ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P4.06 / 609–616
https://doi.org/10.26493/1855-3974.2614.44c
(Also available at http://amc-journal.eu)
The diameter of products of finite simple groups
Daniele Dona *
Einstein Institute of Mathematics, Edmond J. Safra Campus Givat Ram,
The Hebrew University of Jerusalem, 9190401 Jerusalem, Israel
Received 27 April 2021, accepted 14 January 2022, published online 11 August 2022
Abstract
Following partially a suggestion by Pyber, we prove that the diameter of a product
of non-abelian finite simple groups is bounded linearly by the maximum diameter of its
factors. For completeness, we include the case of abelian factors and give explicit constants
in all bounds.
Keywords: Finite simple groups, diameter.
Math. Subj. Class. (2020): 20F69, 20D06
1 Introduction
An important area of research in finite group theory in the last decades has been the produc-
tion of upper bounds for the diameter of Cayley graphs of such groups. For any finite group
G, the maximum diameter over all Cayley graphs defined by symmetric sets of generators
of G (i.e. sets S with S = S−1 and e ∈ S) is called the diameter of G. Arguably the
best known conjecture in the area is Babai’s conjecture [1]: every non-abelian finite simple
group G has diameter ≤ logk |G|, where k is an absolute constant; the conjecture is still
open, despite great progress towards a solution both for alternating groups and for groups
of Lie type.
A more modest question is that of producing bounds for the diameter of direct products
of finite simple groups, depending on the diameter of their factors. This is not an idle
question, for bounds of this sort have been used more than once as intermediate steps
towards the proof of bounds for simple groups themselves: Babai and Seress have done so
in [2, Lemma 5.4], as well as Helfgott more than two decades later in [5, Lemma 4.13]. We
improve on both results in the following theorem, which also features explicit constants.
*The author was partially supported by the European Research Council under Programme H2020-EU.1.1.,
ERC Grant ID: 648329 (codename GRANT). He was also supported by the Israel Science Foundation Grant No.
686/17 of A. Shalev, and by the Emily Erskine Endowment Fund.
E-mail address: daniele.dona@mail.huji.ac.il (Daniele Dona)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
610 Ars Math. Contemp. 22 (2022) #P4.06 / 609–616
Theorem 1.1. Let n ≥ 2. Let G =
∏n
i=1 Ti, where the Ti are finite simple groups.
(a) If the Ti are all abelian (say G =
∏s
j=1(Z/pjZ)ej , where the pj are distinct primes
and ej ≥ 1), then:
diam(G) <
2
3
max{ej |1 ≤ j ≤ s}
s∏
j=1
pj .
(b) If the Ti are all non-abelian, call d = max{diam(Ti)|1 ≤ i ≤ n}; then:
diam(G) <
196
243
n3 max{CA, CL, CS}(4d+ 1) + d,
where:
CA =

max
{
3,
⌊
m
2
⌋}
if there are alternating groups among the Ti
and where m is their maximum degree,
0 if there are no alternating groups among the Ti,
CL =

8(5r + 7) if there are groups of Lie type among the Ti
and where r is their maximum untwisted rank,
0 if there are no groups of Lie type among the Ti,
CS =
{
6 if there are sporadic or Tits groups among the Ti,
0 if there are no sporadic or Tits groups among the Ti.
(c) If there are abelian and non-abelian Ti, write G = GA × GNA, where GA collects
the abelian factors and GNA collects the non-abelian ones; then:
diam(G) ≤ dA + 4dNA,
where dA = diam(GA), dNA = diam(GNA).
The result of part (a) is known and elementary: see [2, Lemma 5.2], where the con-
stant is marginally worse only due to the fact that sets of generators are not required to be
symmetric (cfr. also [5, Lemma 4.14], which treats the case of G = (Z/pZ)e under this as-
sumption). Part (c) is quite natural, given the different (in some sense, opposite) behaviour
of abelian and non-abelian factors, as it can be readily observed in its short proof.
Part (b) is where the novelty of the result resides. Dependence on the maximum of the
diameter of the components, instead of dependence on their product as Schreier’s lemma
(see Lemma 2.1) would naturally give us, was already established in [2, Lemma 5.4]: in
that case, the diameter was bounded as O(d2), where the dependence of the constant on n
was polynomial as in our statement. This result was improved in [5, Lemma 4.13] to O(d),
but only in the case of alternating groups: this was done in part to fix a mistake in the use of
the previously available result in Babai-Seress, which is why only alternating groups were
considered, as permutation subgroups were the sole concern in both papers; a suggestion
by Pyber, reported in Helfgott’s paper, points at the results by Liebeck and Shalev [8] as a
way to prove a bound of O(d) for a product of arbitrary non-abelian finite simple groups.
D. Dona: The diameter of products of finite simple groups 611
Indeed, the general approach that we follow in our proof owes its validity to [8, Theo-
rem 1.6], although we do not explicitly use the statement of that theorem: rather, we closely
follow the proof of [5, Lemma 4.13] and show that the same reasoning applies to groups of
Lie type as well. The way that the lemma is related to Liebeck-Shalev is through the use
of the fact that every element in Alt(m) is a commutator ([5, Lemma 4.12], first proved in
[9, Theorem I]), which is essentially [8, Theorem 1.6] with w = xyx−1y−1 and a c that
is just equal to 1 for Alt(m); the same can be said for all non-abelian finite simple groups
(i.e., c = 1 in general) since Ore’s conjecture [10] was established to be true in [7], a fact
yet unproved at the time of [8].
2 Preliminaries
Before we turn to the proof of Theorem 1.1, we will need a certain number of group-
theoretic results.
Lemma 2.1 (Schreier’s Lemma). Let G be a finite group, let N ⊴G, and let S be a set of
generators of G with e ∈ S = S−1. Then S2d+1∩N generates N , where d = diam(G/N).
Proof. This is a standard result dating back to Schreier [11], written in various fashions
across the literature according to the needs of the user; let us prove here the present version.
Calling π : G → G/N the natural projection, by definition we have π(S)d = G/N ;
this equality means that Sd contains at least one representative for each coset gN in G. For
any coset gN , choose a representative τ(g) ∈ Sd. Then, for any h ∈ N and any way to
write h as a product of elements si ∈ S, we have:
h = s1s2 . . . sk
= (s1τ(s1)
−1) · (τ(s1)s2τ(τ(s1)s2)−1) · · · (τ(τ(τ(. . .)sk−2)sk−1)sk).
Each element of the form τ(x)siτ(τ(x)si)−1 is contained in S2d+1∩N , so the same can be
said about the last element of the form τ(x)sk (since h itself is in N ); therefore S2d+1 ∩N
is a generating set of N .
Proposition 2.2 (Ore’s Conjecture). Let G be a finite non-abelian simple group. Then, for
any g ∈ G, there exist g1, g2 ∈ G such that g = [g1, g2].
Proof. See [7], for references to previously known results and for the proof of the final
case.
Notice that, for any finite non-abelian simple group G, any nontrivial conjugacy class
C must generate the whole G (because ⟨C⟩ would be a normal subgroup). This observation
justifies the following definition.
Definition 2.3. Let G be a finite non-abelian simple group. The conjugacy diameter cd(G)
is the smallest m such that (C ∪C−1 ∪ {e})m = G for all nontrivial conjugacy classes C.
We will need to have bounds for cd(G).
Proposition 2.4. Let G be a finite non-abelian simple group.
(a) If G is an alternating group of degree m, then cd(G) ≤ max
{
3,
⌊
m
2
⌋}
.
(b) If G is a group of Lie type of untwisted rank r, then cd(G) ≤ 8(5r + 7).
612 Ars Math. Contemp. 22 (2022) #P4.06 / 609–616
(c) If G is a sporadic group or the Tits group, then cd(G) ≤ 6.
Proof. First of all, cd(G) is trivially bounded by definition by the covering number of G,
which is defined as cn(G) = min{m|∀C ̸= {e}(Cm = G)}; therefore it suffices to give
bounds for cn(G).
For (a), see [4, Theorem 9.1] (our specific result is credited therein to a manuscript
by J. Stavi). For (b), see [6, Theorem 1]. To prove (c), the sporadic groups all satisfy
cn(G) ≤ 6: this inequality can be checked directly from [13, Table 1]; if G = 2F4(2)′ is
the Tits group, we can show the same inequality using [13, Lemma 3] and the character
values reported in the ATLAS of Finite Groups [3].
Let us also perform a side computation separately from the proof of the main theorem,
so as not to bog down the exposition there.
Lemma 2.5. Let n ≥ 2. Then:
n−1∑
i=1
4⌈log2 i⌉ <
196
243
n3.
Proof. Call m = ⌈log2(n − 1)⌉, and write n − 1 = 2m−1 + l, where 1 ≤ l ≤ 2m−1;
⌈log2 i⌉ = j for all i ∈ (2j−1, 2j ], hence we can rewrite the sum in the statement as:
n−1∑
i=1
4⌈log2 i⌉ = 1 +
m−1∑
j=1
4j2j−1 + 4ml =
1
2
+
1
2
8m − 1
7
+ 4m
(
2log2(n−1) − 2m−1
)
=
3
7
+ 4m2log2(n−1) − 3
7
8m =
3
7
+ 22m
′
(
1− 3
7
2m
′
)
(n− 1)3,
where m′ = m − log2(n − 1) ∈ [0, 1). We have x2
(
1− 37x
)
≤ 196243 for x ∈ [1, 2), and
3
7 <
196
243 (3n
2 − 3n+ 1) for all n ≥ 2, so the result is proved.
3 Proof of the main theorem
Proof of Theorem 1.1(a). Let G = (Z/p1Z)e1×(Z/p2Z)e2×. . .×(Z/psZ)es , with primes
p1 < p2 < . . . < ps; we have:
G = A1A2 . . . As (3.1)
(we are using multiplicative notation even if G is abelian) where the Ai are any sets such
that:
Ai,i =(Z/piZ)ei Ai,j =(0)ej (∀j < i) (3.2)
where Ai,j is the projection of Ai to the j-th component of G.
Let S be a set of generators of G with e ∈ S = S−1: {tp1...pi−1 |t ∈ S} ⊆ Sp1...pi−1 has
elements that are all 0 on the first i− 1 components of G and that still generate the i-th one
since (p1 . . . pi−1, pi) = 1; from now on, let us focus exclusively on the i-th component.
(Z/piZ)ei is also a vector space over Z/piZ, so there must be ei generators that also form
a basis: any element of the space can be written as a linear combination of those generators
with coefficients in
[
−
⌊
pi
2
⌋
,
⌊
pi
2
⌋]
, which corresponds to a word of length ≤ ei
⌊
pi
2
⌋
; thus,
D. Dona: The diameter of products of finite simple groups 613
each set Ai with the properties in (3.2) is covered in ei
⌊
pi
2
⌋
p1 . . . pi−1 steps. This fact and
(3.1) imply that G has diameter bounded by:
s∑
i=1
ei ⌊pi
2
⌋ i−1∏
j=1
pj
 ≤ 1
2
max{ej |1 ≤ j ≤ s}
s∏
j=1
pj ·
s∑
i=1
 s∏
j=i+1
1
pj
 . (3.3)
The sum in (3.3) is maximized when each pj is the j-th prime number: for s = 1 the sum is
1 and for s = 2 it is bounded by 43 ; for s ≥ 3, we use ps ≥ 5 and pj ≥ 3 for all 1 < j < s,
so that the sum is bounded by 1 + 15
1
1− 13
= 1310 . The result follows.
Proof of Theorem 1.1(b). Calling Gj =
∏j
i=1 Ti, we have natural projections πj : G =
Gn → Gj and ρj1,j2 : Gj1 → Tj2 for any j1 ≥ j2. As in (3.1), we write G as a product of
subsets Ai with ρn,i(Ai) = Ti and ρn,j(Ai) = {e} for all j < i, and our aim is to cover
each one of them.
Suppose that we have two subsets X1, X2 of G for which ρn,i(X1) = ρn,i(X2) =
Ti for some fixed i ∈ {1, . . . , n} and that have ρn,j1(X1) = {e} = ρn,j2(X2) for all
j1 ∈ I1, j2 ∈ I2, where I1, I2 are two subsets of indices in {1, . . . , n} \ {i}: then, the set
X = {[x1, x2]|x1 ∈ X1, x2 ∈ X2} has ρn,i(X) = Ti by Proposition 2.2 (Ore’s conjecture)
and ρn,j(X) = {e} for all j ∈ I1∪ I2. Now consider the set of indices I = {1, . . . , i−1}:
if |I| > 1 we can partition I into two parts of size
⌊
|I|
2
⌋
,
⌈
|I|
2
⌉
, then partition each part I ′
with |I ′| > 1 into two new parts again of size
⌊
|I′|
2
⌋
,
⌈
|I′|
2
⌉
, and continue until we reach
a subdivision where all sets have size 1; the tree of partitions that we constructed to reach
this subdivision will have exactly ⌈log2 |I|⌉ layers. Notice that, given any two parts I1, I2
inside the tree, if we have two subsets X1, X2 (as described before) that are covered by a
certain Sa, the resulting set X will be covered by S4a: this observation, together with the
information about the layers, tells us that if we can cover sets Xi,j with ρn,i(Xi,j) = Ti
and ρn,j(Xi,j) = {e} in a steps (for a fixed i > 1 and all j < i) then we are able to cover
a set Ai defined as at the beginning of the proof in 4⌈log2(i−1)⌉a steps as well.
Let us start now with a generating set S with e ∈ S = S−1 and fix two indices i ≥ j:
πi(S) is a set of generators for Gi, and the set πi(S)2d+1 contains generators for the whole
T1 × . . . × Tj−1 × {e} × Tj+1 × . . . × Ti = Gi ∩ ker(ρi,j) by Lemma 2.1 (Schreier’s
lemma), where d is as in the statement. In particular, there is an element x ∈ S2d+1 with
ρn,i(x) ̸= e and ρn,j(x) = e; by hypothesis ρn,i(Sd) = Ti, which means that there is
a set S′ = {yxy−1|y ∈ Sd} ∪ {yx−1y−1|y ∈ Sd} ∪ {e} ⊆ S4d+1 with ρn,i(S′) =
C ∪ C−1 ∪ {e} and ρn,j(S′) = {e}, where C is the conjugacy class of ρn,i(x). By
Proposition 2.4, ρn,i(S′max{3,⌊
mi
2 ⌋}) = Ti if Ti = Alt(mi), ρn,i(S′8(5ri+7)) = Ti if Ti
is of Lie type of untwisted rank ri, and ρn,i(S′6) = Ti otherwise; in all three cases, the
projection to Tj is still {e}, therefore we managed to cover a set Xi,j of the aforementioned
form.
A set A1 is reached in d steps, hence the final count for the whole G following the
reasoning above is:
diam(G) ≤ d+
n∑
i=2
4⌈log2(i−1)⌉xi(4d+ 1),
where xi is either max
{
3,
⌊
mi
2
⌋}
, 8(5ri + 7) or 6, accordingly. The result follows by
Lemma 2.5.
614 Ars Math. Contemp. 22 (2022) #P4.06 / 609–616
A note on the connection between the proof given above and [8]. As mentioned before,
Pyber pointed at [8] as a way to prove linear dependence on d for products of arbitrary non-
abelian finite simple groups. In particular, [8, Theorem 1.6] seems to fit the bill: it states
that for any word w that is not a law in a finite simple group T there is cw ∈ N, depending
on w but not on T , such that any element of T can be written as a product of at most cw
values of w. We use this property, in disguise, when we want to pass from two subsets
being indentically e at indices I1, I2 and filling an entire component Ti to a third subset
that also fills the same component and is e for the whole I1 ∪ I2: the creation of the new
subset is made possible by taking cw values of a word w, so that Ti remains filled, where w
has two distinct letters x1, x2 and presents the same number of xi and x−1i for i ∈ {1, 2},
so that when any one xi is equal to e on a given factor of the product G the result is e
on that factor; in our case, w was the shortest nontrivial word with these characteristics,
namely the commutator [x1, x2] = x1x2x−11 x
−1
2 (not a law for any non-abelian group),
and cw = 1 by Ore’s conjecture. In this sense w = [x1, x2] is also computationally the best
word we can expect, for it yields the lowest possible value of |w|cw, the 4 that we find in
Lemma 2.5.
Proof of Theorem 1.1(c). Define the two projections πA, πNA in the obvious way; for any
generating set S of G, by definition there is a subset XA ⊆ SdA with πA(XA) = GA and
there is a subset XNA ⊆ SdNA with πNA(XNA) = GNA, and then:
G = XA[XNA, XNA] ⊆ SdA+4dNA ,
again by the fact that [T, T ] = T for non-abelian finite simple groups by Ore’s conjecture
and [T, T ] = {e} for abelian groups.
4 Concluding remarks
One could wonder how tight the inequalities in Theorem 1.1 are. The results are essen-
tially in line with what is generally expected from the behaviour of the diameter of fi-
nite groups. The abelian case is tight up to constant: for the group G(x) =
∏
p≤x Z/pZ
(nontrivial for x ≥ 2) one generator s = (1, 1, . . . , 1) is enough, and then the diameter of
Cay(G(x), {s, s−1, e}) is 12 |G(x)|; the fact that abelian groups behave in the worst possible
way, i.e. linearly in the size of the group, should not be a surprise for anyone.
The non-abelian bound of case (b) also matches what is anticipated in general. Babai’s
conjecture posits a polylogarithmic bound on the diameter of finite simple groups: the
natural extension to direct products of such groups would suggest a bound of the form nkd,
which is exactly what we have obtained. Case (c) also fits into the same idea, as a product
|G| = |GA||GNA| becomes a sum of the corresponding diameters.
The dependence on d in Theorem 1.1(b) is almost best possible by definition (we cannot
drop the “almost”, as m, r are not independent from d). It would be more interesting to
understand which power of n is the correct one: here we have proved Om,r,d(n3), and we
can quickly show that the bound is Ωm,r,d(n), as illustrated in the following example.
Example 4.1. If G = (Alt(m))n then diam(G) = Ω(m2n). We prove it for m ≥ 5 odd
and n even, but the proof is analogous for the general case.
Consider the two permutations σ = (1 2 3 . . . m) and τ = (1 2 3 . . . m − 2); they
D. Dona: The diameter of products of finite simple groups 615
generate Alt(m), and the elements:
s0 = (σ, σ, . . . , σ, σ),
s1 = (τ, σ, . . . , σ, σ),
s2 = (σ, τ, . . . , σ, σ),
. . .
sn = (σ, σ, . . . , σ, τ)
generate G. Let S = {e} ∪ {si, s−1i }0≤i≤n: to prove the lower bound on the diameter of
G, we construct a function f : G → N such that there are two elements g1, g2 ∈ G with
|f(g1)− f(g2)| large and such that |f(g)− f(gs)| is small for any g ∈ G, s ∈ S; this is a
known technique to prove lower bounds for the diameter of Sym(m), as shown for instance
in [12, Proposition 3.6].
Call c(g, i, j) = (g(i))(j) the image of j ∈ {1, . . . ,m} under the i-th component of
g ∈ G, for 1 ≤ i ≤ n; define:
f(g) =
m∑
j=1
n∑
i=1
||c(g, i+ 1, j)− c(g, i, j)||Z/mZ,
where ||a||Z/mZ = min{a,m−a} (in the case i = n, c(g, n+1, j) means c(g, 1, j)). First,
f(e) = 0; also, if we call em the identity element in Alt(m) and η =
(
1 m+12
) (
2 m+32
)
· · ·(
m−1
2 m− 1
)
, for g ∈ G that has em at all odd components and η at all even ones we have
f(g) = 12 (m − 1)
2n. Finally, notice that σ simply adds 1 modulo m to all the elements
of {1, . . . ,m}, so that f(g) = f(gs±10 ), while τ is defined so that it adds 1 for m − 3
elements, adds 3 (modulo m) for one element and fixes two elements, which means that
|f(g)− f(gs±1i )| ≤ 10; these facts taken together imply that diam(G,S) ≥ 120 (m− 1)
2n.
The correct (or even expected) order of magnitude for a bound of the form diam(G) =
Om,r(n
kd) for a generic product G is not known to the author, besides knowing that
1 ≤ k ≤ 3 by Theorem 1.1 and Example 4.1.
ORCID iDs
Daniele Dona https://orcid.org/0000-0001-7966-3357
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