# Characteristic Covering Numbers of Finite Simple Groups

Michael Larsen<EMAIL_ADDRESS>Department of Mathematics
Indiana University
Bloomington, IN 47405
U.S.A.  ,  Aner Shalev<EMAIL_ADDRESS>Einstein Institute of
Mathematics
Hebrew University
Givat Ram, Jerusalem 91904
Israel  and  Pham Huu Tiep<EMAIL_ADDRESS>Department of Mathematics
Rutgers University
Piscataway, NJ 08854-8019
U.S.A.

###### Abstract.

We show that, if $w_{1},\ldots,w_{6}$ are words which are not an identity of
any (non-abelian) finite simple group, then $w_{1}(G)w_{2}(G)\cdots
w_{6}(G)=G$ for all (non-abelian) finite simple groups $G$. In particular, for
every word $w$, either $w(G)^{6}=G$ for all finite simple groups, or $w(G)=1$
for some finite simple groups.

These theorems follow from more general results we obtain on characteristic
collections of finite groups and their covering numbers, which are of
independent interest and have additional applications.

###### 1991 Mathematics Subject Classification:

Primary 20D06

ML was partially supported by NSF grant DMS-2001349. AS was partially
supported by ISF grant 686/17 and the Vinik Chair of mathematics which he
holds. PT was partially supported by NSF grant DMS-1840702, the Joshua Barlaz
Chair in Mathematics, and the Charles Simonyi Endowment at the Institute for
Advanced Study (Princeton). The authors were also partially supported by BSF
grant 2016072.

The authors are grateful to Frank Lübeck for kindly providing us with the
character table of the Steinberg group ${}^{3}\\!D_{4}(3)$.

##  1\. Introduction

The theory of word maps on groups, and on finite simple groups in particular,
dates back to Borel [Bo], and has developed significantly in the past 3
decades; see for instance [MZ], [SW], [LiSh1], [La], [Sh1], [LS1], [LS2],
[LOST], [LST1], [GT], as well as the monograph [Se] and the survey paper
[Sh2].

Most of these works are of asymptotic nature. For example, it is shown in
[LST1] (following [LS1, LS2]) that, for every non-identity word $w$ there
exists $N(w)\in{\mathbb{N}}$ such that, if $G$ is a finite simple group of
order at least $N(w)$ then $w(G)^{2}=G$. Recall that a word is an element of a
free group $F_{d}$ of rank $d$, and it defines a word map $w\colon G^{d}\to G$
for every group $G$, induced by substitution. The image of this word map is
denoted by $w(G)$.

Here and throughout this paper, by a finite simple group we mean a non-abelian
finite simple group.

Non-asymptotic results on word maps are often very challenging and require
more tools. See, for instance, [LOST], [LOST2], [GM], [GT] and [GLOST]. In
[LOST] it is shown that the commutator map is surjective on all finite simple
groups, proving the Ore Conjecture. Using a somewhat similar strategy, it was
subsequently proved in [LOST2] that, if $w=x^{p}$ for a prime $p\neq 3,5$,
then $w(G)^{2}=G$ for all finite simple groups $G$. In [GM] a stronger result
is established, yielding the same conclusion for $w=x^{k}$, where $k$ is any
prime power or a power of $6$.

The above result was further extended in [GLOST]. Consider the power words
$w_{1}=x^{k}$ and $w_{2}=x^{\ell}$, where $k=p^{a}q^{b}$ for primes $p,q$ and
$\ell$ is an odd integer. By celebrated theorems of Burnside and of Feit and
Thompson, finite groups satisfying the identity $w_{1}$ or the identity
$w_{2}$ are solvable. In particular, $w_{1},w_{2}$ are not identities of any
finite simple group. It was proved in [GLOST] that, $w_{1}(G)^{2}=G$ and
$w_{2}(G)^{3}=G$ for all finite simple groups $G$.

Our main theorems, stated below, may be regarded as far-reaching extensions of
the above mentioned results. In particular, we derive similar conclusions for
every word $w$ which is not an identity of any finite simple group, sometimes
with somewhat larger exponents. Our strategy is to pose and study a more
general problem.

######  Definition 1.1.

A collection of non-empty subsets $S(G)\subseteq G$, one for each finite group
$G$, is _characteristic_ if for every homomorphism $\varphi\colon G\to H$ we
have $\varphi(S(G))\subseteq S(H)$.

Some comments are in order.

  1. 1.

For any characteristic collection $S$ and for any finite group $G$, $S(G)$
must be a fully invariant – in particular, a characteristic – subset of $G$
(take $\varphi\colon G\to G$); furthermore, $1\in S(G)$ (take $\varphi$ to be
the trivial endomorphism of $G$).

  2. 2.

Trivial examples of characteristic collections are $S(G)=\\{1\\}$ for all $G$
(the minimal collection) and $S(G)=G$ for all $G$ (the maximal collection).

  3. 3.

Given characteristic collections $S,T$, define their product $ST$ by
$ST(G):=S(G)T(G)$, which is easily seen to be a characteristic collection.
This binary operation is associative with an identity element (the minimal
collection). Thus the set of all characteristic collections is a monoid, which
is partially ordered by inclusion (where $S\subseteq T$ if $S(G)\subseteq
T(G)$ for all $G$).

  4. 4.

Let $w\in F_{d}$ be a word, and consider the word maps it defines on finite
groups $G$. Then the collection $S(G):=w(G)$ is obviously characteristic.

  5. 5.

Let $P_{d}$ denote the free profinite group on $x_{1},\ldots,x_{d}$ (namely
the profinite completion of $F_{d}$). For any finite group $G$ and
$g=(g_{1},\ldots,g_{d})\in G^{d}$ there is a unique homomorphism
$\psi_{g}:P_{d}\to G$ satisfying $\psi_{g}(x_{i})=g_{i}$ ($i=1,\ldots,d$). Any
element $W\in P_{d}$ (which may be regarded as a profinite word) gives rise to
a function (a profinite word map) $W:G^{d}\to G$ satisfying
$W(g):=\psi_{g}(W)$. Setting $S(G):=W(G)$, the image of $W$, we obtain a
characteristic collection $S$. It is easy to see that different elements
$W_{1},W_{2}\in P_{d}$ give rise to distinct characteristic collections
$S_{1},S_{2}$. Hence the monoid of characteristic collections has cardinality
$2^{\aleph_{0}}$.

  6. 6.

Any word $w$ in one variable, i.e. $w=x^{k}$ for some integer $k$, defines a
word map $w\colon G\to G$ on each finite group $G$, and the collection of
kernels $S(G):=w^{-1}(1)=\\{g\in G:g^{k}=1\\}$ is characteristic.

######  Definition 1.2.

A characteristic collection $S$ is _ample_ if $|S(G)|\geq 2$ for all finite
simple groups $G$.

For example, let $w$ be a word, and let $S$ be its associated characteristic
collection (i.e. $S(G)=w(G)$). Then the ampleness of $w$ is equivalent to each
of the following conditions.

  1. (a)

$w$ is not an identity of any finite simple group.

  2. (b)

$w$ is not an identity of any minimal finite simple group (these are certain
well known groups of type $\mathrm{PSL}_{2}$ or Suzuki type).

  3. (c)

For all finite groups $G$, $w(G)=1$ implies that $G$ is solvable.

######  Definition 1.3.

We define the _characteristic covering number_ $\mathsf{ccn}(G)$ of a finite
group $G$ to be the smallest integer $n$ such that if $S_{1},\ldots,S_{n}$ are
ample characteristic collections, then $S_{1}(G)\cdots S_{n}(G)=G$. If no such
$n$ exists, we say $\mathsf{ccn}(G)=\infty$.

The main result of this paper is the following:

######  Theorem A.

If $G$ is a finite simple group, then $\mathsf{ccn}(G)\leq 6$.

Applying this for characteristic collections associated with words, we
immediately obtain the following.

######  Theorem B.

If $w_{1},\ldots,w_{6}$ are words in disjoint letters, and none of the $w_{i}$
is an identity on any finite simple group, then the juxtaposition $w_{1}\cdots
w_{6}$ is surjective on every finite simple group.

######  Corollary C.

If $w$ is a word which is not an identity on any finite simple group, then
$w(G)^{6}=G$ for every finite simple group $G$.

Clearly, Theorem B and Corollary C also follow when $w,w_{i}$ are replaced by
profinite words $W$, $W_{i}$.

Note that the condition that none of the $w_{i}$ (or $w$) is an identity on
any finite simple group in Theorem B and Corollary C is necessary. As shown in
[KaN], [GT], for any integer $N$, there exist a word $w$ and a finite simple
group $G$ such that $w(G)\neq 1$ and $w(G)^{N}\neq G$.

As for lower bounds, it follows from [GLOST, 8.9] that $\mathsf{ccn}(G)\geq 3$
for some finite simple groups $G$; indeed, there are odd integers $\ell$ and
finite simple groups $G$ such that $w(G)^{2}\neq G$ for $w=x^{\ell}$. This can
be improved as follows. In the special case that $w=x^{2}$, the collection
$S(G)=w^{-1}(1)$ is ample by Feit-Thompson. The minimal $n$ such that
$S(G)^{n}=G$ is the _involution width_ of $G$, denoted $\mathsf{iw}(G)$. This
invariant has been investigated by several mathematicians (see [Ma] and the
references therein). In particular, Knüppel and Nielsen showed [KnN, 16] that
$\mathsf{iw}(\mathrm{SL}_{n}(q))\geq 4$ if $n\geq 5$ and $q\geq 7$. When, in
addition, $\gcd(n,q-1)=1$, we have $\mathrm{SL}_{n}(q)=\mathrm{PSL}_{n}(q)$,
and this gives a lower bound of $4$ for the characteristic covering number of
infinitely many finite simple groups.

It would be interesting to know whether our upper bound of $6$ for
$\mathsf{ccn}$ can be improved. Malcolm showed [Ma] that $\mathsf{iw}(G)\leq
4$ for all finite simple groups, and it seems quite possible that this is the
optimal bound for $\mathsf{ccn}$ as well. For most finite simple groups we can
prove an upper bound of $4$ or less. Indeed we have

######  Theorem D.

Let $G$ be a finite simple group. Then $\mathsf{ccn}(G)\leq 4$ unless $G$ is a
group of Lie type $X_{r}(q)$ where $q\leq f(r)$ for a suitable function $f$.

Consequently, excluding these possible exceptions, we have $w_{1}(G)\cdots
w_{4}(G)=G$ for all words $w_{1},\ldots,w_{4}$ which are not an identity of
any finite simple group.

Treating the remaining groups seems very challenging.

As $\mathsf{ccn}(G)$ reflects information on all simple subgroups of $G$, it
cannot easily be determined from the character table of $G$, as
$\mathsf{iw}(G)$ and similar invariants such as the covering number and
extended covering number of $G$ can be. We were able to determine it in a few
cases, of which the following results are representative.

######  Proposition 1.4.

If $p\geq 5$ is a prime, then

$\mathsf{ccn}(\mathrm{PSL}_{2}(p))=\begin{cases}2&\text{if $p\equiv
1\pmod{4}$,}\\\ 3&\text{if $p\equiv 3\pmod{4}$.}\end{cases}$

######  Proposition 1.5.

For all sufficiently large $n$, $\mathsf{ccn}({\sf A}_{n})=3$.

######  Proposition 1.6.

If $n\geq 5$ and $q-1$ is sufficiently large and relatively prime to $n$, then
$\mathsf{ccn}(\mathrm{PSL}_{n}(q))=4$.

It would be interesting to find an example where
$\mathsf{ccn}(G)>\mathsf{iw}(G)$; at present, we do not know of any.

##  2\. Preliminaries

######  Proposition 2.1.

Let $G$ be a finite group. If $\mathsf{ccn}(G)<\infty$, then $G$ is perfect.
If $G$ is simple, then $\mathsf{ccn}(G)<\infty$.

###### Proof.

For $G$ finite, $\mathsf{ccn}(G)<\infty$ if and only if $S(G)$ generates $G$
for every ample characteristic collection $S$.

If $G$ is not perfect, then by pulling back a subgroup of prime index from its
abelianization, we obtain a normal subgroup $G_{p}\triangleleft G$ of index
$p$ for some prime $p$. Then $S(H)=\\{h^{p}\mid h\in H\\}$ defines a
characteristic collection of subgroups, and it is ample since no finite simple
group has exponent $p$. However, $S(G)^{n}$ is contained in $G_{p}$ for all
$n\in{\mathbb{N}}$, so $\mathsf{ccn}(G)=\infty$.

If $G$ is simple, let $S$ be an ample characteristic collection. Then $S(G)$
generates a non-trivial normal subgroup of $G$. As $S$ is ample, this must be
$G$ itself. ∎

We recall that the extended covering number $\mathsf{ecn}(G)$ of a finite
simple group $G$ is the smallest integer $n$ such that $C_{1}\cdots C_{n}=G$
if $C_{1},\ldots,C_{n}$ is any sequence of non-trivial conjugacy classes of
$G$.

######  Lemma 2.2.

If $G$ is a finite simple group, then $\mathsf{ccn}(G)\leq\mathsf{ecn}(G)-1$.

###### Proof.

If $\mathsf{ccn}(G)\geq n$, then there exist ample characteristic collections
$S_{i}$ ($i=1,\ldots,n-1$) such that $S_{1}(G)\cdots S_{n-1}(G)\neq G$. As
$S_{1},\ldots,S_{n-1}$ are ample,

$S_{1}(G)\cdots S_{n-1}(G)\supseteq C_{1}\cdots C_{n-1}$

for some sequence $C_{1},\ldots,C_{n-1}$ of non-trivial conjugacy classes of
$G$. Moreover

$1\in S_{1}(G)\cdots S_{n-1}(G).$

Therefore, there exists a non-trivial conjugacy class $D$ such that $D$ is
disjoint from $C_{1}\cdots C_{n-1}$. This implies $1\not\in C_{1}\cdots
C_{n-1}D^{-1}$, so $\mathsf{ecn}(G)\geq n+1$. ∎

######  Lemma 2.3.

Let $q\geq 4$ be a prime power and $C_{1},C_{2},C_{3}$ not necessarily
distinct non-central conjugacy classes in $\mathrm{SL}_{2}(q)$.

  1. (i)

The product $C_{1}C_{2}C_{3}$ contains all non-central elements of
$\mathrm{SL}_{2}(q)$.

  2. (ii)

If $q\neq 5$, then $C_{1}C_{2}$ contains elements of order $q-1$ and order
$q+1$.

  3. (iii)

If $q=5$, then $C_{1}C_{2}$ contains an element of order $4$ and one of order
$3$ or order $6$.

###### Proof.

Let $G$ be a finite group. Recall that, by the extended Frobenius formula, the
number of solutions to the equation $g=x_{1}x_{2}\cdots x_{k}$ for a fixed
$g\in G$ and $x_{i}\in C_{i}=c_{i}^{G}$ ($1\leq i\leq k$) is given by

(2.1)
$\frac{\prod_{i=1}^{k}|C_{i}|}{|G|}\cdot\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(c_{1})\cdots\chi(c_{k})\chi(g^{-1})}{\chi(1)^{k-1}}.$

This implies that $g\in C_{1}\cdots C_{k}$ provided the above expression is
non-zero.

The result now follows from the well known character table of
$G:=\mathrm{SL}_{2}(q)$. ∎

######  Lemma 2.4.

Let $q\geq 4$ be a prime power, $d$ the g.c.d. of $q-1$ and $2$, and
$C_{1},C_{2},C_{3}$ not necessarily distinct non-central conjugacy classes in
$G:=\mathrm{PSL}_{2}(q)$. Let ${\mathcal{C}}_{q}$ denote the set of elements
of order $q$.

  1. (i)

The product $C_{1}C_{2}C_{3}$ equals $G$.

  2. (ii)

The product $C_{1}C_{2}$ contains elements of order $(q-1)/d$ and $(q+1)/d$

  3. (iii)

If $q\equiv 1\pmod{4}$ is prime, then $C_{1}C_{2}$ contains an element of
${\mathcal{C}}_{q}$.

  4. (iv)

If $q\equiv 1\pmod{4}$ is prime, then $C_{1}{\mathcal{C}}_{q}$ contains
$G\setminus\\{1\\}$.

  5. (v)

If $q\equiv 3\pmod{4}$ is prime and $C_{1},C_{2}$ are classes of involutions,
then $C_{1}C_{2}$ is disjoint from ${\mathcal{C}}_{p}$.

###### Proof.

For $q$ even, $\mathrm{PSL}_{2}(q)=\mathrm{SL}_{2}(q)$, but in either case (i)
and (ii) follow immediately from the corresponding statements in Lemma 2.3.
For $q$ an odd prime, ${\mathcal{C}}_{q}$ is the union of the two conjugacy
classes. Conclusions (iii)–(v) follow by computing (2.1) in the relevant cases
using the character table of $G:=\mathrm{PSL}_{2}(q)$. ∎

From this lemma, we easily deduce Proposition 1.4.

###### Proof of Proposition 1.4.

The two conjugacy classes in ${\mathcal{C}}_{q}$ are conjugate to one another
under $\mathrm{PGL}_{2}(q)$, so if $S_{1}(G)S_{2}(G)$ meets
${\mathcal{C}}_{q}$ at all, it contains it. If $S_{1}(G)$ and $S_{2}(G)$ each
contain a non-trivial conjugacy class of $G$, then by Gow’s theorem [G],
$S_{1}(G)S_{2}(G)$ contains all semisimple elements. If $q\equiv 1\pmod{4}$,
then by Lemma 2.4 (iii), $S_{1}(G)S_{2}(G)=G$, so $\mathsf{ccn}(G)\leq 2$. On
the other hand, $\mathsf{ccn}(G)\geq\mathsf{iw}(G)\geq 2$, so
$\mathsf{ccn}(G)=2$. If $q\equiv 3\pmod{4}$, then Lemma 2.4 (i) implies
$\mathsf{ccn}(G)\leq 3$, and Lemma 2.4 (v) implies $\mathsf{iw}(G)\geq 3$, so
$\mathsf{ccn}(G)=3$. ∎

######  Lemma 2.5.

If $q\geq 4$ and $C_{1}$ and $C_{2}$ are non-trivial $\mathrm{PGL}_{2}(q)$
orbits in $\mathrm{PSL}_{2}(q)$, then $C_{1}C_{2}$ covers all split regular
semisimple conjugacy classes in $\mathrm{PSL}_{2}(q)$.

###### Proof.

Again, this follows from easy character table computations. ∎

######  Lemma 2.6.

  1. (i)

If $C_{1},C_{2},C_{3}$ are non-trivial conjugacy classes in
$\mathrm{PSL}_{3}(q)$, then $C_{1}C_{2}C_{3}$ contains all non-trivial
elements of $\mathrm{PSL}_{3}(q)$.

  2. (ii)

If $2<q\equiv-1(\bmod\ 3)$ and $C_{1},C_{2},C_{3}$ are non-trivial conjugacy
classes in $\mathrm{PSU}_{3}(q)$, then $C_{1}C_{2}C_{3}$ contains all non-
trivial elements of $\mathrm{PSU}_{3}(q)$.

  3. (iii)

If $3\nmid(q+1)$ and $C_{1},C_{2},C_{3}$ are non-trivial conjugacy classes in
$\mathrm{SU}_{3}(q)$, then $C_{1}C_{2}C_{3}$ contains all elements of order
$q^{2}-q+1$ in $\mathrm{SU}_{3}(q)$. Moreover,
$\mathsf{ecn}(\mathrm{SU}_{3}(q))=5$.

###### Proof.

Parts (i) and (ii) follow immediately from the fact [O, Corollary 1.9] that
$\mathsf{ecn}(\mathrm{PSL}_{3}(q))=4$, and that
$\mathsf{ecn}(\mathrm{PSU}_{3}(q))=4$ for those $q$.

(iii) In this case,
$\mathrm{GU}_{3}(q)=\mathrm{SU}_{3}(q)\times\mathbf{Z}(\mathrm{GU}_{3}(q))$,
and so each $C_{i}=g_{i}^{\mathrm{SU}_{3}(q)}$ is a non-central
$\mathrm{GU}_{3}(q)$-conjugacy class; furthermore, any element $g$ of order
$q^{2}-q+1$ belongs to a class $C_{8}^{(k)}$ in [O, Table 1]. By assumption,
$\det(g_{i})=1$ for $1\leq i\leq 3$, and thus condition (2) of [O, Theorem
1.3] holds. Choosing $m=4$, we see that condition (3) of [O, Theorem 1.3] also
holds; see the first remark on [O, p. 221]. Now [O, Theorem 1.3] shows that
$1\in C_{1}C_{2}C_{3}\cdot(g^{-1})^{\mathrm{SU}_{3}(q)}$, and so $g\in
C_{1}C_{2}C_{3}$. The last statement is contained in [O, Corollary 1.9]. ∎

######  Lemma 2.7.

Let $G$ be a finite simple group of Lie type, and let $s,t\in G$ be two
semisimple elements such that

  1. (i)

For any pair of ample characteristic collections $S_{1},S_{2}$, the conjugacy
class of $s$ belongs to $S_{1}(G)S_{2}(G)$,

  2. (ii)

The product of the conjugacy class of $s$ and the conjugacy class of $t$
contains $G\smallsetminus\\{1\\}$,

  3. (iii)

The element $s$ is regular.

Then $\mathsf{ccn}(G)\leq 6$.

###### Proof.

By the theorem of Gow [G], every semisimple element in $G$, in particular $t$,
can be written as a product of two conjugates of $s$. Now, if
$S_{3},\ldots,S_{6}$ are ample characteristic collections, then
$s^{G}\subseteq S_{3}S_{4}$ and $s^{G}\subseteq S_{5}S_{6}$, whence $t^{G}$
lies in $S_{3}(G)S_{4}(G)S_{5}(G)S_{6}(G)$. Therefore

$G\smallsetminus\\{1\\}\subset s^{G}\cdot t^{G}\subseteq S_{1}(G)\cdots
S_{6}(G),$

and it follows that $\mathsf{ccn}(G)\leq 6$. ∎

######  Lemma 2.8.

Let $G$ be a finite group, and let $\mathfrak{d}(G)$ the lowest degree of a
non-trivial character of $G$. Let $q\geq 4$ and
$\varphi\colon\mathrm{SL}_{2}(q)\to G$ be a non-trivial homomorphism. Suppose
that

(2.2) $|\mathbf{C}_{G}(\varphi(s))|^{3/2}\leq\mathfrak{d}(G)$

for any $s\in\mathrm{SL}_{2}(q)$ of order $q-1$, or for any
$s\in\mathrm{SL}_{2}(q)$ of order $q+1$ if $q\neq 5$, or for any
$s\in\mathrm{SL}_{2}(q)$ of order $3$ or $6$ if $q=5$. Then
$\mathsf{ccn}(G)\leq 6$.

###### Proof.

By Lemma 2.4, if $S_{1},S_{2}$ are ample characteristic collections, then
$g:=\varphi(s)\in S_{1}(G)S_{2}(G)$, where $s\in\mathrm{SL}_{2}(q)$ can be
chosen to have order $q-1$, or $q+1$ if $q\neq 5$, or either $3$ or $6$ when
$q=5$. Let $C:=g^{G}$. By the Frobenius formula, $C^{3}=G$ if

$\sum_{1_{G}\neq\chi\in\mathrm{Irr}(G)}\frac{|\chi(g)|^{3}}{\chi(1)}<1.$

For any $\chi$ in the summation, $|\chi(g)|\leq|\mathbf{C}_{G}(g)|^{1/2}$ by
the centralizer bound for character values, and $\chi(1)\geq\mathfrak{d}(G)$.
Hence, by the second orthogonality relation,

$\sum_{1_{G}\neq\chi\in\mathrm{Irr}(G)}\frac{|\chi(g)|^{3}}{\chi(1)}<\frac{|\mathbf{C}_{G}(g)|^{1/2}}{\mathfrak{d}(G)}\cdot\sum_{\chi\in\mathrm{Irr}(G)}|\chi(g)|^{2}=\frac{|\mathbf{C}_{G}(g)|^{3/2}}{\mathfrak{d}(G)},$

and we can deduce the needed inequality from (2.2). ∎

We conclude this section with two results which allow us to prove that
$\mathsf{ccn}(G)\leq 4$ when $G$ is of Lie type and is sufficiently large
compared to its rank.

######  Proposition 2.9.

Let ${\underline{G}}$ denote a connected, simply connected simple algebraic
group of rank $r$ over ${\mathbb{F}}_{q}$ and $G$ the finite simple group
obtained by taking the quotient of ${\underline{G}}({\mathbb{F}}_{q})$ by its
center. Let $r=r_{1}+\cdots+r_{k}$ be a partition. If there exists a
homomorphism
$\phi\colon\mathrm{SL}_{2}(q^{r_{1}})\times\cdots\times\mathrm{SL}_{2}(q^{r_{k}})\to
G$ with central kernel and if $q$ is sufficiently large in terms of $r$, then
$\mathsf{ccn}(G)\leq 4$.

###### Proof.

The number of ${\mathbb{F}}_{q}$-points on a torus of rank $s$ is at most
$(q+1)^{s}$. The regular elements in a maximal torus ${\underline{T}}$ of
${\underline{G}}$ lie in a union of proper subtori of ${\underline{T}}$
indexed by positive roots, so the number of elements in
${\underline{T}}({\mathbb{F}}_{q})$ which are not regular semisimple is less
than $2r^{2}(q+1)^{r-1}$.

We write $H:=\prod_{i=1}^{k}\mathrm{SL}_{2}({q^{r_{i}}})/Z$, where
$Z:=\ker\phi$ is some subgroup of the product of the centers of the
$\mathrm{SL}_{2}({\mathbb{F}}_{q^{r_{i}}})$ and is therefore of order at most
$2^{r}$. The inclusion of $H$ in $G$ gives a homomorphism

$\prod_{i=1}^{k}{\mathbb{F}}_{q^{r_{i}}}^{\times}\to G.$

The image under this homomorphism of elements which are non-trivial in each
coordinate is at least

$2^{-r}(1-2q^{-1})^{r}q^{r}.$

However, the image lies in a maximal torus $T$ of $G$, i.e., the image of the
${\mathbb{F}}_{q}$-points of a maximal torus ${\underline{T}}$ of
${\underline{G}}$ in $G$. Lemma 2.5 therefore gives a lower bound of the form
$c_{r}q^{r}$ for the cardinality of the set of regular semisimple elements of
$T$ which lie in $S_{1}(H)S_{2}(H)$ where $S_{1}$ and $S_{2}$ are ample
characteristic collections. The conjugacy class of each such element meets $T$
in at most $|W|$ elements, where $W$ denotes the Weyl group of
${\underline{G}}$ with respect to ${\underline{T}}$, whose order is bounded in
terms of $r$. Each conjugacy class of a regular semisimple element in
${\underline{G}}({\mathbb{F}}_{q})$ has at least
$(q+1)^{-r}|{\underline{G}}({\mathbb{F}}_{q})|$ elements, so overall, we get a
positive lower bound, depending only on $r$, for the proportion of elements of
$G$ which lie in $S_{1}(G)S_{2}(G)$. Since $S_{1}(G)S_{2}(G)$ is a normal
subset of $G$, [LST2, Theorem A] implies that
$S_{1}(G)S_{2}(G)S_{3}(G)S_{4}(G)$ contains all non-trivial elements of $G$. ∎

Note that the same argument works also for Suzuki and Ree groups.

######  Lemma 2.10.

Let ${\underline{G}}$ be a connected, simply connected simple algebraic group
of rank $r$ over ${\mathbb{F}}_{q}$ and $G$ the quotient of
${\underline{G}}({\mathbb{F}}_{q})$ by its center. Let
$\phi\colon\mathrm{SL}_{2}^{k}\to{\underline{G}}$ be a homomorphism of
algebraic groups over ${\mathbb{F}}_{q}$ whose kernel lies in the center of
$\mathrm{SL}_{2}^{k}$. Let ${\mathbb{G}}_{m}^{k}$ denote a split maximal torus
of $\mathrm{SL}_{2}^{k}$, and suppose $\phi({\mathbb{G}}_{m}^{k})$ is a
maximal split torus of ${\underline{G}}$ which contains regular
${\mathbb{F}}_{q}$-points. Then if $q$ is sufficiently large in terms of $r$,
we have $\mathsf{ccn}(G)\leq 4$.

Note that if ${\underline{G}}$ is split, then $\phi({\mathbb{G}}_{m}^{k})$ is
a maximal torus, so it contains regular elements over ${\mathbb{F}}_{q}$ if
$q$ is sufficiently large. If ${\underline{G}}$ is not split, a maximal split
torus is not a maximal torus, so $k<r$.

###### Proof.

For $q$ sufficiently large, there exists
$(a_{1},\ldots,a_{k})\in({\mathbb{G}}_{m}({\mathbb{F}}_{q})\smallsetminus\\{1\\})^{k}$
such that $\phi(a_{1},\ldots,a_{k})$ is regular. By Lemma 2.5,
$(a_{1},\ldots,a_{k})\in
S_{1}(\mathrm{SL}_{2}(q)^{k})S_{2}(\mathrm{SL}_{2}(q)^{k})$ if $S_{1}$ and
$S_{2}$ are ample characteristic collections. By a theorem of Ellers and
Gordeev [EG, Theorem 1] the product of $G$-conjugacy classes of any two
regular semisimple elements lying in maximal split tori of $G$ covers
$G\smallsetminus\\{1\\}$. ∎

##  3\. Classical groups

For every positive integer $n$, we denote by $\Phi_{n}(x)$ the $n$th
cyclotomic polynomial. We recall [Zs] that for $n>2$ and $(n,q)\neq(6,2)$,
$\Phi_{n}(q)$ is always divisible by a Zsigmondy prime $\ell$, meaning that
$\ell$ does not divide $q$, which implies and the order of $q$ (mod $\ell$) is
exactly $n$. In particular, $\ell\equiv 1\pmod{n}$.

######  Theorem 3.1.

If $G$ is a finite simple classical group, then $\mathsf{ccn}(G)\leq 6$.

###### Proof.

We consider all the six types $A_{r}$, ${}^{2}\\!A_{r}$, $B_{r}$, $C_{r}$,
$D_{r}$, and ${}^{2}\\!D_{r}$. Let $S_{1},\ldots,S_{6}$ be ample
characteristic collections and set $d:=\gcd(2,q-1)$.

Case $G=\mathrm{PSL}_{n}(q)$ with $n\geq 2$ and $(n,q)\neq(2,2)$, $(2,3)$.
Note that $\mathsf{ecn}(\mathrm{PSL}_{2}(q))=4$ for $q\geq 4$ [AH, p. 2], and
$\mathsf{ecn}(\mathrm{PSL}_{3}(q))=4$ [O, Corollary 1.9], so we may assume
$n\geq 4$. For $(n,q)=(6,2)$, $(7,2)$, we have $\mathrm{SL}_{2}(8)<G$, and any
element of order $7$ lies in $S_{1}(G)S_{2}(G)$ for $S_{i}$ ample
characteristic collections. Using the character tables for these two groups in
the GAP library, we check that the square of this conjugacy class covers the
non-trivial elements of the group, so $\mathsf{ccn}(G)\leq 4$.

We may therefore assume that $(n,q)\neq(6,2)$, $(7,2)$. Write $q=p^{f}$ for a
prime $p$, and $m:=\lfloor n/2\rfloor$ so that $m\geq 2$ and
$n\in\\{2m,2m+1\\}$. By [Zs] and the assumption on $(n,q)$, we can find a
Zsigmondy prime $\ell$ for $(2mf,p)$. By Lemma 2.4,
$S_{1}(\mathrm{PSL}_{2}(q^{m}))S_{2}(\mathrm{PSL}_{2}(q^{m}))$ contains an
element of order $(q^{m}+1)/d$. Since
$\mathrm{SL}_{2}(q^{m})\leq\mathrm{SL}_{2m}(q)\leq\mathrm{SL}_{n}(q)$, it
follows that $S_{1}(G)S_{2}(G)$ contains an element $s$ of order divisible by
$(q^{m}+1)/d$ which is divisible by $\ell$, and so $s$ is regular semisimple.
As discussed in [GT, §2.2.1], one can find a semisimple element $t\in G$ such
that $s^{G}\cdot t^{G}\supseteq G\smallsetminus\\{1\\}$. Hence
$\mathsf{ccn}(G)\leq 6$ by Lemma 2.7. Case $G=\mathrm{PSU}_{n}(q)$ with $n\geq
3$ and $(n,q)\neq(3,2)$. Note that $\mathsf{ecn}(\mathrm{PSU}_{3}(q))\leq 5$
[O, Corollary 1.9], so we may assume $n\geq 4$. If $G=\mathrm{PSU}_{6}(2)$,
then $\mathrm{SL}_{2}(8)<G$, and we easily check that the square of the class
in $G$ of any element of order $7$ covers the nontrivial elements of $G$, so
$\mathsf{ccn}(G)\leq 4$.

We therefore assume that $(n,q)\neq(6,2)$. Again write $m:=\lfloor
n/2\rfloor$, so that $m\geq 2$ and $n\in\\{2m,2m+1\\}$. By [Zs] and the
assumption on $(n,q)$, we can find a Zsigmondy prime $\ell$ for $(2mf,p)$ when
$2|m$, and for $(mf,p)$ when $2\nmid m$; note that $\ell$ divides $q^{2m}-1$,
but not $\prod^{2m-1}_{i=1}((-q)^{i}-1)$. By Lemma 2.4,
$S_{1}(\mathrm{PSL}_{2}(q^{m}))S_{2}(\mathrm{PSL}_{2}(q^{m}))$ contains an
element of order $(q^{m}+(-1)^{m})/d$. Note that

$\mathrm{SL}_{2}(q^{m})\leq\mathrm{SL}_{m}(q^{2})<\mathrm{SU}_{2m}(q)\leq\mathrm{SU}_{n}(q)$

when $2|m$, and

$\mathrm{SL}_{2}(q^{m})\cong\mathrm{SU}_{2}(q^{m})\leq\mathrm{SU}_{2m}(q)\leq\mathrm{SU}_{n}(q)$

when $2\nmid m$. It follows that $S_{1}(G)S_{2}(G)$ contains an element $s$ of
order divisible by $(q^{m}+(-1)^{m})/d$ which is divisible by $\ell$, and so
$s$ is regular semisimple (as one can see by checking the eigenvalues of $s$).
As discussed in [GT, §2.2.2], one can find a semisimple element $t\in G$ such
that $s^{G}\cdot t^{G}\supseteq G\smallsetminus\\{1\\}$. Hence
$\mathsf{ccn}(G)\leq 6$ by Lemma 2.7. Case $G=\mathrm{PSp}_{2n}(q)$ with
$n\geq 2$ and $(n,q)\neq(2,2)$. (Note that
$\mathrm{Sp}_{4}(2)^{\prime}\cong\mathrm{PSL}_{2}(9)$). By Lemma 2.4,
$S_{1}(\mathrm{PSL}_{2}(q^{n}))S_{2}(\mathrm{PSL}_{2}(q^{n}))$ contains an
element $\overline{s}$ of order $(q^{n}-1)/d$ and
$S_{3}(\mathrm{PSL}_{2}(q^{n}))S_{4}(\mathrm{PSL}_{2}(q^{n}))$ an element
$\overline{t}$ of order $(q^{n}+1)/d$. Since

$\mathrm{SL}_{2}(q^{n})\cong\mathrm{Sp}_{2}(q^{n})<\mathrm{Sp}_{2n}(q),$

$S_{1}(G)S_{2}(G)$ contains regular semisimple elements $s$ of order divisible
by $(q^{n}-1)/d$, and $S_{3}(G)S_{4}(G)$ contains a regular semisimple element
of order divisible by $(q^{n}+1)/2$. As discussed in [GT, §2.2.3], $s^{G}\cdot
t^{G}\supseteq G\smallsetminus\\{1\\}$. Hence
$S_{1}(G)S_{2}(G)S_{3}(G)S_{4}(G)\supseteq G\smallsetminus\\{1\\}$, and
$\mathsf{ccn}(G)\leq 4$. Case $G=P\Omega^{\epsilon}_{n}(q)$ with $n\geq 7$ and
$\epsilon=\pm$. First assume that $n=7$ and $2\nmid q$. Note that

$\Omega_{7}(q)>\mathrm{SO}^{+}_{6}(q)>\mathrm{SL}_{3}(q),~{}\Omega_{7}(q)>\mathrm{SO}^{-}_{6}(q)>\mathrm{SU}_{3}(q).$

Hence, by Lemma 2.6, $S_{1}(G)S_{2}(G)S_{3}(G)$ contains a regular semisimple
element $s$ of order $q^{2}+q+1$ and $S_{4}(G)S_{5}(G)S_{6}(G)$ contains a
regular semisimple element $t$ of order divisible by
$(q^{2}-q+1)/\gcd(3,q+1)$. As discussed in [GT, §2.2.3], $s^{G}\cdot
t^{G}\supseteq G\smallsetminus\\{1\\}$. Hence $\mathsf{ccn}(G)\leq 6$.

We may now assume $n\geq 8$, so that $m:=\lfloor n/4\rfloor\geq 2$. Note that

$\mathrm{PSL}_{2}(q^{2m})\cong\Omega^{-}_{4}(q^{m})\leq\Omega^{-}_{4m}(q),$

and $S_{1}(\mathrm{PSL}_{2}(q^{2m}))S_{2}(\mathrm{PSL}_{2}(q^{2m}))$ contains
an element $\tilde{s}$ of order $(q^{2m}+1)/d$. Such an element $s$ is regular
semisimple in each of the terms of

$\Omega^{-}_{4m}(q)<\Omega_{4m+1}(q)<\Omega^{\pm}_{4m+2}(q).$

Hence, if
$G\in\\{P\Omega^{-}_{4m}(q),\Omega_{4m+1}(q),P\Omega^{\pm}_{4m+2}(q)\\}$, then
$S_{1}(G)S_{2}(G)$ contains a regular semisimple element $s$ of order
$(q^{2m}+1)/d$. As discussed in [GT, §§2.2.3, 2.2.4], one can find a
semisimple element $t\in G$ such that $s^{G}\cdot t^{G}\supseteq
G\smallsetminus\\{1\\}$. Hence $\mathsf{ccn}(G)\leq 6$ by Lemma 2.7. In
particular, we are done with type ${}^{2}\\!D_{r}$.

Consider the case of $G=P\Omega^{+}_{4m}(q)$, and note that

$\mathrm{PSL}_{2}(q^{2})\times\mathrm{PSL}_{2}(q^{2m-2})\leq\Omega^{-}_{4}(q)\times\Omega^{-}_{4m-4}(q)<\Omega^{+}_{4m}(q).$

Hence, if $m\geq 3$, then applying Lemma 2.4 we see that $S_{1}(G)S_{2}(G)$
contains an element $s=(x,y)$, with $x\in S_{1}(\mathrm{PSL}_{2}(q^{2}))$ of
order $(q^{2}+1)/d$ and $y\in S_{2}(\mathrm{PSL}_{2}(q^{2m-2}))$ of order
$(q^{2m-2}+1)/d$. This element $s$ is regular semisimple of type
$T^{-,-}_{2,2m-2}$, and, as shown in [LST1, Proposition 7.1.1] and [GM,
Theorem 7.6], there exists a semisimple element $t\in G$ such that $s^{G}\cdot
t^{G}\supseteq G\smallsetminus\\{1\\}$. Hence $\mathsf{ccn}(G)\leq 6$ by Lemma
2.7. Suppose now that $m=2$, but $q\geq 3$. Then

$\mathrm{SL}_{3}(q)<\mathrm{SO}^{+}_{6}(q)\hookrightarrow\Omega^{+}_{8}(q),~{}\mathrm{SU}_{3}(q)<\mathrm{SO}^{-}_{6}(q)\hookrightarrow\Omega^{+}_{8}(q).$

By Lemma 2.6, $S_{1}(G)S_{2}(G)S_{3}(G)$ contains an element $s$ of order
divisible by $(q^{2}+q+1)/\gcd(3,q-1)$, and such an element is regular
semisimple of type $T^{+,+}_{3,1}$ in $G$, see [GT, §2.1]. Likewise,
$S_{4}(G)S_{5}(G)S_{6}(G)$ contains an element $t$ of order divisible by
$(q^{2}-q+1)/\gcd(3,q+1)$, and such an element is regular semisimple of type
$T^{-,-}_{3,1}$ in $G$ (since $q\geq 3$). By [GT, Lemma 2.4], $s^{G}\cdot
t^{G}=G\smallsetminus\\{1\\}$, whence $\mathsf{ccn}(G)\leq 6$. In the case
$G=\Omega^{+}_{8}(2)$, one can check directly that $\mathsf{ccn}(G)\leq 4$,
using an element of order $7$ in $\mathrm{SL}_{2}(8)<G$.

Consider the case of $G=\Omega_{4m+3}(q)$ with $m\geq 2$, and note that

$\mathrm{PSL}_{2}(q^{2m})<\Omega^{-}_{4m}(q)\times\Omega_{3}(q)<\Omega^{+}_{4m+3}(q).$

Again using Lemma 2.4 we see that $S_{1}(G)S_{2}(G)$ contains an element $s\in
S_{1}(\mathrm{PSL}_{2}(q^{2m}))$ of order $(q^{2m}+1)/d$. Note that

$\mathbf{C}_{\mathrm{SO}_{4m+3}(q)}(s)=T\times\mathrm{SO}_{3}(q),$

where $T<\mathrm{SO}^{-}_{4m}(q)$ has order $q^{2m}+1$. Since
$\mathrm{SO}_{3}(q)\cap G=\Omega_{3}(q)$ has index $2$ in
$\mathrm{SO}_{3}(q)$, it follows that

$|\mathbf{C}_{G}(s)|=(q^{2m}+1)(q^{3}-q)/2.$

Let $B:=q^{8m-4}$. By [LOST, Corollary 5.8], $\Omega$ has exactly $q+4$
nontrivial irreducible characters of degree $\leq B$, which are the characters
$D^{\circ}_{\alpha}$ with $\alpha\in\mathrm{Irr}(\mathrm{Sp}_{2}(q))$, listed
in [LOST, Proposition 5.7]. The proof of Lemma 2.8 shows that

(3.1)
$\sum_{\chi\in\mathrm{Irr}(G),~{}\chi(1)>B}\frac{|\chi(s)|^{3}}{\chi(1)}<\frac{|\mathbf{C}_{G}(s)|^{3/2}}{B}\leq\frac{\bigl{(}q^{2m}+1)(q^{3}-q)/2\bigr{)}^{3/2}}{q^{8m-4}}<\frac{2^{-3/2}}{q^{5m-17/2}}<0.07.$

The degrees of $D^{\circ}_{\alpha}$ are listed in [LOST, Table I], showing
that two of them have $\ell$-defect $0$ for $\ell$ a Zsigmondy prime for
$(n-3,q)$ and so vanish at $s$. Next we estimate $|\chi(s)|$ for the remaining
$q+2$ characters. Note that, in the action of $s$ on the natural $G$-module
${\mathbb{F}}_{q}^{n}$, $s$ has a unique eigenvalue $\lambda$ that belongs to
${\mathbb{F}}_{q^{2}}$, and this eigenvalue is $\lambda=1$ and has
multiplicity $3$. Consider the action of $x\otimes s$ on
$V:={\mathbb{F}}_{q}^{2}\otimes{\mathbb{F}}_{q}^{n}$ for any
$x\in\mathrm{Sp}_{2}(q)$ and ${\mathbb{F}}_{q}^{2}$ being the natural module
for $\mathrm{Sp}_{2}(q)$. Then the fixed point subspace of $x\otimes s$ on $V$
has dimension $6$ if $x=1$, $3$ if $1\neq x$ is unipotent, and $0$ otherwise.
It follows from the formula [LOST, Lemma 5.5] for $D_{\alpha}$ that

(3.2)
$|D_{\alpha}(s)|\leq\frac{\alpha(1)}{q(q^{2}-1)}\bigl{(}q^{3}+q^{3/2}(q^{2}-1)+q(q^{2}-1)-q^{2}\bigr{)}=\alpha(1)(q^{1/2}+2q/(q+1)).$

Note that $D^{\circ}_{\alpha}=D_{\alpha}$, unless $\alpha(1)=(q+1)/2$ in which
case $D^{\circ}_{\alpha}=D_{\alpha}-1_{G}$. Now, if $q=3$, then $\alpha(1)\leq
3$, and so (3.2) shows that $|D^{\circ}_{\alpha}(s)|<10.5$. Since
$D^{\circ}_{\alpha}(1)\geq 7260$, it follows that

$\sum_{\chi\in\mathrm{Irr}(G),~{}1<\chi(1)\leq
B}\frac{|\chi(s)|^{3}}{\chi(1)}\leq\frac{5\cdot(10.5)^{3}}{7260}<0.8.$

If $q\geq 5$, then since $\alpha(1)\leq q+1$, (3.2) shows that
$|D^{\circ}_{\alpha}(s)|\leq 2q+q^{1/2}(q+1)$. As
$D^{\circ}_{\alpha}(1)>q^{8}$, it follows that

$\sum_{\chi\in\mathrm{Irr}(G),~{}1<\chi(1)\leq
B}\frac{|\chi(s)|^{3}}{\chi(1)}\leq\frac{(q+2)\cdot\bigl{(}2q+q^{1/2}(q+1)\bigr{)}^{3}}{q^{8}}<0.2.$

Together with (3.1), we have shown that

$\sum_{1_{G}\neq\chi\in\mathrm{Irr}(G)}\frac{|\chi(s)|^{3}}{\chi(1)}<0.87,$

and so $(s^{G})^{3}\supseteq G\smallsetminus\\{1\\}$ by the Frobenius formula.
Consequently, $\mathsf{ccn}(G)\leq 6$. ∎

######  Proposition 3.2.

If $G$ is a classical group and the order of $G$ is sufficiently large in
terms of the rank of $G$, then $\mathsf{ccn}(G)\leq 4$.

###### Proof.

We do a case analysis.

Case $G=\mathrm{PSL}_{n}(q)$. Let $m=\lfloor n/2\rfloor$. The obvious
homomorphism $\mathrm{SL}_{2}(q)^{m}\to\mathrm{SL}_{n}(q)$ maps
$(x_{1},\ldots,x_{m})$ to a regular semisimple element in a split maximal
torus of $\mathrm{SL}_{n}$ over ${\mathbb{F}}_{q}$ and hence to a split
regular semisimple element of $G$, whenever the $x_{i}$ are regular semisimple
elements with eigenvalues $\lambda_{i}^{\pm 1}$,
$\lambda_{i}\in{\mathbb{F}}_{q}^{\times}$, and
$\lambda_{i}\neq\lambda_{j}^{\pm 1}$ for all $i,j$. This case now follows from
Lemma 2.10.

Case $G=\mathrm{PSU}_{n}(q)$. Let $m=\lfloor n/2\rfloor$. We proceed exactly
as before, using the obvious homomorphism from
$\mathrm{SL}_{2}(q)^{m}=\mathrm{SU}_{2}(q)^{m}$ to $\mathrm{SU}_{n}(q)$. An
$m$-tuple of regular semisimple elements such that
$\lambda_{i}\neq\lambda_{j}^{\pm 1}$ for all $i,j$ maps to an element in an
$m$-dimensional split torus of $G$, and this is a maximal split torus [T,
Table 2].

Case $G=\mathrm{PSp}_{2n}(q)$. We proceed as before, using the obvious
homomorphism from $\mathrm{SL}_{2}(q)^{n}\cong\mathrm{Sp}_{2}(q)^{n}$ to
$\mathrm{Sp}_{2n}(q)$, which maps any split maximal torus of
$\mathrm{SL}_{2}^{n}$ to a split maximal torus of $\mathrm{Sp}_{2n}$.

Case $G=P\Omega_{2n+1}(q)$. If $n=2m$, we proceed as before, using composition
of the obvious homomorphisms from $\mathrm{SL}_{2}(q)^{n}$ to

$\mathrm{Spin}_{5}(q)*\underbrace{\mathrm{Spin}^{+}_{4}(q)*\ldots*\mathrm{Spin}^{+}_{4}(q)}_{m-1}\cong\mathrm{Sp}_{4}(q)*\underbrace{\mathrm{SL}_{2}(q)*\ldots*\mathrm{SL}_{2}(q)}_{n-2}$

and
$\mathrm{Spin}_{5}(q)*\underbrace{\mathrm{Spin}^{+}_{4}(q)*\ldots*\mathrm{Spin}^{+}_{4}(q)}_{m-1}\to\mathrm{Spin}_{2n+1}(q)$.
If $n=2m+1$, we map $\mathrm{SL}_{2}(q)^{n}$ onto
$\mathrm{Spin}_{3}(q)*\underbrace{\mathrm{Spin}^{+}_{4}(q)*\ldots*\mathrm{Spin}^{+}_{4}(q)}_{m}$
and maps the latter to $\mathrm{Spin}_{2n+1}(q)$.

Case $G=P\Omega_{2n}^{+}(q)$. If $n=2m$, we map $\mathrm{SL}_{2}(q)^{n}$ onto
$\underbrace{\mathrm{Spin}^{+}_{4}(q)*\ldots*\mathrm{Spin}^{+}_{4}(q)}_{m}$
and map the latter to $\mathrm{Spin}_{2n}^{+}(q)$. If $n=2m+1$, we map
$\mathrm{SL}_{2}(q)^{n-1}$ to

$\mathrm{SL}_{4}(q)*\underbrace{\mathrm{SL}_{2}(q)*\ldots*\mathrm{SL}_{2}(q)}_{n-3}\cong\mathrm{Spin}_{6}^{+}(q)*\underbrace{\mathrm{Spin}_{4}^{+}(q)*\ldots*\mathrm{Spin}_{4}^{+}(q)}_{m-1},$

which embeds in the usual way in $\mathrm{Spin}_{2n}^{+}(q)$.

Case $G=P\Omega_{2n}^{-}(q)$. If $n=2m+1$, we map
$\mathrm{SL}_{2}(q)^{n-1}=\mathrm{SU}_{2}(q)^{2}\times\mathrm{SL}_{2}(q)^{n-3}$
onto

$\mathrm{SU}_{4}(q)*\underbrace{\mathrm{SL}_{2}(q)*\ldots*\mathrm{SL}_{2}(q)}_{n-3}\cong\mathrm{Spin}_{6}^{-}(q)*\underbrace{\mathrm{Spin}_{4}^{+}(q)*\ldots*\mathrm{Spin}^{+}_{4}(q)}_{m-1},$

which embeds in the usual way in $\mathrm{Spin}_{2n}^{-}(q)$. Note that an
ordered pair of regular semisimple elements of $\mathrm{SU}_{2}(q)$ with
distinct eigenvalue pairs gives a regular semisimple element of
$\mathrm{SU}_{4}(q)$, and it follows that the image of a maximal torus of
$\mathrm{SL}_{2}^{n-1}$ in $\mathrm{Spin}_{2n}^{-}(q)$ meets the regular
semisimple locus. As $\mathrm{Spin}_{2n}^{-}$ is not split, it cannot have a
split torus of rank $n$, so any split torus of rank $n-1$ must be a maximal
split torus. It follows that the image of a split maximal torus of
$\mathrm{SL}_{2}^{n-1}$ is a maximal split torus of $\mathrm{Spin}_{2n}^{-}$.

If $n=2m$, we identify $\mathrm{Spin}_{4}^{-}(q)$ with
$\mathrm{SL}_{2}(q^{2})$ and map
$\mathrm{SL}_{2}(q^{2})\times\mathrm{SL}_{2}(q)^{n-2}$ into
$\mathrm{Spin}_{2n}^{-}(q)$. This maps $(y,x_{1},\ldots,x_{n-2})$ to a regular
semisimple element when $y$ and the $x_{i}$ are regular semisimple, the
eigenvalues of all $x_{i}$ lie in ${\mathbb{F}}_{q}$, the eigenvalues of $y$
do not lie in ${\mathbb{F}}_{q}$, and no two $x_{i}$ have an eigenvalue in
common. We can no longer use the Ellers-Gordeev method, but Proposition 2.9
applies, so we still obtain $\mathsf{ccn}(G)\leq 4$. ∎

##  4\. Exceptional groups of Lie type

The following lemma will be useful for us.

######  Lemma 4.1.

Let $G$ be an exceptional group of Lie type and $s$ an element of $G$ whose
order is divisible by a prime $\ell$ which is Zsigmondy for $(n,q)$. Then

  1. (i)

If $G=F_{4}(q)$ and $n=8$, then $s$ is regular semisimple with centralizer
order $q^{4}+1$.

  2. (ii)

If $G={}^{2}\\!F_{4}(q)$ with $q\geq 8$, $n=4$, and $\ell>5$, then $s$ is
regular semisimple with centralizer order $q^{2}+1$.

  3. (iii)

If $G=E_{6}(q)$ or $G={}^{2}\\!E_{6}(q)$, and $n=8$, then $s$ is regular
semisimple with centralizer order $(q^{2}-1)(q^{4}+1)$.

  4. (iv)

If $G=E_{7}(q)$ and $n=7$, then $s$ is regular semisimple with centralizer
order $q^{7}-1$.

  5. (v)

If $G=E_{8}(q)$ and $n=7$, then $s$ has centralizer order dividing
$(q^{3}-q)(q^{7}-1)$.

###### Proof.

In all five cases, $\ell-1$ has a divisor $\geq 5$, so $\ell\geq 11$. Let $t$
denote a power of $s$ of exact order $\ell$, so $t$ is semisimple. By [MT,
Lemma 2.2], the connected center $T$ of the centralizer of $t$ has order
divisible by $\ell$. It suffices to prove the stated claims for $t$, since the
centralizer of $t$ contains the centralizer of $s$. In general, $\ell$ divides
$\Phi_{m}(q)$ only if $m$ is $n$ times a power of $\ell$. The order of every
torus $T$ in $G$ can be written $\prod_{j}\Phi_{i_{j}}(q)$, where
$\sum\phi(i_{j})$ is the rank of the torus; in particular, $\phi(i_{j})\leq
8$, which implies that $i_{j}<n\ell$, so if $\ell$ divides $|T$, then
$i_{j}=n$ for some $i_{j}$.

In the $F_{4}$ case, $\phi(8)=4$ so $T$ must have order $\Phi_{8}(t)$, so $t$
must be regular semisimple. The same argument applies to ${}^{2}\\!F_{4}(q)$
(which is ${}^{2}\\!F_{4}(r^{2})$ with $r=q^{1/2}$). In the $E_{6}$ and
${}^{2}E_{6}$ cases, we consult the connected centralizers in the $E_{6}$
table in [FJ1] and conclude that $t$ must be regular semisimple and associated
to $w_{19}$, implying the stated centralizer order. The $E_{7}$-table in [FJ1]
has a $\Phi_{7}(q)$ factor in the connected centralizer of $t$ only for the
regular semisimple case associated to $w_{39}$. The $E_{8}$-table in [FJ2] has
a $\Phi_{7}(q)$ factor in three cases: the regular semisimple classes
associated to $w_{37}$ and $w_{55}$, with centralizer orders $(q-1)(q^{7}-1)$
and $(q+1)(q^{7}-1)$ respectively and the $(A_{1},w_{51})$ class, with
centralizer order $(q^{3}-q)(q^{7}-1)$. ∎

######  Theorem 4.2.

If $G$ is a finite simple group of exceptional Lie type, then
$\mathsf{ccn}(G)\leq 6$.

###### Proof.

We consider each of the ten possibilities. Case ${}^{2}B_{2}(q)$. As $G$ is a
finite simple group, we have $q\geq 8$. By [AH, p. 2], $\mathsf{ecn}(G)=4$, so
by Lemma 2.2, $\mathsf{ccn}(G)\leq 3$. Case $G_{2}(q)^{\prime}$. The character
table for $q=2$ is in the GAP library, and we easily compute that the extended
covering number of this group is $5$, so $\mathsf{ccn}(G)\leq 4$. Otherwise,
$q\geq 3$, so $G_{2}(q)^{\prime}=G_{2}(q)$. By [LSS, Table 5.1], we have
$\mathrm{SL}_{3}(q)<G_{2}(q)$. Let $e:=\gcd(3,q-1)=\gcd(3,q^{2}+q+1)$. By
Lemma 2.6, for all $S_{1},S_{2},S_{3}$, every element $\overline{s}$ of order
$\frac{q^{2}+q+1}{e}$ in $\mathrm{PSL}_{3}(q)$ belongs to
$S_{1}(\mathrm{PSL}_{3}(q))S_{2}(\mathrm{PSL}_{3}(q))S_{3}(\mathrm{PSL}_{3}(q))$,
and let $s$ be any lift of $\overline{s}$ to $\mathrm{SL}_{3}(q)$. By [GM,
Table 10], every non-trivial element of $G$ is a product of two such elements
$s$, so $\mathsf{ccn}(G_{2})\leq 6$. Case ${}^{2}G_{2}(q)^{\prime}$. When
$q=3$, this is $\mathrm{PSL}_{2}(8)$, so we have $\mathsf{ccn}(G)\leq 3$. When
$q\geq 27$, ${}^{2}G_{2}(q)$ is already simple. By [LSS, Table 5.1],
$\mathrm{PSL}_{2}(q)<G$, so by Lemma 2.4, any $S_{1}(G)S_{2}(G)$ contains an
element $s$ of order $\frac{q+1}{2}$, and by [GM, Theorem 7.1], every element
of $G\smallsetminus\\{1\\}$ is a product of two conjugates of $s$. Thus
$\mathsf{ccn}(G)\leq 4$. Case ${}^{3}D_{4}(q)$. We compute

$\displaystyle\mathsf{ecn}({}^{3}D_{4}(2))=7,~{}\mathsf{ecn}({}^{3}D_{4}(3))=6,$

using the character table in the GAP library for the former and Frank Lübeck’s
character table, computed using the generic character table for ${}^{3}D_{4}$
in CHEVIE [GHLMP], for the latter. So we may assume $q\geq 4$.

By [LSS, Table 5.1], we have
$\mathrm{SL}_{2}(q)\times\mathrm{SL}_{2}(q^{3})<{}^{3}D_{4}(q)$ if $q$ is even
and the central product $\mathrm{SL}_{2}(q)\ast\mathrm{SL}_{2}(q^{3})$ if $q$
is odd. In either case, by Lemma 2.3, given ample characteristic collections
$S_{1}$ and $S_{2}$, this subgroup contains an element $s$ of
$S_{1}(G)S_{2}(G)$ of order $(q-1)(q^{3}+1)/d$ where $d:=\gcd(2,q-1)$. This
element is regular semisimple of type $s_{11}$ in the Deriziotis-Michler
classification [DM, Table 2.1]; it follows that the order of its centralizer
is $(q-1)(q^{3}+1)<q^{4}$. From Lübeck’s table of degrees for ${}^{3}D_{4}(q)$
[Lu], we know that except for the trivial character and a character $\chi_{1}$
of degree $q\Phi_{12}(q)$, all other irreducible characters of
${}^{3}\\!D_{4}(q)$ have degree greater than $q^{8}/2$ when $q\geq 4$. By [Sp,
Table 2], the value of $\chi_{1}$ at a regular semisimple element with
centralizer $(q-1)(q^{3}+1)$ is $1$. Certainly,
$\sum_{\chi\in\mathrm{Irr}(G)}|\chi(s)|^{2}=|\mathbf{C}_{G}(s)|<q^{4}$, and
$|\chi(s)|<((q-1)(q^{3}+1))^{1/2}<q^{2}$ for all
$1_{G}\neq\chi\in\mathrm{Irr}(G)$. We conclude that

$\displaystyle\sum_{\chi\neq 1}\frac{|\chi(s)|^{3}}{\chi(1)}$
$\displaystyle<\frac{1}{q\Phi_{12}(q)}+\sum_{\chi(1)>\epsilon_{1}(1)}\frac{|\chi(s)|^{3}}{\chi(1)}$
$\displaystyle<\frac{2}{q^{5}}+\frac{q^{2}}{q^{8}/2}\sum_{\chi}|\chi(s)|^{2}$
$\displaystyle<\frac{2}{q^{5}}+\frac{2}{q^{2}}<1,$

so the Frobenius formula implies that $(s^{G})^{3}=G$. Case $F_{4}(q)$. By
[LSS, Table 5.1], $\mathrm{Sp}_{4}(q^{2})<F_{4}(q)$. Therefore,
$\mathrm{SL}_{2}(q^{4})<F_{4}(q)$. We are therefore guaranteed elements $s$ of
order $q^{4}+1$ in $S_{1}(F_{4}(q))S_{2}(F_{4}(q))$. By Lemma 4.1, such
elements are regular semisimple. By [GM, Table 11], If $t$ is any element of
order $\Phi_{12}(t)$, then the product of the conjugacy class of $s$ and the
conjugacy class of $t$ covers $F_{4}(q)\smallsetminus\\{1\\}$. Thus, Lemma 2.7
implies $\mathsf{ccn}(F_{4}(q))\leq 6$.

Case ${}^{2}F_{4}(q)^{\prime}$. For $q=2$, this is the Tits group. In this
case, we can use the character table in GAP to compute the extended covering
number, which is $5$, so $\mathsf{ccn}(G)\leq 4$. We may therefore assume
$q\geq 8$, so ${}^{2}F_{4}(q)$ is already perfect. By [LSS, Table 5.1],
$\mathrm{SL}_{2}(q^{2})<\mathrm{Sp}_{4}(q)<{}^{2}F_{4}(q)$. Thus,
${}^{2}F_{4}(q)$ contains regular semisimple elements $s$ of order $q^{2}+1$
by Lemma 4.1. The centralizer of $s$ in ${}^{2}F_{4}(q)$ has order $q^{2}+1$,
and by [LS] we have $\mathfrak{d}(G)\geq 2^{-1/2}(1-q^{-1})q^{11/2}$. Lemma
2.8 now implies that $\mathsf{ccn}(G)\leq 6$ for $q\geq 8$.

Cases $E_{6}(q)$ and ${}^{2}E_{6}(q)$. As $F_{4}(q)<G$, we have
$\mathrm{SL}_{2}(q^{4})<G$, and proceeding as in the $F_{4}(q)$ case, we are
guaranteed elements $s$ of order $q^{4}+1$ in $S_{1}(G)S_{2}(G)$. These
elements are regular semisimple by Lemma 4.1. By [GM, Table 11], there exists
a semisimple element $t$ such that the product of the conjugacy classes of $s$
and $t$ cover $G\smallsetminus\\{1\\}$. Thus, $\mathsf{ccn}(G)\leq 6$ by Lemma
2.7.

Case $E_{7}(q)$. By [LSS, Table 5.1], $E_{7}(q)$ contains
$\mathrm{PSL}_{2}(q^{7})$. If $S_{1}$ and $S_{2}$ are ample characteristic
collections, then
$S_{1}(\mathrm{PSL}_{2}(q^{7}))S_{2}(\mathrm{PSL}_{2}(q^{7}))$ contains an
element $s$ of order $q^{7}-1$ or $\frac{q^{7}-1}{2}$. By Lemma 4.1, this
element is regular semisimple. In [GM, Table 11] is it shown that there exists
an order $x_{2}$ prime to $q$ such that the product of any conjugacy class of
order $x_{1}=q^{7}-1$ and any conjugacy class of order $x_{2}$ covers
$G\smallsetminus\\{1\\}$. The proof depends only on the divisibility of
$x_{1}$ by a Zsigmondy prime and therefore goes through unchanged if
$x_{1}=\frac{q^{7}-1}{2}$. Lemma 2.7 now implies that
$\mathsf{ccn}(E_{7}(q))\leq 6$.

Case $E_{8}(q)$. By [LSS, Table 5.1], $E_{8}(q)$ contains a
$E_{7}^{\mathrm{sc}}(q)$, which, in turn, contains a perfect central extension
$\tilde{H}$ of $H=\mathrm{PSL}_{2}(q^{7})$. We can therefore regard
$\tilde{H}$ as a central quotient of $\mathrm{SL}_{2}(q^{7})$. If $S_{1}$ and
$S_{2}$ are ample characteristic collections, then
$S_{1}(\tilde{H})S_{2}(\tilde{H})$ contains an element $s$ of order $q^{7}-1$
or $\frac{q^{7}-1}{2}$. This element is therefore semisimple, and by Lemma
4.1, the order of its centralizer divides $(q^{3}-q)(q^{7}-1)$. By [LS],
$\mathfrak{d}(G)\geq q^{27}(q^{2}-1)\geq q^{28}$. As
$(q^{10})^{3/2}=q^{15}<q^{28}$, Lemma 2.8 implies $\mathsf{ccn}(E_{8}(q))\leq
6$. ∎

In most cases, we can prove the improved bound of $4$.

######  Proposition 4.3.

If $G$ is a sufficiently large finite simple group of exceptional Lie type,
then $\mathsf{ccn}(G)\leq 4$.

###### Proof.

For each series of exceptional groups of Lie type with the exception of
${}^{2}B_{2}$ (which is already covered by Theorem 4.2), $E_{6}$ and
${}^{2}E_{6}$, [LSS, Table 5.1] gives a subgroup of the type for which
Proposition 2.9 applies.

Suppose $G$ is of type $E_{6}$, i.e., it is the quotient of
${\underline{G}}({\mathbb{F}}_{q})$ by its center, where ${\underline{G}}$ is
the split simply connected group group over ${\mathbb{F}}_{q}$ of type
$E_{6}$. As $A_{1}\times A_{5}$ can be obtained from the extended Dynkin
diagram of $E_{6}$ by deleting a vertex, it follows that there exists a
homomorphism $\mathrm{SL}_{2}\times\mathrm{SL}_{6}\to{\underline{G}}$ of
algebraic groups over ${\mathbb{F}}_{q}$, and a maximal torus of the former
maps to a maximal torus of the later. Thus, there exists a homomorphism
$\phi\colon\mathrm{SL}_{2}^{4}\to{\underline{G}}$ which factors through
$\mathrm{SL}_{2}\times\mathrm{SL}_{6}\to{\underline{G}}$, which maps a split
maximal torus into a split maximal torus, and whose image contains regular
elements if $q$ is sufficiently large. The proposition now follows from Lemma
2.10.

Finally suppose $G$ is of type ${}^{2}E_{6}$. Let ${\underline{G}}$ be a
simply connected non-split group of type $E_{6}$ over ${\mathbb{F}}_{q}$. By
[T, Table 2], it is of rank $4$ over ${\mathbb{F}}_{q}$. The rank of
$\mathrm{SL}_{2}\times\mathrm{SU}_{6}$ over ${\mathbb{F}}_{q}$ is also $4$, so
there is a homomorphism
$\mathrm{SL}_{2}\times\mathrm{SU}_{6}\to{\underline{G}}$ mapping a maximal
split torus to a maximally split maximal torus. Using
$\mathrm{SL}_{2}^{3}=\mathrm{SU}_{2}^{3}\to\mathrm{SU}_{6}$, we obtain a
morphism of algebraic groups over ${\mathbb{F}}_{q}$ from
$\mathrm{SL}_{2}^{4}$ to ${\underline{G}}$ which sends any split maximal torus
of $\mathrm{SL}_{2}^{4}$ to a maximal split torus of ${\underline{G}}$ and
whose image in ${\underline{G}}$ contains regular semisimple elements. The
proposition again follows from Lemma 2.10. ∎

From this, we easily deduce Proposition 1.6.

###### Proof of Proposition 1.6.

Because $n$ and $q-1$ are relatively prime,
$\mathrm{PSL}_{n}(q)=\mathrm{SL}_{n}(q)$, so

$4=\mathsf{iw}(\mathrm{SL}_{n}(q))=\mathsf{iw}(\mathrm{PSL}_{n}(q))\leq\mathsf{ccn}(\mathrm{PSL}_{n}(q))\leq
4.$

∎

##  5\. Alternating groups and sporadic groups

######  Proposition 5.1.

For $n\geq 5$ we have $\mathsf{ccn}({\sf A}_{n})\leq 4$.

###### Proof.

By [AH, p. 1], we have $\mathsf{ecn}({\sf A}_{n})\leq 5$ for $n\leq 9$, so by
Lemma 2.2, we may assume $n\geq 10$.

If $p$ is a prime, the permutation representation on
${\mathbb{P}}^{1}({\mathbb{F}}_{p})$ embeds $\mathrm{PSL}_{2}(p)$ in ${\sf
A}_{p+1}$. Every element of order $p$ maps to a $p$-cycle in ${\sf A}_{p+1}$
and therefore to a $p$-cycle in ${\sf A}_{n}$ for all $n\geq p+1$. By Lemma
2.4, if $p\equiv 1\pmod{4}$, then there exist elements

$s\in S_{1}(\mathrm{PSL}_{2}(p))S_{2}(\mathrm{PSL}_{2}(p)),\ t\in
S_{3}(\mathrm{PSL}_{2}(p))S_{4}(\mathrm{PSL}_{2}(p))$

of order $p$, so every element in ${\sf A}_{n}$ which is a product of two
$p$-cycles lies in $S_{1}({\sf A}_{n})\cdots S_{4}({\sf A}_{n})$.

By a theorem of Bertrand [B], if $\lfloor\frac{3n}{4}\rfloor\leq p\leq n$,
then every element of ${\sf A}_{n}$ is a product of two $p$-cycles. Therefore,
$\mathsf{ecn}({\sf A}_{n})\leq 4$ whenever there exists a prime which is $1$
(mod $4$) in $[\lfloor 3n/4\rfloor,n-1]$. Applying this for $p=13$ and $p=17$,
we get the desired inequality for $n$ in $\\{14,15,16,17\\}$ and
$\\{18,19,20,21,22,23\\}$ respectively. The following table gives primes which
together cover all values of $n\in[30,1.3\cdot 10^{10}]$.

29 | 37 | 41 | 53 | 61
---|---|---|---|---
73 | 97 | 113 | 149 | 197
257 | 337 | 449 | 593 | 773
1021 | 1361 | 1801 | 2393 | 3181
4241 | 5653 | 7537 | 10037 | 13381
17837 | 23773 | 31657 | 42209 | 56269
75017 | 99989 | 133277 | 177677 | 236897
315857 | 421133 | 561461 | 748613 | 998117
1330789 | 1774373 | 2365829 | 3154433 | 4205909
5607853 | 7477121 | 9969457 | 13292593 | 17723449
23631253 | 31508329 | 42011093 | 56014789 | 74686357
99581809 | 132775693 | 177034217 | 236045497 | 314727293
419636389 | 559515161 | 746020213 | 994693597 |

By a theorem of Ramaré and Rumely [RR], if $n\geq 10^{10}$, then the sum of
$\log p$ over all $p<n$ which are $1$ (mod $4$) lies between $.495n$ and
$.505n$. It follows that there is at least one such prime in the interval
$[n,1.1n]$ for all $n>10^{10}$, and this is enough to show $\mathsf{ccn}({\sf
A}_{n})\leq 4$ for all $n>10^{10}$.

If $q$ is any odd prime power, then for $C_{1}$ and $C_{2}$ non-trivial
conjugacy classes in $\mathrm{PSL}_{2}(q)$, there is an element of order
$\frac{q+1}{2}$ in $C_{1}C_{2}$. This maps into a product of two disjoint
$\frac{q+1}{2}$ cycles. A theorem of Xu [X] asserts that if $n-1\leq r+s\leq
n$, then every element of ${\sf A}_{n}$ is a product of two elements, each
consisting of two disjoint cycles of length $r$ and $s$. Applying this for
$q=9$, $q=11$, $q=23$, $q=25$, and $q=27$ we get $\mathsf{ccn}({\sf
A}_{n})\leq 4$ for $n$ in $\\{10,11\\}$, $\\{12,13\\}$, $\\{24,25\\}$,
$\\{26,27\\}$, and $\\{28,29\\}$ respectively, so all cases are covered. ∎

To deal with ${\sf A}_{n}$ for large $n$, it is useful to have the following
lemma:

######  Lemma 5.2.

Let $p$ be a prime, $k$ and $n$ positive integers, $n\geq kp$, and
$u_{1},\ldots,u_{k},v_{1},\ldots,v_{k}\in\\{1,\ldots,n\\}$ such that
$u_{i}\neq u_{j}$, $v_{i}\neq v_{j}$, and $u_{i}\neq v_{j}$ when $i\neq j$.
Then there exists an element $\sigma\in{\sf S}_{n}$ such that
$\sigma(u_{i})=v_{i}$ for $1\leq i\leq k$ and $\sigma^{p}$ is the identity.

###### Proof.

For each $i$ such that $u_{i}=v_{i}$, let $X_{i}=\\{u_{i}\\}$, and for each
$i$ such that $u_{i}\neq v_{i}$, let $X_{i}$ be a $p$-element set containing
$u_{i}$ and $v_{i}$ and such that the $X_{i}$ are disjoint from one another.
We define $\sigma$ to be the identity on $\\{1,\ldots,n\\}\setminus\bigcup
X_{i}$ and to cyclically permute the elements of each $X_{i}$ such that
$\sigma(u_{i})=v_{i}$. ∎

We can now prove Proposition 1.5.

###### Proof of Proposition 1.5.

First we note by [TZ, Theorem 1.2] that $\mathsf{iw}({\sf A}_{n})\geq 3$ when
$n\geq 15$: indeed, not every element in ${\sf A}_{n}$ with $n\geq 15$ is a
product of two involutions. Hence it suffices to show that $\mathsf{ccn}({\sf
A}_{n})\leq 3$ when $n$ is large enough.

We say a characteristic collection $S$ if of _type_ $r$ if $r$ is the smallest
prime such that $S({\sf A}_{5})$ contains an element of order $r$. If $S$ is
ample it must be of type $2$, type $3$, or type $5$. For any positive integer
$n$ and $0\leq m\leq\lfloor n/5\rfloor$, the embedding ${\sf A}_{5}^{m}<{\sf
A}_{n}$ guarantees that $S({\sf A}_{n})$ contains an element of cycle type
$1^{5-4m}2^{2m}$ if $S$ is of type $2$, $1^{5-3m}3^{m}$ if $S$ is of type $3$,
and $1^{5-5m}5^{m}$, if $S$ is of type $5$.

For $p\equiv\pm 1\pmod{10}$, ${\sf A}_{5}<\mathrm{PSL}_{2}(p)<{\sf A}_{p+1}$,
and every element of ${\sf A}_{5}$ of order $r$, when regarded as an element
$x$ of ${\sf A}_{p+1}$, has at most $2$ fixed points and otherwise consists
entirely of $r$-cycles. For $p=59$, $x$ is therefore a permutation of the form
$r^{60/r}$. If $S$ is of type $2$, the embedding ${\sf A}_{60}^{i}\times{\sf
A}_{n-60i}<{\sf A}_{n}$ gives $S({\sf A}_{n})$ elements of cycle type
$1^{5-4m}2^{2m}$ whenever $m$ belongs to

$I_{i}=\biggl{[}15i,15i+\bigl{\lfloor}\frac{n-60i}{5}\bigr{\rfloor}\biggr{]}.$

When $75+60k\leq n<75+60(k+1)$, the consecutive intervals in the sequence
$I_{0},\ldots,I_{k}$ overlap, so for $0\leq m\leq 15k$, all elements of cycle
type $1^{5-4m}2^{2m}$ belong to $S({\sf A}_{n})$. This means that every even
permutation of order $2$ with at least $B_{2}=75$ fixed points lies in $S({\sf
A}_{n})$. By the same reasoning, for $B_{3}=100$ and $B_{5}=60$, all elements
of order $3$ and $5$ in ${\sf A}_{n}$ with at least $B_{3}$ and $B_{5}$ fixed
points respectively lie in $S({\sf A}_{n})$ if $S$ is respectively of type $3$
or $5$.

By [LS1, Theorem 1.2(ii)], if $x$ is an element of ${\sf S}_{n}$ of prime
order $r$ and a bounded number of fixed points, then for every irreducible
character $\chi$ of ${\sf S}_{n}$, $|\chi(x)|\leq\chi(1)^{1/r+o(1)}$.
Therefore, if $n$ is sufficiently large and $x_{1},x_{2},x_{3}$ are three such
elements in ${\sf A}_{n}$, of orders $r_{1},r_{2},r_{3}$ respectively, and

$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}<1,$

then the product of the ${\sf S}_{n}$-conjugacy classes of the $x_{i}$ cover
${\sf A}_{n}\smallsetminus\\{1\\}$. (Indeed, note that we have
$1/r_{1}+1/r_{2}+1/r_{3}\leq 41/42$ in such a case, and so, by [LiSh2, Theorem
2.6],

$\sum_{\chi\in\mathrm{Irr}({\sf
S}_{n}),~{}\chi(1)>1}\frac{|\chi(x_{1})\chi(x_{2})\chi(x_{3})|}{\chi(1)}<\sum_{\chi\in\mathrm{Irr}({\sf
S}_{n}),~{}\chi(1)>1}\frac{1}{\chi(1)^{1/42}}\to 0$

when $n\to\infty$.) We may assume without loss of generality that $r_{1}\leq
r_{2}\leq r_{3}$, so we need only consider the cases $(r_{1},r_{2})=(2,2)$,
$(r_{1},r_{2})=(2,3)$, and $r_{1}=r_{2}=r_{3}=3$. We proceed by case analysis.

Case $(2,2,r_{3})$. For any $k\geq 1$, let

$\sigma_{k}=(1,2)\cdots(2k-1,2k)\in{\sf
S}_{2k},~{}\tau_{k}=(2,3)\cdots(2k,2k+1)\in{\sf S}_{2k+1}.$

Thus, $\sigma_{k}\tau_{k}$ is a $(2k+1)$-cycle, while $\sigma_{k}\tau_{k-1}$
is a $2k$-cycle. Applying this to any cycle, we see that every element in
${\sf S}_{n}$ can be written as a product of two involutions. Regarding ${\sf
S}_{n-2}$ as the stabilizer in ${\sf A}_{n}$ of $\\{n-1,n\\}$, we see that
every element of ${\sf A}_{n}$ which fixes $n-1$ and $n$ can be written as a
product of two involutions in ${\sf A}_{n}$. More generally, every element of
${\sf A}_{n}$ with at least $2+B_{2}$ fixed points can be written as a product
of even involutions, each of which has at least $B_{2}$ fixed points and hence
belongs to $S({\sf A}_{n})$.

We claim that if $n$ is sufficiently large, for every $\rho\in{\sf A}_{n}$ and
$p\in\\{2,3,5\\}$ there exists $\pi\in{\sf A}_{n}$ of order dividing $p$ with
at least $B_{p}$ fixed points such that $\pi\rho$ has at least $2+B_{2}$ fixed
points. We choose a sequence $x_{1},\ldots,x_{2+B_{2}}$ of distinct elements
of $\\{1,\ldots,n\\}$ such that $\rho(x_{i})\neq x_{j}$ for $i\neq j$ and then
a sequence $y_{1},\ldots,y_{B_{p}}$ of distinct elements of
$\\{1,\ldots,n\\}\smallsetminus\\{\rho(x_{1}),\ldots,\rho(x_{2+B_{2}})\\}$. By
Lemma 5.2, if $n$ is sufficiently large, there exists an even permutation
$\pi$ of order $1$ or $p$ which fixes each $y_{i}$ and maps each $\rho(x_{j})$
to $x_{j}$. By the above discussion, $\pi\rho\in S({\sf A}_{n})S({\sf A}_{n})$
and $\pi^{-1}\in S({\sf A}_{n})$, hence $\rho\in S({\sf A}_{n})S({\sf
A}_{n})S({\sf A}_{n})$.

Case $(2,3,r_{3})$. Let

$\displaystyle\sigma_{k}$ $\displaystyle=(1,2)(4,5)\cdots(3k-2,3k-1),$
$\displaystyle\sigma^{\prime}_{k}$ $\displaystyle=(3,4)(6,7)\cdots(3k,3k+1),$
$\displaystyle\tau_{k}$ $\displaystyle=(2,3,4)(5,6,7)\cdots(3k-1,3k,3k+1),$
$\displaystyle\tau^{\prime}_{k}$
$\displaystyle=(1,2,3)(4,5,6)\cdots(3k-2,3k-1,3k).$

Then $\sigma_{k}\tau_{k}$ is a $(3k+1)$-cycle, $\sigma_{k+1}\tau_{k}$ is a
$(3k+2)$-cycle, and $\sigma^{\prime}_{k}\tau^{\prime}_{k+1}$ is a
$(3k+3)$-cycle. Thus, every element of ${\sf S}_{n}$ is a product of an
involution $\sigma$ and an element $\tau$ of order $1$ or $3$. Since any such
$\tau$ lies in ${\sf A}_{n}$, the same statement holds in ${\sf A}_{n}$, and
if the product has at least $\max(B_{2},B_{3})$ fixed points, the same can be
assumed of $\sigma$ and $\tau$. Applying Lemma 5.2 as before, we can guarantee
for each $\rho\in{\sf A}_{n}$ the existence of an element $\pi\in{\sf A}_{n}$
of order dividing $r_{3}$ with at least $B_{r_{3}}$ fixed points such that
$\pi\rho$ has at least $\max(B_{2},B_{3})$ fixed points, and then finish as in
the previous case.

Case $(3,3,3)$. It suffices to prove that every element of ${\sf A}_{n}$ can
be written as a product of two elements of order dividing $3$. Let

$\displaystyle\sigma_{k,l}$
$\displaystyle=(1,2,3)(5,6,7)\cdots(4k-7,4k-6,4k-5)(4k-3,4k-2,4k-1)$
$\displaystyle\qquad\qquad\qquad(4k,4k+1,4k+2)(4k+4,4k+5,4k+6)\cdots(4l,4l+1,4l+2),$
$\displaystyle\tau_{k,l}$
$\displaystyle=(3,4,5)(7,8,9)\cdots(4k-5,4k-4,4k-3)(4k-1,4k,4k+1)$
$\displaystyle\qquad\qquad\qquad(4k+2,4k+3,4k+4)(4k+6,4k+7,4k+8)\cdots(4l+2,4l+3,4l+4),$

where the first line in each expression is omitted if $k=0$ and the second
line is omitted if $l=k-1$. Then $\sigma_{k,k-1}\tau_{k-1,k-2}$ is a
$(4k-1)$-cycle, and $\sigma_{k,k-1}\tau_{k,k-1}$ is a $(4k+1)$-cycle, so all
odd cycles can be written as a product of two elements of order dividing $3$.
For $l\geq k\geq 0$ and $k+l\geq 1$, $\sigma_{k,l}\tau_{k,l}$ is a disjoint
product of a $4k$-cycle and a $(4l+4-4k)$-cycle; $\sigma_{k,l}\tau_{k-1,l}$ is
a disjoint product of a $(4k-2)$-cycle and a $(4l+6-4k)$-cycle;
$\sigma_{k,l}\tau_{k,l-1}$ is a disjoint product of a $4k$-cycle and a
$(4l+2-4k)$-cycle; and $\sigma_{k,l}\tau_{k-1,l-1}$ is a disjoint product of a
$(4k-2)$-cycle and a $(4l+4-4k)$-cycle. Thus, all possible permutations that
are products of two disjoint even-length cycles can be written as a product of
two permutations of order dividing $3$. ∎

######  Proposition 5.3.

For all sporadic finite simple groups we have $\mathsf{ccn}(G)\leq 4$.

###### Proof.

By a theorem of Zisser [Zi], we have $\mathsf{ecn}(G)\leq 5$ for all sporadic
groups except $\mathrm{Fi}_{22}$ and $\mathrm{Fi}_{23}$, so it suffices to
consider these two cases. It is known [ATLAS, pp. 74, 163] that there are
inclusions $\mathrm{PSL}_{2}(25)<{}^{2}F_{4}(2)^{\prime}<\mathrm{Fi}_{22}$. By
Lemma 2.4, if $S_{1}$ and $S_{2}$ are ample characteristic collections, then
$S_{1}(\mathrm{PSL}_{2}(25))S_{2}(\mathrm{PSL}_{2}(25))$ has an element of
order $13$, and a machine computation shows that the product of any two
conjugacy classes of elements of order $13$ in $\mathrm{Fi}_{22}$ contains
$\mathrm{Fi}_{22}\smallsetminus\\{1\\}$. It is also known [ATLAS, p. 177] that
$\mathrm{PSL}_{2}(17)<\mathrm{Fi}_{23}$. By Lemma 2.4, the product
$S_{1}(\mathrm{PSL}_{2}(17))S_{2}(\mathrm{PSL}_{2}(17))$ contains an element
of order $17$; a machine computation shows that the square of the unique
conjugacy class of order $17$ in $\mathrm{Fi}_{23}$ is the whole group. ∎

Together with Theorems 3.1 and 4.2, Propositions 5.1 and 5.3 complete the
proof of Theorem A. Theorem D follows from results 5.1, 5.3, 3.2 and 4.3.

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