts the assumption that
DEF GHD −1 G−1 F −1 E −1 H −1 is cyclically reduced.
Note that pairs (W, p) such that ni = 0 for some i ∈ {1, 2, 3} are counted in the above cases.
This justifies our assumption of n1 , n2 , n3 > 0 in our earlier counting of the main term by passing to
the task of counting viable pairs of length X, which a priori only accounts for Wicks commutators
ABCA−1 B −1 C −1 such that |A|, |B|, |C| > 0. We have thus shown the following.
Lemma 2.2. The total number of Wicks commutators with length 2X is given by

(2r − 2)2 (2r − 1)X−1
X 2 + Or (X) .
4r

2.2. A conjugacy class of commutators contains six Wicks commutators on average.
We need to count the number of conjugacy classes containing at least one Wicks commutator.
Consider the conjugacy class C of the Wicks commutator W ′ = ABCA−1 B −1 C −1 arising from
(W, p), where p = (n1 , n2 , n3 ) with n1 , n2 , n3 > 0; recall that the Wicks commutator not satisfying
this latter hypothesis are negligible. Note that the minimum-length elements in a conjugacy class
are precisely the cyclically reduced words, and that two cyclically reduced words are conjugate if and
only if they are cyclically conjugate. The Wicks commutators BCA−1 B −1 C −1 A, CA−1 B −1 C −1 AB,
A−1 B −1 C −1 ABC, B −1 C −1 ABCA−1 , and C −1 ABCA−1 B −1 are conjugates of W ′ . We show that
the number of other Wicks commutators in C is on average negligible.
For an arbitrary 1 ≤ ℓ ≤ n3 /2 denoting the number of letters of the conjugation, let C = DEF
be a decomposition without cancellation such that |D| = |F | = ℓ. Label the letters of W ′ by
A = a1 · · · an1 , B = b1 · · · bn2 , D = d1 · · · dℓ , E = e1 · · · en3 −2ℓ , and F = f1 · · · fℓ . Consider the cyclic
conjugate W ′′ ··= D −1 ABDEF A−1 B −1 F −1 E −1 of W ′ . We wish to show that on average, W ′′ is
not a Wicks commutator. Suppose the contrary, that there exists a partition p′ = (m1 , m2 , m3 ) of

CONJUGACY GROWTH OF COMMUTATORS

13

X into three parts such that
W ′′ = D −1 ABDEF A−1 B −1 F −1 E −1 = w1 w2 w3 w1−1 w2−1 w3−1
for subwords w1 , w2 , and w3 of lengths m1 , m2 , and m3 .
Label the letters of A from left to right as a1 , . . . , an1 , and label the letters of B, C, D, E, and F
similarly. We have that w1 , w2 , and w3 are subwords comprised of the letters
(2.1)

−1
d−1
ℓ , . . . , d1 , a1 , . . . , an1 , b1 , . . . , bn2 , d1 , . . . , dℓ , e1 , . . . , en3 −2ℓ .

Then, note that the second half of W ′ can be considered in two forms:
F A−1 B −1 F −1 E −1 = w1−1 w2−1 w3−1 .
Equivalently, this equality can be written as
EF BAF −1 = w3 w2 w1 ,

(2.2)

where we reiterate that we substitute the letters of (2.1), in the correct order, for w3 , w2 , and w1 .
Consider the function g mapping the ordered set of symbols of the left-hand side,
A ··= {e1 , . . . , en3 −2ℓ , f1 , . . . , fℓ , b1 , . . . , bn2 , a1 , . . . , an1 , fℓ−1 , . . . , f1−1 }

to the ordered set B of symbols of the right-hand side of (2.2), which are just the symbols of (2.1)
reordered appropriately. Specifically, g maps the ith leftmost letter of the left-hand side of (2.2) to
the ith leftmost letter of the right-hand side.
First, suppose g has no fixed points (i such that g(i) = i). Then, use an algorithm similar to the
previous one to conclude that there are ≤ X/2 degrees of freedom for ABDEF , so W must be one
of only O((2r − 1)X/2 ) choices (for each choice of ℓ and p′ ).
Now, suppose that there exists an i such that g(i) = i. Such fixed points i must be letters of A,
B, or E. We first consider the case that all the fixed points are letters of only one of A, B, and E.
In this case, we consider the following subcases for W , p, and p′ :
Case 1. Suppose that the fixed points are letters of E. Then, all of the fixed points must be in
one of w2 and w3 ; they cannot be in w1 , since this would mean that w1 contains e1 , but e1 is then
necessarily located at different positions in the left-hand side and right-hand side of (2.2). Suppose
that the fixed points of E are in w3 . Then, in order for the letters of E to match, we require that
w3 = E. This means that g(f1 ) is the first letter of w2 , which is adjacent to the last letter of w1 .
But the last letter of w1 is g(f1−1 ), which shows that we have adjacent letters that are inverses.
This contradicts the fact that W ′ is cyclically reduced.
Next, suppose all the fixed points are in w2 . Then, we must have that m3 = n3 − 2ℓ − (m2 + m3 )
so that the first letter of w2 is at the same position in both the left-hand and right-hand side.
Thus, m2 + 2m3 = n3 − 2ℓ, which means there are ≤ (n3 − 2ℓ)/2 choices for p′ parametrized by
m3 ≤ (n3 − 2ℓ)/2 . For each such choice of p′ , there are m2 = n3 − 2ℓ − 2m3 fixed letters, from
em3 +1 to en3 −2ℓ−m3 , and (X − (n3 − 2ℓ − 2m3 ))/2 non-fixed letters. Counting across all choices
of values for the letters, p, p′ , and ℓ, we have that the number of additional Wicks commutators
arising from this case is bounded from above by
X X−n
X1
X

n1 =0 n2 =0

j

X−n1 −n2
2

X
ℓ=1

kj

n3 −2ℓ
2

X

m3 =0

k

(2r − 1)

X+n3 −2ℓ−2m3
2

≪ (2r − 1)X ,

which is dominated by our error term.
Case 2. Suppose that the fixed points are letters of B. Then, all of the fixed points must be
in one of w1 , w2 , and w3 . First, suppose they are in w1 . Then, note that g(fℓ ) = an1 , but we
also have g(an1 ) is next to g(fℓ−1 ), which leads to the contradiction that a letter cannot equal its
inverse. Second, suppose the fixed letters are in w3 . Then, g(d1 ) = a1 , but a1 is adjacent to d−1
1 , a
contradiction.

14

PETER S. PARK

Thus, the fixed points of B must be in w2 . We consider three subcases: n2 > m2 , n2 < m2 , and
n2 = m2 . If n2 > m2 , then in order for the letters of B to match, we require that the leftmost
fixed letter of B is b n2 −m2 +1 . But then b n2 −m2 is both equal to f1−1 (since g(f1−1 ) = b n2 −m2 ) and
2

2

2

en3 −2ℓ (since g(b n2 −m2 ) = en3 −2ℓ ) as letters of G, which contradicts the fact that f1 and en3 −2ℓ are
2

adjacent. If n2 < m2 , then g(fℓ ) = an1 , but also g(an1 ) is the letter in D −1 A that is left of the
letter g(fℓ−1 ), giving us the contradiction that the letter in G in the position fℓ is adjacent to the
letter in the position fℓ−1 . This implies that n2 = m2 , from which we can use an argument similar
to that in Case 1 of the previous casework (showing that W ′ on average can be only decomposed
as a commutator in one way) to conclude that A is a power of D −1 and E, a power of D. It follows
that our original (W, p) is one of the pairs falling under Case 1 of the previous casework, which are
negligible.
Case 3. Suppose that the fixed points are letters of A. Then, all of the fixed points must be in
one of w1 and w2 ; they cannot be in w3 , since then there must be more than ℓ letters right of A.
Suppose the fixed letters of A are in w1 . Then, we must have g(bn2 ) = d−1
1 , which contradicts the
fact that bn2 is adjacent to d1 . Therefore, the fixed points are necessarily in w2 . This requires that
m1 − ℓ = n1 + ℓ − (m1 − m2 ) in order for the letters of A to be in matching positions. Thus, we
have m2 = n1 + 2ℓ − 2m1 . Note then that p′ is parametrized by m1 ≤ (n1 + 2ℓ)/2. For each choice
of p′ , we have n1 fixed letters (and (X + n1 )/2 ≤ (X + n1 − m1 + ℓ)/2 overall degrees of freedom)
if m1 ≤ ℓ, and n1 − (m1 − ℓ) fixed letters (and (X + n1 − m1 + ℓ)/2 overall degrees of freedom) if
m1 > ℓ. Thus, counting across all choices of values for the letters, p, p′ , and ℓ, we have that the
number of additional Wicks commutators arising from this case is bounded from above by
X X−n
X1
X

n1 =0 n2 =0

j

X−n1 −n2
2

X
ℓ=1

kj

n1 +2ℓ
2

X

m1 =0

k

(2r − 1)

X+n1 +ℓ−m1
2

≪ (2r − 1)X ,

which is dominated by our error term.
Next, consider the case where the fixed points are in two of A, B, and E. It is necessary that
the fixed letters inside these two subwords must respectively be in two distinct subwords among
w1 , w2 , and w3 . However, we have shown above that the subwords w1 and w3 cannot contain fixed
points, a contradiction. Finally, the fixed letters cannot be in all of A, B, and E. Indeed, if this
were true, then in order for the letters of A and E to match, we require m1 = n1 + 2ℓ and m3 = n3 .
But then the letters of B cannot possibly match, a contradiction.
If ℓ ≥ n3 /2, then we can think of our commutator as a cyclic conjugate of C −1 ABCA−1 B −1 such
that the letters are moved from left to right. A symmetric argument like above gives us the same
conclusion for this case. We have thus shown that the number of conjugacy classes of commutators
with length 2X is given by

1 (2r − 2)2 (2r − 1)X−1
·
X 2 + Or (X)
6
4r

(2r − 2)2 (2r − 1)X−1
X 2 + Or (X) ,
=
24r
as needed.
3. Proof of Theorem 1.4
3.1. Counting the Wicks commutators of G1 ∗ G2 . In addition to his theorem classifying
commutators of free groups, Wicks [28] also proved the following analogous theorem characterizing
all commutators of a free product of arbitrary groups.
Theorem 3.1 (Wicks). A word in ∗i∈I Gi is a commutator if and only if it is a conjugate of one
of the following fully cyclically reduced products:

CONJUGACY GROWTH OF COMMUTATORS

15

(a) a word comprised of a single letter that is a commutator in its factor Gi ,
(b) Xα1 Xα−1
2 , where X is nontrivial and α1 , α2 belong to the same factor Gi as conjugate
elements,
(c) Xα1 Y α2 X −1 α3 Y −1 α4 , where X and Y are both nontrivial, α1 , α2 , α3 , α4 belong to the same
factor Gi , and α4 α3 α2 α1 is trivial,
(d) XY ZX −1 Y −1 Z −1 ,
(e) XY α1 ZX −1 α2 Y −1 Z −1 α3 , where Y and at least one of X and Z is nontrivial, α1 , α2 , α3
belong to the same factor Gi , and α3 α2 α1 is trivial,
(f ) Xα1 Y β1 Zα2 X −1 β2 Y −1 α3 Z −1 β3 , where α1 , α2 , α3 belong to the same factor Gi and β1 , β2 , β3 ,
to Gj , α3 α2 α1 = β3 β2 β1 = 1, and either α1 , α2 , α3 , β1 , β2 , β3 are not all in the same factor
or X, Y, Z are all nontrivial.
Note that in the above, the Greek letters are assumed to be nontrivial. This convention is used
later in our proof, where when a Greek letter α is said to satisfy α ∈ Gi , we mean that α is a
nontrivial element of Gi .
Equipped with a complete classification of commutators in an arbitrary free product, we proceed
with our proof of Theorem 1.4. Let G1 ∗ G2 be the free product of two nontrivial finite groups G1
and G2 . The letters of a cyclically reduced word of G must alternate between equal numbers of
elements of G1 \ {1} and elements of G2 \ {1}, and thus must be of even length. Let C be a fully
reduced commutator of G1 ∗ G2 . If C is of the form (d) listed in Theorem 3.1 with none of X, Y,
and Z trivial, then the last letter of X is in different factors compared to the first letter of Y and
the last letter of Z, which must also be in different factors, a contradiction. If C is of the form (e),
then α1 , α2 , and α3 must be in the same free-product factor, but this would imply that the last
letter of Y is in different free factors compared to the first letter of Z and the first letter of X; this
contradicts the similar implication that the first letter of Z and the first letter of X are in different
factors. Thus, C must be of the form (a), (b), (c), (f ), or XY X −1 Y −1 (i.e., of the form (d) with
|Z| = 0). However, if C is of the form (f ), then Gi and Gj must be the same free-product factor,
since otherwise α1 would be adjacent to letters of distinct free-product factors, a contradiction.
By Wicks’ theorem for free products, we need to count cyclic conjugacy classes of Wicks commutators of G1 ∗ G2 . It follows from the discussion above that every commutator of W is conjugate to
one of the fully cyclically reduced forms listed under Definition 1.3. These general forms, labeled
from (1) to (9), have without loss of generality been taken to have the letters of odd position be in
G1 and those of even position in G2 . Throughout this proof, we will regularly use the terminology
Wicks commutators of the form (i), where 1 ≤ i ≤ 9, to refer to the corresponding general form
listed under Definition 1.3.
Consider Wicks commutators of G1 ∗ G2 with length k, where k is a multiple of 4. Let X = k/4,
so that the left-half subword of W contains X letters of the G1 factor and X letters of the G2 factor,
which are placed in alternating order. The commutators of the forms (1) and

 (2) are O(1) in number
and all have length 1. The commutators of the forms (3) and (4) are O (|G1 | − 1)X (|G2 | − 1)X
in number, since there are ≤ X degrees of freedom among
the letters of each free-product
factor.


X
X
in number, since
The commutators of the forms (5), (6), and (7) are O X(|G1 | − 1) (|G2 | − 1)
there are ≤ X degrees of freedom for the X letters of each free-product factor and O(X) possible
pairs of values for |A| and |B|, by an argument similar to that used in Section 2.
The number of Wicks commutators of the form (8) is
(X − 5)(X − 4)
(|G2 | − 2)2 (|G1 | − 1)X (|G2 | − 1)X−1 .
2
Indeed, we have (|G2 | − 1) (|G2 | − 2) distinct choices of the triple (α1 , α2 , α3 ), since there are |G2 |−1
choices for α1 , |G2 | − 2 choices for α2 6= α−1
1 , and α3 is uniquely determined from the previous
choices. Likewise, there are (|G2 | − 1) (|G2 | − 2) distinct choices of the triple (β1 , β2 , β3 ). Finally,

16

PETER S. PARK

there are (X − 5)(X − 4)/2 partitions of X − 3 into three nontrivial parts (n1 , n2 , n3 ) such that
|A| = 2n1 + 1, |B| = 2n2 + 1, and |C| = 2n3 + 1; X degrees of freedom for choosing the G1 -letters
of A, B, and C; and X − 3 degrees of freedom for choosing the G2 -letters of A, B, and C. By an
analogous argument, the number of Wicks commutators of the form (9) is
(X − 5)(X − 4)
(|G1 | − 2)2 (|G1 | − 1)X−1 (|G2 | − 1)X
2
We define a generic Wicks commutator of G1 ∗ G2 to be one of the form (8) or (9). These comprise
the main term of the total number of Wicks commutators of length k = 4X, since we have seen
above that the non-generic Wicks commutators, those of the forms (1) − (7), comprise a negligible
subset. Overall, we have shown the following.
Lemma 3.2. The total number of Wicks commutators with length 4X is given by

1
(|G1 | − 1) (|G2 | − 2)2 + (|G1 | − 2)2 (|G2 | − 1) X 2 (|G1 | − 1)X−1 (|G2 | − 1)X−1
2



+O X(|G1 | − 1)X (|G2 | − 1)X .

3.2. A conjugacy class of commutators contains six Wicks commutators on average. We
need to count the number of conjugacy classes containing at least one generic Wicks commutator. As
before, let C be the conjugacy class of the Wicks commutator W ··= Aα1 Bβ1 Cα2 A−1 β2 B −1 α3 C −1 β3 ,
with (n1 , n2 , n3 ) a partition of X − 3 and |A| = 2n1 + 1, |B| = 2n2 + 1, and |C| = 2n3 +
1. We wish to show that on average, C does not contain generic Wicks commutators other
than the six obvious ones: W , Bβ1 Cα2 A−1 β2 B −1 α3 C −1 β3 Aα1 , Cα2 A−1 β2 B −1 α3 C −1 β3 Aα1 Bβ1 ,
A−1 β2 B −1 α3 C −1 β3 Aα1 Bβ1 Cα2 , B −1 α3 C −1 β3 Aα1 Bβ1 Cα2 A−1 β2 , and C −1 β3 Aα1 Bβ1 Cα2 A−1 β2 B −1 α3 .
We wish to show that any other cyclic conjugate W ′ of W is not on average a Wicks commutator.
Suppose the contrary. One of the ways this can happen is if there is such a W ′ that is of the
form (8), i.e, there exists a partition p′ = (m1 , m2 , m3 ) of X − 3 into three parts such that
W ′ = w1 α′1 w2 β1′ w3 α′2 w1−1 β2′ w2−1 α′3 w3−1 β3′

for subwords w1 , w2 , and w3 of lengths 2m1 +1, 2m2 +1, and 2m3 +1 given by p′ , and α′1 , α′2 , α′3 , β1′ , β2′ , β3′ ∈
G2 . We will see that the argument showing that these exceptions are negligible also shows that the
exceptions such that W ′ is of the other Wicks-commutator forms are also negligible.
Suppose the number of letters of the conjugation is 2ℓ, where ℓ ≤ (n3 + 1)/2 is arbitrary. For
the desired uniformity of our presented argument, we assume that ℓ > 0, although the exceptions
in the ℓ = 0 case can be bounded similarly. We decompose C = DEF without cancellation so that
|D| = |F | = 2ℓ − 1. Label the letters of W by A = a1 ā1 a2 ā2 · · · ān1 an1 +1 , B = b1 b̄1 b2 b̄2 · · · b̄n2 bn2 +1 ,
D = d1 d¯1 · · · d¯ℓ−1 dℓ , E = ē1 e2 ē2 · · · en3 −2ℓ+2 ēn3 −2ℓ+2 , and F = f1 f¯1 · · · f¯ℓ−1 fℓ ; note that the barred
letters denote the G2 -letters and the non-barred letters, the G1 -letters. Consider the cyclic conjugate W ′ ··= D −1 β3 Aα1 Bβ1 DEF α2 A−1 β2 B −1 α3 F −1 E −1 of W . We wish to show that on average,
W ′ is not a Wicks commutator.
The subwords w1 , w2 , and w3 are comprised of the letters
(3.1)

−1
d−1
ℓ , . . . , d1 , β3 , a1 , . . . , an1 +1 , α1 , b1 , . . . , bn2 +1 , β1 , d1 , . . . , dℓ , ē1 , . . . , ēn3 −2ℓ+2 ,

except three of these letters instead correspond to α′1 , α′2 , and α′3 and thus omitted. Then, note
that the second half of W ′ can be considered in two forms:
F α2 A−1 β3 B −1 α3 F −1 E −1 = w1−1 β2′ w2−1 α′3 w3−1 β3′ .
Equivalently, this equality can be written as
(3.2)

−1
−1 −1
′−1
EF α−1
= β3′−1 w3 α′−1
3 Bβ3 Aα2 F
3 w2 β2 w1 ,

where we reiterate that we substitute the appropriate letters of (3.1), in the correct order, for
w3 , w2 , and w1 .

CONJUGACY GROWTH OF COMMUTATORS

17

Consider the function g mapping the ordered set of symbols of the left-hand side of (3.2),
−1
−1
−1
−1
A ··= {ē1 , . . . , ēn3 −2ℓ+2 , f1 , . . . , fℓ , α−1
3 , b1 , . . . , bn2 +1 , β3 , a1 , . . . , an1 +1 , α2 , fℓ , . . . , f1 }

to the ordered set B of symbols of the right-hand side of (3.2), which when reordered are comprised of
the letters of (3.1) except we replace the (2m1 +2)th, (2m1 +2m2 +4)th, and (2m1 +2m2 +2m3 +6)th
′−1
leftmost letters in (3.1) with β2′−1 , α′−1
3 , and β3 ; note that the (2m1 + 2m2 + 2m3 + 6)th letter is
always ēn3 −2ℓ+2 . Specifically, g maps the ith leftmost letter of the left-hand side of (3.2) to the ith
leftmost letter of the right-hand side.
First, suppose g has no fixed points (i such that g(i) = i). Then, use an algorithm similar to the
one used in Section 2 to conclude that there are ≤ X/2 degrees of freedom
for the G1 -letters of


ABDEF , and likewise for the G2 -letters. So, W must be one of only O (|G1 | − 1)X/2 (|G2 | − 1)X/2

choices (for each choice of ℓ and p′ ).
Now, suppose that there exists an i such that g(i) = i. Such fixed points i must be letters of A,
B, or E. We first consider the case that all the fixed points are letters of only one of A, B, and E.
In this case, we consider the following subcases for W , p, and p′ :
Case 1. Suppose that the fixed points are letters of E. Then, all of the fixed points must be
in one of w2 and w3 ; they cannot be in w1 since this would mean that w1 contains e1 , but e1 is
necessarily located at different positions in the left-hand side and right-hand side of (3.2). If all the
fixed points are in w2 , then we require that
1 + (2m3 + 1) + 1 = (2n3 − 4ℓ + 3) − ((2m2 + 1) + 1 + (2m3 + 1) + 1) ,

in order for the first letter of w2 to be at the same position in both the left-hand and right-hand
side. Hence, we have m2 + 2m3 = n3 − 2ℓ − 2, which means there are ≤ (n3 − 2ℓ − 2)/2 choices for
p′ parametrized by m3 ≤ (n3 − 2ℓ − 2)/2 . For each such choice of p′ , there are
2n3 − 4ℓ + 3 − 2 ((2m3 + 1) + 1 + 1) = 2n3 − 4ℓ − 4m3 − 3

fixed letters excluding the first 2m3 + 3 and the last 2m3 + 3 letters of E, with n3 − 2ℓ − 2m3 − 1
of them in G1 and n3 − 2ℓ − 2m3 − 2 in G2 . There are ≤ (X − (n3 − 2ℓ − 2m3 − 1)) /2 degrees
of freedom for the non-fixed letters in G1 and ≤ (X − (n3 − 2ℓ − 2m3 − 2)) /2 degrees of freedom
for the fixed letters in G2 . Thus, counting across all choices of values for the letters, p, p′ , and ℓ,
we have that the number of additional Wicks commutators arising from this case is bounded from
above by
X−3
X X−3−n
X 1

n1 =0

n2 =0

j

X−2−n1 −n2
2

X

kj

n3 −2ℓ−2
2

X

m3 =0

ℓ=1

k

(|G1 | − 1)

X+n3 −2ℓ−2m3 −1
2

(|G2 | − 1)

X+n3 −2ℓ−2m3 −2
2

≪ (|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, suppose the fixed letters are in w3 . Then, it is necessary that g(ē1 ) = β3′−1 and w3 is given
by E with the first and last letters omitted. Thus, we have that m3 = n3 − 2ℓ, so the number of
possible choices for p′ is at most the number of partitions of X − 3 − n3 + 2ℓ into two nontrivial
parts, which is X − 3 − n3 + 2ℓ. For each choice of p′ , there are n3 − 2ℓ + 1 fixed letters in G1 and
n3 − 2ℓ fixed letters in G2 . Additionally, there are ≤ (X − (n3 − 2ℓ + 1)) /2 degrees of freedom for
the non-fixed letters in G1 and ≤ (X − (n3 − 2ℓ)) /2 degrees of freedom for the non-fixed letters in
G2 . Counting across all choices of values for the letters, p, p′ , and ℓ, we have that the number of
additional Wicks commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

X−2−n1 −n2
2

X
ℓ=1

k

(X − 3 − n3 + 2ℓ)(|G1 | − 1)

X−n3 +2ℓ−1
2

(|G2 | − 1)

X−n3 +2ℓ
2

18

PETER S. PARK

≪ (|G1 | − 1)X (|G2 | − 1)X ,
which is dominated by our error term.
Case 2. Suppose that the fixed points are letters of B. Then, all of the fixed points must be in
one of w1 , w2 , and w3 . First, suppose they are in w1 . This requires that w1 = D −1 β3 Aα1 Bβ1 V ,
−1 ). All letters
where V is the left subword of DE having length 2n1 + 2ℓ + 1 (the length of Aα−1
2 F
of B are thus included in w1 . We have
2m1 + 1 = (2ℓ − 1) + 1 + (2n1 + 1) + 1 + (2n2 + 1) + 1 + (2n1 + 2ℓ + 1) = 4n1 + 2n2 + 4ℓ + 5,
so m1 = 2n1 + n2 + 2ℓ + 2. Thus, the number of possible choices for p′ is at most the number of
partitions of X − 3 − 2n1 − n2 − 2ℓ − 2 into two nontrivial parts, which is X − 5 − 2n1 − n2 − 2ℓ ≤
X − 3 − n1 − n2 − 2ℓ (the latter is guaranteed to be nonnegative for any choice of p). The fixed
letters are precisely the letters of B, so for each choice of p′ , we have n2 + 1 fixed G1 -letters, n2
fixed G2 -letters, ≤ (X − n2 − 1)/2 degrees of freedom for the non-fixed G1 -letters, and ≤ (X − n2 )/2
degrees of freedom for the non-fixed G2 -letters. Counting across all choices of values for the letters,
p, p′ , and ℓ, we have that the number of additional Wicks commutators arising from this case is
bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

X−2−n1 −n2
2

k

X

(X − 3 − n1 − n2 − 2ℓ)(|G1 | − 1)

ℓ=1

X+n2 +1
2

(|G2 | − 1)

X+n2
2

≪ (|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, suppose that the fixed points are in w2 . Then, we require that the difference between the
−1
−1 −1
lengths of EF α−1
(length 2n1 + 2ℓ + 2) is the same as that
3 (length 2n3 − 2ℓ + 3) and β3 Aα2 F
′−1
′−1
′−1
between β3 w3 α3 (length 2m3 + 3) and β2 w1 (length 2m1 + 2). Furthermore, the number of
(fixed) letters of B in w2 is 2n2 + 1 if
j ··=

2n3 − 2ℓ + 3 − (2m3 + 3)
2n1 + 2ℓ + 2 − (2m1 + 2)
=
2
2

is negative and 2(n2 − j) + 1 if j ≥ 0. First, suppose that j ≥ 0. In this case, p′ is determined by
the choice of j ≤ n2 /2, for which there are n2 − 2j + 1 fixed G1 -letters and n2 − 2j fixed G2 -letters
of B. There are ≤ (X − (n2 − 2j + 1)) /2 degrees of freedom for the non-fixed letters of G1 and
≤ (X − n2 + 2j)/2 degrees of freedom for the non-fixed letters of G2 , so overall, we can count
across all choices of values for the letters, p, j, and ℓ to get that the number of additional Wicks
commutators arising from this case is bounded from above by
n3
X−3
2 ⌋
X 1 ⌊X
X X−3−n

n1 =0

n3 =0

ℓ=1

j

X−3−n1 −n3
2

X
j=0

k

(|G1 | − 1)

X+n2 −2j+1
2

(|G2 | − 1)

X+n2 −2j
2

≪ (|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Now, suppose that j < 0. In this case, −j = m1 −n1 −ℓ = m3 −n3 +ℓ is a positive integer satisfying
(n1 +ℓ−j)+(n3 −ℓ−j) = m1 +m3 ≤ X−3, which gives the condition −j ≤ (X−3−n1 −n3 )/2 = n2 /2.
Note that p′ is determined by the choice of −j, for which there are n2 + 1 fixed G1 -letters and n2
fixed G2 -letters, all of which are in B. The non-fixed letters in G1 have ≤ (X − n2 − 1)/2 degrees
of freedom and those in G2 have ≤ (X − n2 )/2 degrees of freedom. Overall, we can count across
all choices of values for the letters, p, −j, and ℓ to get that the number of additional Wicks

CONJUGACY GROWTH OF COMMUTATORS

19

commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

j

X−2−n1 −n2
2

n2 =0

X
ℓ=1

k

n2
⌊X
2 ⌋

−j=1

(|G1 | − 1)

X+n2 −2j+1
2

(|G2 | − 1)

X+n2 −2j
2

≪(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Finally, suppose that the fixed points are in w3 . This requires that w3 is given by V Bβ1 DE with
the last letter omitted, where V is the right subword of D −1 β3 Aα1 having length (2n3 − 4ℓ + 3) +
(2ℓ − 1) = 2n3 − 2ℓ + 2, one less than that of EF α−1
3 . All letters of B are thus included in w3 . We
have that
2m3 + 1 = (2n3 − 2ℓ + 2) + (2n2 + 1) + 1 + (2ℓ − 1) + (2n3 − 4ℓ + 2) = 2n2 + 4n3 − 4ℓ + 5,

so m3 = n2 + 2n3 − 2ℓ + 2. Thus, the number of possible choices for p′ is at most the number of
partitions of X − 5 − n2 − 2n3 + 2ℓ into two nontrivial parts, which is X − 5 − n2 − 2n3 + 2ℓ ≤
X − 3 − n2 + 2ℓ (the latter is guaranteed to be nonnegative for any choice of p). The fixed letters
are precisely the letters of B, so there are n3 + 1 fixed letters of G1 and n3 fixed letters of G2 , with
the non-fixed G1 -letters having ≤ (X − n2 − 1)/2 degrees of freedom and the non-fixed G2 -letters
having ≤ (X − n2 )/2 degrees of freedom. Counting across all choices of values for the letters, p, p′ ,
and ℓ, we have that the number of additional Wicks commutators arising from this case is bounded
from above by For each choice of p′ ,
X−3
X 1
X X−3−n

n1 =0

j

X−2−n1 −n2
2

X

n2 =0

ℓ=1

k

(X − 3 − n2 + 2ℓ)(|G1 | − 1)

X+n2 +1
2

(|G2 | − 1)

X+n2
2

≪(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Case 3. Suppose that the fixed points are letters of A. Then, all of the fixed points must be in
one of w1 and w2 ; they cannot be in w3 , since then there would be more than 2ℓ letters right of A.
First, suppose they are in w1 . This requires that w1 = D −1 β3−1 AV , where V is the left subword
−1
−1
of α−1
1 Bβ1 DE having length 2ℓ, the length of α2 F . The fixed letters are then precisely the
letters of A, so there are n1 + 1 fixed letters of G1 and n1 fixed letters of G2 . The non-fixed G1 letters have ≤ (X − n1 − 1)/2 degrees of freedom and the non-fixed G2 -letters have ≤ (X − n1 )/2
degrees of freedom.The number of possible choices for p′ is at most the number of partitions of
X − 3 − m1 = X − 3 − (2ℓ + 2n1 + 1 + 2ℓ) = X − 4 − 2ℓ − n1 into two nontrivial parts, which
is ≤ X − 3 − 2ℓ − n1 (the latter is guaranteed to be nonnegative for any choice of p). Counting
across all choices of values for the letters, p, p′ , and ℓ, we have that the number of additional Wicks
commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

X−2−n1 −n2
2

X
ℓ=1

k

(X − 3 − 2ℓ − n1 )(|G1 | − 1)

X+n1 +1
2

(|G2 | − 1)

X+n1
2

≪(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Finally, suppose the fixed points are in w2 . Since the length difference 2m1 − 2ℓ + 2 between the
lengths of D −1 and w1 must be equal to the length difference 2n1 + 2ℓ − 2m1 − 2m2 − 2 between
the lengths of D −1 β3 A and w1 α′1 w2 , we require that m2 = n1 + 2ℓ − 2m1 − 2. Note then that p′
is parametrized by m1 ≤ (n1 + 2ℓ − 2)/2. For a given choice of p′ , if m1 ≤ ℓ, then we have n1 + 1
fixed letters of G1 and n1 fixed letters of G2 , with ≤ (X − n1 − 1)/2 degrees of freedom for the

20

PETER S. PARK

non-fixed G1 -letters and ≤ (X − n1 )/2 degrees of freedom for the non-fixed G2 -letters. This gives
≤ (X + n1 + 1)/2 ≤ (X + n1 − m1 + ℓ + 1)/2 overall degrees of freedom for the G1 -letters, as well
as ≤ (X + n1 − m1 + ℓ)/2 ones for the G2 -letters. If m1 > ℓ, then we have 2n1 − 2(m1 − ℓ) + 1 fixed
letters of G1 and 2n1 −2(m1 −ℓ) fixed letters of G2 , so analogously we have ≤ (X +n1 −m1 +ℓ+1)/2
overall degrees of freedom for the G1 -letters, as well as ≤ (X + n1 − m1 + ℓ)/2 ones for for the
G2 -letters. Thus, counting across all choices of values for the letters, p, p′ , and ℓ, we have that the
number of additional Wicks commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

j

X−2−n1 −n2
2

kj

X

n2 =0

n1 +2ℓ−2
2

X

m1 =0

ℓ=1

k

(|G1 | − 1)

X+n1 −m1 +ℓ+1
2

(|G2 | − 1)

X+n1 −m1 +ℓ
2

≪(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, we suppose that the fixed letters of g are in two of the three subwords A, B, and E.
Consider the following subcases:
Case 1. Suppose the fixed letters of g are in A and B. It is necessary that the fixed letters
of A and those of B are in wi and wj , respectively, such that i < j; otherwise, the fixed letters
of A would come before the fixed letters of B, a contradiction. First, suppose that the fixed
letters of B are in w2 , which implies that the fixed letters of A are in w1 . Then, we require
that w1 = D −1 β3 AV , where V is the left subword of α1 Bβ1 DE having length 2ℓ. This gives
2m1 + 1 = 2n1 + 4ℓ + 1, i.e, m1 = n1 + 2ℓ. Furthermore, since we have fixed letters of B, we require
that V does not include all of B, i.e., 2n2 + 2 > 2ℓ, or equivalently, n2 ≥ ℓ. Next, for the fixed
letters of B to match in position, we require that w2 ends at the letter b2n2 −2ℓ+1 , which gives us
2m2 + 1 = (2n2 − 2ℓ + 1) − (2ℓ + 1) + 1 = 2n2 − 4ℓ + 1, i.e., m2 = n2 − 2ℓ ≥ 0. Then, m3 is
automatically determined, and in particular, w3 is the subword of bn2 −ℓ+2 · · · bn2 +1 β1 DE omitting
the rightmost letter. For this p′ corresponding to p, we have 2n1 + 1 fixed letters of A (n1 + 1
letters of G1 and n1 letters of G2 ) and 2n2 − 4ℓ + 1 fixed letters of B (n2 − 2ℓ + 1 letters of G1 and
n2 − 2ℓ letters of G2 ). Next, we bound the degrees of freedom of the non-fixed letters. Note that
′−1
g maps the letters of EF α−1
3 b1 · · · bℓ−1 b̄ℓ−1 to those of β3 bn2 −ℓ+2 · · · bn2 +1 β1 DE in order, but g
−1
also maps fℓ−1 , . . . , f¯2−1 f1−1 to b1 , . . . , b̄ℓ−1 , bℓ and bn2 −ℓ+2 , . . . , bn2 +1 , to d−1
ℓ , . . . , d1 . Thus, arguing
inductively by translation, we see that that choosing the letters of F determines the letters of E,
and thus also determines those of D, thereby determining all non-fixed letters (while not caring
about the constant number of αi and βi letters). There are ℓ G1 -letters and ℓ − 1 G2 -letters in F .
Counting across all choices of values for the letters, p, and ℓ, we have that the number of additional
Wicks commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

X−2−n1 −n2
2

X
ℓ=1

k

(|G1 | − 1)(n1 +1)+(n2 −2ℓ+1)+ℓ (|G2 | − 1)n1 +(n2 −2ℓ)+(ℓ−1)
≪X(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, suppose that the fixed letters of B are in w3 . Then, we require that w3 = V Bβ1 DE, where
V is the right subword of D −1 β3 Aα1 having length 2n3 −2ℓ+2, the length of EF . Furthermore, since
we have fixed letters of A, we require that V does not include all of A, i.e., 2n3 − 2ℓ + 2 < 2n1 + 2, or
in other words, n3 < n1 +ℓ. Next, for the fixed letters of A to match in position, we require that they
are in w2 , and specifically that w2 = an3 −ℓ+1 · · · an1 −n3 +ℓ . This gives us 2m2 +1 = 2n1 −4n3 +4ℓ−1,
and thus m2 = n1 − 2n3 + 2ℓ − 1 > 0. It follows that w1 = D −1 β3 a1 · · · an3 −ℓ . In particular, m1
is automatically determined, and for this p′ corresponding to p, we have n2 + 1 fixed G1 -letters
and n2 fixed G2 -letters of B, as well as n1 − 2n3 + 2ℓ fixed G1 -letters and n1 − 2n3 + 2ℓ − 1 fixed

CONJUGACY GROWTH OF COMMUTATORS

21

G2 -letters of A. Next, we bound the degrees of freedom of the non-fixed letters. Note that g maps
−1 to those of D −1 β a · · · a
the letters of an1 −n3 +ℓ+1 · · · an1 α−1
3 1
n3 −ℓ in order. However, g maps
2 F
d1 , . . . , dℓ , ē1 , . . . , en3 −2ℓ+1 to a1 , ā1 , . . . , an3 −ℓ+1 , ān3 −ℓ+1 . Also, g maps an1 −n3 +ℓ+1 , · · · , an1 +1 , to
e2 , . . . , ēn3 −2ℓ+2 , f1 , . . . , fℓ . Thus, arguing inductively by translation, we see that that choosing the
letters of F determines the letters of E, and thus also determines those of D, thereby determining
all non-fixed letters (while not caring about the constant number of αi and βi letters). Counting
across all choices of values for the letters, p, and ℓ, we have that the number of additional Wicks
commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

k

X−2−n1 −n2
2

(|G1 | − 1)(n1 −2n3 +2ℓ)+(n2 +1)+ℓ (|G2 | − 1)(n1 −2n3 +2ℓ−1)+n2 +(ℓ−1)

X
ℓ=1

≪X(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Case 2. Suppose the fixed letters of g are in B and E. Similarly to before, it is necessary that
the fixed letters of B and those of E are in wi and wj , respectively, such that i < j. First, suppose
that the fixed letters of B are in w1 . Then, we require that w1 = D −1 β3 Aα1 BV , where V is the
−1 . Then, w must start
left subword of β1 DE having length 2n1 + 2ℓ + 2, the length of β3−1 Aα−1
2
2 F
with en1 +3 , which means in order to have the letters of E match, we must have 2m3 + 3 = 2n1 + 3,
i.e., m3 = n1 . Since (2n1 + 2) + 1 + (2m2 + 1) + 1 + (2m3 + 1) + 1 = 2n3 − 4ℓ + 3 (counting the
letters of E in two ways), we have m2 = n3 − 2n1 − 2ℓ − 2 > 0. There are n3 − 2n1 − 2ℓ − 1
fixed G1 -letters and n3 − 2n1 − 2ℓ − 2 fixed G2 -letters of E in w2 . Also, there are n2 + 1 fixed
G1 -letters and n2 fixed G2 -letters of B. Next, we bound the degrees of freedom of the non-fixed
−1 to those of Dē · · · ē
letters. Note that g maps the letters of Aα−1
1
n1 +1 en1 +2 in order. Likewise,
2 F
by observing the left end of w1 , we see that g maps the letters of en3 −n1 −2ℓ+2 · · · ēn3 −2ℓ+2 F to those
of D −1 β3 A in order. However, we also have that g maps the letters of ē1 · · · ēn1 +1 en1 +2 to those of
β3′−1 en3 −n1 −2ℓ+2 · · · ēn3 −2ℓ+1 en3 −2ℓ+2 in order, which overall gives us the ordered equality (by g) of
letters
−1
ēn3 −2ℓ+2 F = Dē1 · · · en1 +2 ēn3 −2ℓ+2 F
Aα−1
2 F

= Dβ3′−1 en3 −n1 −2ℓ+2 · · · en3 −2ℓ+2 ēn3 −2ℓ+2 F = Dβ3′−1 D −1 β3 A.

Thus, arguing inductively by translation, we see that that choosing the letters of F determines the
letters of A, and thus also determines those of D, thereby determining all non-fixed letters (while
not caring about the constant number of exceptional letters). Counting across all choices of values
for the letters, p, and ℓ, we have that the number of additional Wicks commutators arising from
this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

n2 =0

j

X−2−n1 −n2
2

X
ℓ=1

k

(|G1 | − 1)(n2 +1)+(n3 −2n1 −2ℓ−1)+ℓ (|G2 | − 1)n2 +(n3 −2n1 −2ℓ−2)+(ℓ−1)
≪X(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, suppose that the fixed letters of B are in w2 , which implies the fixed letters of E are
in w3 . Then, we require that w3 = e2 · · · en3 −2ℓ+2 . Furthermore, in order for the letters of B to
match in position, we must have that w2 = V α1 Bβ1 D, where V is the right subword of D −1 β3 A
having length 2ℓ − 1. We thus have n3 − 2ℓ + 1 fixed G1 -letters and n3 − 2ℓ fixed G2 -letters in
E, as well as n2 + 1 fixed G1 -letters and n2 fixed G2 -letters in B. Now, we bound the degrees of
freedom of the non-fixed letters. First, suppose that n1 > ℓ. Note then that g maps the letters

22

PETER S. PARK

−1 to those of
of F to those of an1 −ℓ+2 · · · an1 +1 in order, and likewise maps the letters of Aα−1
2 F
′−1 −1
Dβ2 D β3 a1 · · · an1 −ℓ+1 in order. Thus, we have the ordered equality (by g) of letters
−1
−1
a1 · · · an1 −ℓ+1 ān1 −ℓ+1 F α−1
= a1 · · · an1 −ℓ+1 ān1 −ℓ+1 (an1 −ℓ+2 · · · an1 +1 )α−1
2 F
2 F

= Dβ2′−1 D −1 β3 a1 · · · an1 −ℓ−1 .

Thus, arguing inductively by translation, we see that that choosing the letters of F determines
the letters of a1 · · · an1 −ℓ−1 , and thus also determines those of the rest of A and of D, thereby
determining all non-fixed letters (while not caring about the constant number of exceptional letters).
In the other case of n1 ≤ ℓ, the notation above for a1 · · · an1 −ℓ−1 becomes inviable, but nevertheless
we can use a similar argument as above to conclude that choosing the letters of F determines all
the non-fixed letters. Counting across all choices of values for the letters, p, and ℓ, we have that
the number of additional Wicks commutators arising from this case is bounded from above by
X−3
X 1
X X−3−n

n1 =0

j

X−2−n1 −n2
2

k

(|G1 | − 1)(n2 +1)+(n3 −2ℓ+1)+ℓ (|G2 | − 1)n2 +(n3 −2ℓ)+(ℓ−1)

X

n2 =0

ℓ=1

≪X(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Case 3. Finally, suppose the fixed letters of g are in A and E. Similarly to before, it is necessary
that the fixed letters of A and those of E are in wi and wj , respectively, such that i < j. First, suppose that the fixed letters of E are in w3 . Then, we require that w3 = e2 · · · en3 −2ℓ+2 . Furthermore,
in order for the letters of A to match in position, we need that the fixed letters of A are contained
in w1 , and in particular, that w1 = D −1 β3 Aα1 V , where V is the left subword of Bβ1 DE having
length 2ℓ − 1. We thus have n3 − 2ℓ + 1 fixed G1 -letters and n3 − 2ℓ fixed G2 -letters in E, as well as
n1 + 1 fixed G1 -letters and n1 fixed G2 -letters in A. Now, we bound the degrees of freedom of the
non-fixed letters. First, suppose that n2 > ℓ. Note then that g maps the letters of F −1 to those
′−1 −1
of b1 · · · bℓ in order, and likewise maps the letters of F α−1
3 B to those of bℓ+1 · · · bn2 +1 β1 Dβ2 D .
Thus, we have the ordered equality (by g) of letters
′−1 −1
−1
F α−1
b̄ℓ bℓ+1 · · · bn2 +1 = F α′−1
3 F
3 (b1 · · · bℓ )b̄ℓ bℓ+1 · · · bn2 +1 = bℓ+1 · · · bn2 +1 β1 Dβ2 D

Arguing inductively by translation, we see that that choosing the letters of F determines the letters
of B of D, thereby determining all non-fixed letters. In the other case of n2 ≤ ℓ, the notation above
for b1 · · · bℓ becomes inviable, but nevertheless we can use a similar argument as above to conclude
that F determines all the non-fixed letters. Counting across all choices of values for the letters, p,
and ℓ, we have that the number of additional Wicks commutators arising from this case is bounded
from above by
X−3
X X−3−n
X 1

n1 =0

n2 =0

j

X−2−n1 −n2
2

X
ℓ=1

k

(|G1 | − 1)(n1 +1)+(n3 −2ℓ+1)+ℓ (|G2 | − 1)n1 +(n3 −2ℓ)+(ℓ−1)
≪X(|G1 | − 1)X (|G2 | − 1)X ,

which is dominated by our error term.
Next, suppose that the fixed letters of E are in w2 . Then, the fixed letters of A are contained in
w1 , which requires that w1 = D −1 β3 Aα1 V , where V is the left subword of Bβ1 DE having length
2ℓ − 1. But then w2 must start on a letter not in E, which makes it impossible for the letters of E
to match in position.
Finally, if there are fixed letters in A, B, and C, then it is necessarily that ℓ = 0 and p = p′ ,
which does not need to be considered.

CONJUGACY GROWTH OF COMMUTATORS

23

If ℓ > (n3 +1)/2, then we can think of our commutator as a cyclic conjugate of C −1 ABCA−1 B −1
such that the letters are moved from left to right. A symmetric argument like above gives us the
conclusion that W ′ = D −1 β3 Aα1 Bβ1 DEF α2 A−1 β2 B −1 α3 F −1 E −1 is on average not a generic
Wicks commutator of the form w1 α′1 w2 β1′ w3 α′2 w1−1 β2′ w2−1 α′3 w3−1 β3′ . Note that this entire argument
can then be repeated mutatis mutandis to show that W ′ is on average also not a Wicks commutator
of the other generic form (9) or of the other possible forms (3) − (7), since the only difference from
the previous case is a constant number of exceptional letters and possibly setting one or more
of the subwords w1 , w2 , and w3 to be trivial, which overall can only affect error bounds by at
worst a multiplicative constant. Likewise, a similar argument shows that W ′ is also not a Wicks
commutator for the case of ℓ = 0, again since the only difference from the previous case is a constant
number of exceptional letters. Finally, a similar argument shows that even if W is taken to be of
the generic form (9) rather than (8), W is on average only decomposable as a Wicks commutator
in one prescribed way (i.e., with respect to a unique partition of X − 3), and any cyclic conjugate
of W is on average not a Wicks commutator.
We have thus shown that the number of conjugacy classes of commutators with length 4X is
given by


1 1
(|G1 | − 1) (|G2 | − 2)2 + (|G1 | − 2)2 (|G2 | − 1) X 2 (|G1 | − 1)X−1 (|G2 | − 1)X−1
6 2


X
X
+O X(|G1 | − 1) (|G2 | − 1)
=


1 
(|G1 | − 1) (|G2 | − 2)2 + (|G1 | − 2)2 (|G2 | − 1) X 2 (|G1 | − 1)X−1 (|G2 | − 1)X−1
12



+O X(|G1 | − 1)X (|G2 | − 1)X ,

as needed.

4. Concluding Remarks
There are a number of directions in which Theorems 1.2 and 1.4 can be generalized. First,
one can ask: how many conjugacy classes of commutators with word length k are in an arbitrary
finitely-generated free product G = G1 ∗ · · · ∗ Gn , with each Gi nontrivial and having symmetric
(i)
(i)
(i)
(i)
generating set Si = {g1 . . . . , gmi , (g1 )−1 . . . . , (gmi )−1 }? While one can define the word length in
this context to be with respect to an arbitrary generating set S, a natural notion of length to use
in this setting would be with respect to the symmetric generating set
n
[
Si .
S ··=
i=1

We note that in the case that Gi is finite for all 1 ≤ i ≤ n, we can take Si = Gi \ {1}, for which S
is consistent with our choice of the set S of generators for G1 ∗ G2 in the statement of Theorem 1.4.
Counting conjugacy classes of commutators by word length for groups in this more general form
would have analogous geometric consequences as those discussed in Corollary 1.5. For example,
Hecke Fuchsian groups H(λ) for λ ≥ 2 have the presentation [13]
hS, Rλ : S 2 = Ii ∼
= Z/2Z ∗ Z,
and a desire for geometric corollaries akin to those discussed in Section 1 motivates a result analogous to Theorem 1.2 and Theorem 1.4 in the case of the free product of a nontrivial finite group
and an infinite cyclic group.
To describe a second potential direction for generalization, let the n-commutators of a given
group be defined by the elements with trivial abelianization and commutator length n. A second

24

PETER S. PARK

direction for generalizing Theorems 1.2 and 1.4 is to, for any n, count the number of conjugacy
classes of n-commutators with word length k in a group G taken to be either the free group Fr
or a finitely generated free product. This is natural to ask, given that Culler [7] has classified the
possible forms of n-commutators for a free group and Vdovina [26] has done this for an arbitrary free
product. In fact, Culler has also classified the possible forms that a product of n square elements
can take for a free group, so an analogous question can be asked for the number of conjugacy
classes comprised of products of n square elements. If one obtained the asymptotic number of ncommutators with length k (respectively, of n-square-element-products with length k), this would
also give a corollary analogous to Corollary 1.5. Specifically, given a connected CW-complex X
with fundamental group G, one would obtain the asymptotic number of free homotopy classes of
loops γ : S 1 → X with length k (in the generators of S) such that there exists a genus-n orientable
surface Y with one boundary component (respectively, a connected sum Y of n real projective
planes, with one boundary component) and a continuous map f : Y → X satisfying f (∂Y ) = Im γ,
but also that this statement does not hold when replacing n with any m ≤ n. We expect our
combinatorial method to also work in this generalized setting, as long as one has, for the given
group, an explicit list of the possible Wicks forms of n-commutators (or of products of n square
elements).
Acknowledgments
This work began at Princeton University as part of the author’s senior thesis advised by Peter
Sarnak, whom the author would like to thank for providing the number-theoretic motivation for
this problem, invaluable discussions/references, and constant encouragement.
The generalization of Theorem 1.4 to an arbitrary free product of two nontrivial finite groups
was done at Harvard University and was supported by the National Science Foundation Graduate
Research Fellowship Program (grant number DGE1745303). The author would like to thank Bena
Tshishiku for reading over the manuscript and providing many valuable suggestions. He would also
like to thank Keith Conrad, Noam Elkies, Curt McMullen, Hector Pasten, Xiaoheng Wang, and
Boyu Zhang for very helpful discussions.
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Department of Mathematics, Harvard University, Cambridge, MA 02138
E-mail address: pspark@math.harvard.edu

