# Schur’s exponent conjecture II

Michael Vaughan-Lee

(December 2021)

###### Abstract

Primoz Moravec published a very important paper in 2007 where he proved that
if $G$ is a finite group of exponent $n$ then the exponent of the Schur
multiplier of $G$ can be bounded by a function $f(n)$ depending only on $n$.
Moravec does not give a value for $f(n)$, but actually his proof shows that we
can take $f(n)=ne$ where $e$ is the order of $b^{-n}a^{-n}(ab)^{n}$ in the
Schur multiplier of $R(2,n)$. (Here $R(2,n)$ is the largest finite two
generator group of exponent $n$, and we take $a,b$ to be the generators of
$R(2,n)$.) It is an easy hand calculation to show that $e=n$ for $n=2,3$, and
it is a straightforward computation with the $p$-quotient algorithm to show
that $e=n$ for $n=4,5,7$. The groups $R(2,8)$ and $R(2,9)$ are way out of
range of the $p$-quotient algorithm, even with a modern supercomputer. But we
are able to show that $e\geq n$ for $n=8,9$. Moravec’s proof also shows that
if $G$ is a finite group of exponent $n$ with nilpotency class $c$, then the
exponent of the Schur multiplier of $G$ is bounded by $ne$ where $e$ is the
order of $b^{-n}a^{-n}(ab)^{n}$ in the Schur multiplier of the class $c$
quotient $R(2,n;c)$ of $R(2,n)$. If $q$ is a prime power we let $e_{q,c}$ be
the order of $b^{-q}a^{-q}(ab)^{q}$ in the Schur multiplier of $R(2,q;c)$. We
are able to show that $e_{p^{k},p^{2}-p-1}$ divides $p$ for all prime powers
$p^{k}$. If $k>2$ then $e_{2^{k},c}$ equals 2 for $c<4$, equals 4 for $4\leq
c\leq 11$, and equals $8$ for $c=12$. If $k>1$ then $e_{3^{k},c}$ equals 1 for
$c<3$, equals 3 for $3\leq c<12$, and equals 9 for $c=12$.

We also investigate the order of $[b,a]$ in a Schur cover for $R(2,q;c)$.

##  1 Introduction

There is a long-standing conjecture attributed to I. Schur that the exponent
of the Schur multiplier, $M(G)$, of a finite group $G$ divides the exponent of
$G$. It is easy to show that this conjecture holds true for groups of exponent
2 and exponent 3, but a counterexample in exponent 4 was found by Bayes,
Kautsky and Wamsley [1] in 1974. The conjecture remained open for odd exponent
until 2020, when I found counterexamples of exponent 5 and exponent 9 [6]. It
seems certain that there are counterexamples to this conjecture for all prime
powers greater than 3, but this leaves open the question of what bounds on the
exponent of $M(G)$ might hold true.

The usual definition of the Schur multiplier of a finite group $G$ is the
second homology group $H_{2}(G,\mathbb{Z})$. For computational purposes we use
the Hopf formulation of the Schur multiplier, which is as follows. We write
$G=F/R$, where $F$ is a free group, and then

$M(G)=(R\cap F^{\prime})/[F,R].$

If we let $H=F/[F,R]$ then $H$ is an infinite group, and the quotient
$H/H^{\prime}$ is a free abelian group with rank equal to the rank of $F$ as a
free group. However the centre of $H$ contains $R/[F,R]$ and has finite index
in $H$. So the derived group $H^{\prime}=F^{\prime}/[F,R]$ is finite, and
$M(G)$ is finite. It is known that the exponent of $M(G)$ divides the order of
$G$. Furthermore if $G$ has exponent $n$ and if $h\in H^{\prime}$ then
$h^{n}\in M(G)$. So the quotient $H^{\prime}/M(G)$ has exponent dividing $n$.
In particular, if $G$ is a finite $p$-group then $H^{\prime}$ is a finite
$p$-group.

Primoz Moravec published a very important paper [5] in 2007 in which he proved
that if $G$ is a finite group of exponent $n$ then the exponent of $M(G)$ is
bounded by a function $f(n)$ depending only on $n$. Moravec does not give an
explicit formula for $f(n)$, but his proof of this theorem actually shows that
if $G$ is a finite group of exponent $n$ then the exponent of $M(G)$ divides
$ne$ where $e$ is the order of $b^{-n}a^{-n}(ab)^{n}$ in the Schur multiplier
of $R(2,n)$. (We let $R(2,n)$ denote the largest finite $2$ generator of
exponent $n$, and we take the generators of $R(2,n)$ to be $a,b$.)

######  Theorem 1 (Moravec, 2007)

Let $b^{-n}a^{-n}(ab)^{n}\in M(R(2,n))$ have order $e$. If $G$ is any finite
group of exponent $n$ then the exponent of $M(G)$ divides $ne$.

We give a short proof of this theorem in Section 2. The same proof gives the
following theorem.

###### Theorem 2

Let $R(2,n;c)$ be the nilpotent of class $c$ quotient of $R(2,n)$, and let
$e_{n,c}$ be the order of $b^{-n}a^{-n}(ab)^{n}\in M(R(2,n;c))$. If $G$ is any
finite group of exponent $n$ with class $c$, then the exponent of $M(G)$
divides $ne_{n,c}$.

Our proof of Moravec’s theorem also gives the following corollary.

###### Corollary 3

Let $R(d,n)$ be the largest finite $d$ generator group of exponent $n$. Then
if $d\geq 2$ the exponent of $M(R(d,n))$ is the order of
$b^{-n}a^{-n}(ab)^{n}\in M(R(2,n))$.

Theorem 1 led me to investigate the order $e_{q}$ of $b^{-q}a^{-q}(ab)^{q}\in
M(R(2,q))$ for prime power exponents $q$. (We restrict ourselves to groups of
prime power exponent since if $G$ is any finite group and $p$ is any prime
then the Sylow $p$-subgroup of $M(G)$ is a subgroup of the Schur multiplier of
the Sylow $p$-subgroup of $G$.) It is an easy hand calculation to show that
$e_{q}=q$ for $q=2,3$. And it is a straightforward computation with the
$p$-quotient algorithm [4] to show that $e_{q}=q$ for $q=4,5,7$. Computing the
groups $R(2,8)$ or $R(2,9)$ (or their Schur covers) is way out of the range of
the $p$-quotient algorithm, even with a modern supercomputer. But I am able to
show that $e_{q}\geq q$ for $q=8,9$.

Theorem 2 led me to investigate the order $e_{q,c}$ of
$b^{-q}a^{-q}(ab)^{q}\in M(R(2,q;c))$ for prime power exponents $q$ and
various $c$.

###### Theorem 4

If $q$ is a power of the prime $p$ then $e_{q,p^{2}-p-1}$ divides $p$.

###### Theorem 5

If $k>2$ then $e_{2^{k},c}$ equals $2$ for $c<4$, equals $4$ for $4\leq c\leq
11$, and equals $8$ for $c=12$.

###### Theorem 6

If $k>1$ then $e_{3^{k},c}$ equals $1$ for $c<3$, equals $3$ for $3\leq c<12$,
and equals $9$ for $c=12$.

The counterexamples to Schur’s conjecture found in [1] and [6] are based on
the following construction. Let $H(q,c)$ be the largest four generator group
of exponent $q$ and nilpotency class $c$ which is generated by $a,b,c,d$ and
subject to the relation

$[b,a][d,c]=1.$

The Bayes, Kautsky and Wamsley example in [1] is $H(4,4)$ which has Schur
multiplier of exponent 8. The element $[b,a][d,c]$ has order 8 in the Schur
multiplier, and in fact $[b,a]$ has order $8$ in a Schur cover of $R(2,4;4)$.
Similarly my examples in [6] are based on $H(5,9)$ and $H(9,9)$. In the Schur
multipliers of these two groups the elements $[b,a][d,c]$ have order 25 and
27. The examples “work” because $[b,a]$ has order 25 and 27 in Schur covers of
$R(2,5;9)$ and $R(2,9;9)$. It seems plausible that more generally the exponent
of $M(H(q,c))$ is the order of $[b,a]$ in a Schur cover of $R(2,q;c)$, though
I have no idea how to prove this in general. For this reason I have looked at
the order of $[b,a]$ in Schur covers of $R(2,q;c)$ for various $q,c$. For
example, $[b,a]$ has order 32 in a Schur cover of $R(2,8;12)$ so it seems to
me to be extremely likely that $M(H(8,12))$ has exponent 32, though I cannot
prove it (yet!). This is an interesting example because all $p$-group
counterexamples $G$ to Schur’s conjecture found so far have
exp$\,M(G)=p\,$exp$\,G$.

I am able to prove that $M(H(q,4))$ has exponent $2q$ whenever $q\geq 4$ is a
power of 2. For $q=9,27$ $M(H(q,9))$ has exponent $3q$, and it seems very
likely that $M(H(q,9))$ has exponent $3q$ for all $q>3$ which are powers of 3.
Similarly for $q=5,25$ $M(H(q,9))$ has exponent $5q$, and it seems very likely
that $M(H(q,9))$ has exponent $5q$ for all $q$ which are powers of 5.

###### Theorem 7

If $q>4$ is a power of $2$, and if we let $f$ be the order of $[b,a]$ in a
Schur cover of $R(2,q;c)$ then $f=q$ for $c<4$, $f=2q$ for $4\leq c<12$, and
$f=4q$ for $c=12$. If $G$ is a finite $2$-group with nilpotency class less
than $4$ then exp$\,M(G)$ divides exp$\,G$.

###### Theorem 8

If $q>3$ is a power of $3$, and if we let $f$ be the order of $[b,a]$ in a
Schur cover of $R(2,q;c)$ then $f=q$ for $c<9$, and $f=3q$ for $c=9$. If $G$
is any group of exponent $q$ with class less than $9$ then exp$\,M(G)$ divides
$q$.

###### Theorem 9

Let $q$ be a power of $5$. If we let $f$ be the order of $[b,a]$ in a Schur
cover of $R(2,q;c)$ then $f=q$ for $c<9$, and $f=5q$ for $c=9$. If $G$ is any
group of exponent $q$ with class less than $9$ then exp$\,M(G)$ divides $q$.

###### Theorem 10

Let $q$ be a power of $7$. If we let $f$ be the order of $[b,a]$ in a Schur
cover of $R(2,q;c)$ then $f=q$ for $c<13$, and $f=7q$ for $c=13$. If $G$ is
any group of exponent $q$ with class less than $13$ then exp$\,M(G)$ divides
$q$.

Theorem 9 follows from the fact that in a Schur cover of $R(2,5)$ the element
$[b,a]^{5}$ is a product of commutators with entries $a$ or $b$, where at
least 5 of the entries are $a$’s and at least 5 of the entries are $b$’s.
Similarly Theorem 10 follows from the fact that in a Schur cover of $R(2,7)$
the element $[b,a]^{7}$ is a product of commutators with at least 7 entries
$a$ and at least 7 entries $b$. If the same pattern is repeated for higher
primes $p$, then I would expect the order of $[b,a]$ in $M(R(2,p^{k};c))$ to
be $p^{k}$ for $c<2p-1$ and to be $p^{k+1}$ when $c=2p-1$.

The theorems above omit the exponents 2,3,4. It is easy to see that the Schur
multiplier of a group of exponent 2 or exponent 3 has exponent 2 or 3
(respectively). Moravec [5] proves that the Schur multiplier of a group of
exponent 4 has exponent dividing 8.

##  2 Proof of Theorem 1

We write $R(2,n)=F/R$ where $F$ is the free group generated by $a,b$, and let
$H=F/[F,R]$. (We are not assuming here that $n$ is a prime power.) Then
$b^{-n}a^{-n}(ab)^{n}\in M(R(2,n))$. Let $b^{-n}a^{-n}(ab)^{n}$ have order
$e$. We show that $e$ is the exponent of $M(R(d,n))$ for all $d\geq 2$, and
that if $G$ is any finite group of exponent $n$ then exp$(M(G))$ divides $ne$.

So let $G$ be _any_ finite group of exponent $n$, let $G=F/R$ where $F$ is a
free group, and let $H=F/[F,R]$. (Apologies for using the same notation for
the covering group of $G$ as I used for the covering group of $R(2,n)$.) Let
$a,b$ be _any_ two elements in $H$. Then the subgroup of $H$ generated by $a$
and $b$ is a homomorphic image of the cover of $R(2,n)$, and so
$b^{-n}a^{-n}(ab)^{n}\in H^{\prime}$ lies in the centre of $H$ and has order
dividing $e$. Let $F$ be freely generated by the set $X$ and let $K$ be the
subgroup of $H$ generated by the elements $x^{n}[F,R]$ ($x\in X$). Then $K$ is
a free abelian group which intersects $H^{\prime}$ trivially. If $w$ is an
arbitrary element of $F$ we can write $w=x_{1}x_{2}\ldots x_{k}$ for some $k$
and some $x_{1},x_{2},\ldots,x_{k}\in X\cup X^{-1}$. Letting $a=x_{1}$ and
$b=x_{2}x_{3}\ldots x_{k}$ we see that

$w^{n}=(x_{1}x_{2}\ldots x_{k})^{n}=x_{1}^{n}(x_{2}\ldots
x_{k})^{n}b^{-n}a^{-n}(ab)^{n}.$

Repeating this argument we see that

$w^{n}=x_{1}^{n}x_{2}^{n}\ldots x_{k}^{n}c,$

where $c$ is a product of terms of the form $b^{-n}a^{-n}(ab)^{n}$ with
$a,b\in F$. So we see that if $h\in H$ then $h^{n}$ is a product of an element
in $K$ and an element in $H^{\prime}$ which lies in the centre of $H$ and has
order dividing $e$. It follows that any product of $n^{th}$ powers in $H$ can
be expressed in the same form. Since $K\cap H^{\prime}=\\{1\\}$ we see that
this implies that any product of $n^{th}$ powers in $H$ which lies in
$H^{\prime}$ has order dividing $e$. So exp$(M(R(d,n)))=e\,$ for all $d\geq
2$. If $h\in H^{\prime}$, then $h^{n}$ is an $n^{th}$ power which lies in
$H^{\prime}$, and so $h^{ne}=1$. So $H^{\prime}$ has exponent dividing $ne$,
and this implies that exp$(M(G))$ divides $ne$.

##  3 Some commutator calculus

Let $F$ be the free group of rank 2 generated by $a$ and $b$. If we are
working in the nilpotent quotient $F/\gamma_{k+1}(F)$ for some $k$ then we
pick a fixed ordered set of basic commutators of weight at most $k$. See [3,
Chapter 11]. The first few basic commutators in our sequence are

$a,\,b,\,[b,a],\,[b,a,a],\,[b,a,b],\,[b,a,a,a],\,[b,a,a,b],\,[b,a,b,b],\,[b,a,a,[b,a]].$

If $c_{1},c_{2},\ldots,c_{m}$ is our list of basic commutators of weight at
most $k$ then every element of $F/\gamma_{k+1}(F)$ can be written uniquely in
the form

$c_{1}^{n_{1}}c_{2}^{n_{2}}\ldots c_{m}^{n_{m}}\gamma_{k+1}(F)$

for some integers $n_{1},n_{2},\ldots,n_{m}$. From the theory of Hall
collection (see [3, Theorem 12.3.1]), if $n$ is any positive integer then in
$F/\gamma_{k+1}(F)$

$(ab)^{n}=a^{n}b^{n}[b,a]^{\binom{n}{2}}[b,a,a]^{\binom{n}{3}}c_{5}^{n_{5}}\ldots
c_{m}^{n_{m}}$ (1)

where the exponents $n,\binom{n}{2},\binom{n}{3},n_{5},\ldots,n_{m}$ take a
very special form. If $c_{r}$ has weight $w$ then $n_{r}$ is an integral
linear combination of the binomial coefficients
$n,\binom{n}{2},\binom{n}{3},\ldots,\binom{n}{w}$. Furthermore the integer
coefficients which arise in these linear combinations are positive, and are
independent of $n$.

The exponents $n,\binom{n}{2},\binom{n}{3},n_{5},\ldots,n_{m}$ which arise in
equation (1) are all polynomials in $n$ which take integer values when $n$ is
an integer. The formula

$\binom{n}{r}=\frac{n(n-1)\ldots(n-r+1)}{r!}$

also makes sense when $n$ is negative. We let $P=\mathbb{Q}[t]$ be the ring of
polynomials in an indeterminate $t$ over the rationals $\mathbb{Q}$. An
_integer-valued polynomial_ is a polynomial $f(t)\in P$ which takes integer
values whenever $f(t)$ is evaluated at an integer $n$. The set of integer-
valued polynomials is a subring of $P$, and is a free abelian group with basis

$1,\,t,\,\frac{t(t-1)}{2!},\,\ldots,\frac{t(t-1)\ldots(t-d+1)}{d!},\,\ldots.$

The exponents $n_{i}$ which arise in equation (1) all take the form
$n_{i}=f_{i}(n)$ where $f_{i}(t)$ is an integer-valued polynomial of degree at
most wt$\,c_{i}$ which does not depend on $n$. The polynomials $f(t)\in P$
that arise in this way in equation (1) also satisfy $f(0)=0$.

We rewrite equation (1) in the following form

$(ab)^{n}=a^{n}b^{n}[b,a]^{\binom{n}{2}}[b,a,a]^{\binom{n}{3}}c_{5}^{f_{5}(n)}\ldots
c_{m}^{f_{m}(n)}$ (2)

where the integer-valued polynomials $f_{i}(t)$ are independent of $n$. The
key properties of these polynomials to keep in mind are that $f_{i}(0)=0$ and
$\deg f_{i}(t)\leq\text{wt}\,c_{i}$.

We use equation (2) to get an expansion of $[y^{n},x]$ for $x,y\in F$.
Equation (2) gives

$y^{n}x=x(y[y,x])^{n}=xy^{n}[y,x]^{n}[y,x,y]^{\binom{n}{2}}(c_{4}\alpha)^{f_{4}(n)}\ldots(c_{m}\alpha)^{f_{m}(n)}\text{
modulo }\gamma_{k+1}(F)$

where $\alpha$ is the endomorphism of $F$ mapping $a,b$ to $y,[y,x]$. This
equation gives

$[y^{n},x]=[y,x]^{n}[y,x,y]^{\binom{n}{2}}(c_{4}\alpha)^{f_{4}(n)}\ldots(c_{m}\alpha)^{f_{m}(n)}\text{
modulo }\gamma_{k+1}(F).$ (3)

##  4 Proof of Theorem 4

We want to prove that if $q$ a power of the prime $p$ then the order of
$b^{-q}a^{-q}(ab)^{q}$ in $M(R(2,q;p^{2}-p-1)$ divides $p$. The case $p=2$ is
covered by Theorem 5, and so we assume that $p>2$. We write
$R(2,q;p^{2}-p-1)=F/R$, where $F$ is the free group with free generators
$a,b$. Let $H=F/[F,R]$. So $H$ is nilpotent of class at most $p^{2}-p$, and
(as we noted in the introduction) $H^{\prime}$ is a finite $p$-group. We let
$c_{1},c_{2},\ldots,c_{m}$ be our list of basic commutators of weight at most
$p^{2}-p$ as described in Section 3. Let $x,y\in H$ and set $n=q$ in equation
(3) from Section 3. Using the fact that $H$ is nilpotent of class $p^{2}-p$ we
obtain a relation

$[y,x]^{q}[y,x,y]^{\binom{q}{2}}[y,x,y,y]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{m}\alpha)^{f_{m}(q)}=1,$
(4)

where $\alpha$ is the homomorphism from $F$ to $H$ mapping $a,b$ to $y,[y,x]$.
Recall that if wt$\,c_{i}=w$ then $\deg f_{i}(t)\leq w$. Note that if
wt$\,c_{i}=w$ then $c_{i}\alpha$ is a commutator in $x,y$ with $w$ entries
$y$. Also note that wt$\,c_{i}<p^{2}$ so that $f_{i}(q)$ is divisible by
$\frac{q}{p}$ for all $i$. And if wt$\,c_{i}<p$ then $f_{i}(q)$ is divisible
by $q$.

Since $H$ is nilpotent of class at most $p^{2}-p$, if $y\in\gamma_{p-1}(H)$
then $c_{i}\alpha=1$ whenever wt$\,c_{i}\geq p$. So if $y\in\gamma_{p-1}(H)$
then relation (4) shows that $[y,x]^{q}$ is a product of $q^{th}$ powers of
commutators in $x,y$ with higher weight. First let
$y\in\gamma_{p^{2}-p-1}(H)$. Then equation (4) gives $[y,x]^{q}=1$. Since
elements $[y,x]$ of this form generate $\gamma_{p^{2}-p}(H)$ this implies that
$\gamma_{p^{2}-p}(H)$ has exponent $q$. Next let $y\in\gamma_{p^{2}-p-2}(H)$.
Then equation (4) gives $[y,x]^{q}\in\gamma_{p^{2}-p}(H)^{q}=\\{1\\}$. So we
see that $\gamma_{p^{2}-p-1}(H)$ has exponent $q$. We continue in this way,
successively proving that $\gamma_{p^{2}-p-2}(H)$, $\gamma_{p^{2}-p-3}(H)$, …
have exponent $q$. Finally we let $y\in\gamma_{p-1}(H)$ and prove that
$\gamma_{p}(H)$ has exponent $q$.

Now set $n=pq$ in equation (3), and we obtain the relation.

$[y,x]^{pq}[y,x,y]^{\binom{pq}{2}}[y,x,y,y]^{\binom{pq}{3}}(c_{5}\alpha)^{f_{5}(pq)}\ldots(c_{m}\alpha)^{f_{m}(pq)}=1$

where all the exponents $f_{i}(pq)$ are divisible by $q$, and where
commutators with less than $p$ entries $y$ have exponents divisible by $pq$.
Since all commutators in $H$ with weight at least $p$ have order dividing $q$,
this implies that $[x,y]^{pq}$ is a product of $(pq)^{th}$ powers of
commutators of higher weight in $x,y$. So $[x,y]^{pq}=1$, and hence all
commutators in $H$ have order dividing $pq$. Since $H$ has class less than
$p^{2}$ and $\gamma_{p}(H)$ has exponent $q$ this implies that $H^{\prime}$
has exponent dividing $pq$.

Finally consider $(b^{-q}a^{-q}(ab)^{q})^{p}=b^{-pq}a^{-pq}(ab)^{pq}$.
Equation (2) from Section 3 gives

$(b^{-q}a^{-q}(ab)^{q})^{p}=[b,a]^{\binom{pq}{2}}[b,a,a]^{\binom{pq}{3}}c_{5}^{f_{5}(pq)}\ldots
c_{m}^{f_{m}(pq)}.$

All the exponents in this product are divisible by $q$, and the exponents of
commutators of weight less than $p$ in the product are divisible by $pq$. So
$(b^{-q}a^{-q}(ab)^{q})^{p}=1$.

The same argument (though easier) shows that in $M(R(2,q;p-2)$ the element
$b^{-q}a^{-q}(ab)^{q}=1$.

##  5 Proof of Theorem 5 and Theorem 7

Let $q=2^{k}$ $(k>2)$, and let $e_{q,c}$ be the order of
$b^{-q}a^{-q}(ab)^{q}$ in $M(R(2,q;c))$. We want to prove that $e_{q,c}$
equals $2$ for $c<4$, equals $4$ for $4\leq c\leq 11$, and equals $8$ for
$c=12$. We also want to prove that if $f$ is the order of $[b,a]$ in a Schur
cover of $R(2,q;c)$ then $f=q$ for $c<4$, $f=2q$ for $4\leq c<12$, and $f=4q$
when $c=12$.

Let $R(2,q;12)=F/R$ where $F$ is the free group of rank two generated by
$a,b$, and let $H=F/[F,R]$. Then $H$ is an infinite group, but the subgroup
$\langle a^{q},b^{q}\rangle\leq H$ is a central subgroup with trivial
intersection with $H^{\prime}$. We can factor this subgroup out, without
impacting $M(R(2,q;12))$, and we now have a finite $2$-group. We used the
$p$-quotient algorithm to compute this quotient for $q=8,16,32$. (These were
quite easy computations.) The computations showed that $e_{q,c}$ takes the
values given in Theorem 5 for $q=8,16,32$, and that $f$ takes the values given
in Theorem 7 for $q=8,16,32$. We show that the fact that Theorem 5 and Theorem
7 hold true for $q=2^{5}$ implies that they hold true for all exponents
$q=2^{k}$ with $k\geq 5$.

We let $q=2^{k}$ where $k\geq 5$, and let $c_{1},c_{2},\ldots,c_{m}$ be our
list of basic commutators of weight at most $13$ as described in Section 3. As
in the proof of Theorem 4 we let $x,y\in H$ and obtain a relation

$[y^{q},x]=[y,x]^{q}[y,x,y]^{\binom{q}{2}}[y,x,y,y]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{m}\alpha)^{f_{m}(q)}$
(5)

where $\alpha$ is the homomorphism from $F$ to $H$ mapping $a,b$ to $y,[y,x]$.
If wt$\,c_{i}=w$ then $c_{i}\alpha$ is a commutator in $x$ and $y$, with $w$
entries $y$, and $\deg f_{i}\leq w$. The binomial coefficients $\binom{q}{2}$
and $\binom{q}{3}$ are both divisible by $\frac{q}{2}$, the binomial
coefficients $\binom{q}{d}$ for $d<8$ are divisible by $\frac{q}{4}$, and the
binomial coefficients $\binom{q}{d}$ for $d<16$ are divisible by
$\frac{q}{8}$.

If we let $y\in\gamma_{7}(H)$ then $[y,x,y]\in\gamma_{15}(H)=\\{1\\}$, so we
see that $[y,x]^{q}=1$. Now $\gamma_{8}(H)$ is generated by elements $[y,x]$
with $y\in\gamma_{7}(H)$, and $\gamma_{8}(H)$ is abelian. So $\gamma_{8}(H)$
has exponent $q$.

Now let $y\in\gamma_{4}(H)$, and replace $q$ by $2q$ in equation (5). Using
the fact that $\gamma_{8}(H)$ has exponent $q$ we see that $[y,x]^{2q}=1$. So
$\gamma_{5}(H)$ is generated by elements of order $2q$. Furthermore
$\gamma_{5}(H)$ is nilpotent of class 2, and
$\gamma_{5}(H)^{\prime}\leq\gamma_{10}(H)$ has exponent $q$. So
$\gamma_{5}(H)$ has exponent $2q$.

Next let $y\in H^{\prime}$, and replace $q$ by $4q$ in equation (5). We obtain
$[y,x]^{4q}=1$. So $\gamma_{3}(H)$ is generated by elements of order $4q$, and
using the fact that $\gamma_{5}(H)$ has exponent $2q$ and $\gamma_{8}(H)$ has
exponent $q$, we see that $\gamma_{3}(H)$ has exponent $4q$.

Finally replace $q$ by $8q$ in equation (5) and we obtain $[y,x]^{8q}=1$ for
all $x,y$. Using facts that $\gamma_{3}(H)$ has exponent $4q$, $\gamma_{5}(H)$
has exponent $2q$, and $\gamma_{8}(H)$ has exponent $q$, we see that
$H^{\prime}$ has exponent $8q$.

Let $N$ be the normal subgroup

$\gamma_{2}(F)^{8q}\gamma_{3}(F)^{4q}\gamma_{5}(F)^{2q}\gamma_{8}(F)^{q}\gamma_{14}(F)<F.$

Then $H=F/M$ where $M=\langle[y^{q},x]\,:\,x,y\in F\rangle N$.

Next we let $K$ be the normal subgroup

$\gamma_{2}(F)^{q}\gamma_{3}(F)^{\frac{q}{2}}\gamma_{5}(F)^{\frac{q}{4}}\gamma_{8}(F)^{\frac{q}{8}}\gamma_{14}(F)<F.$

(The relevance of $K$ is that if $q\geq 8$ is a power of 2 then $[y^{q},x]\in
K$ for all $x,y\in F$.) We show that every element $k\in K$ can be written
uniquely modulo $\gamma_{14}(F)$ in the form

$k=[b,a]^{qm_{3}}c_{4}^{\frac{q}{2}m_{4}}\ldots
c_{8}^{\frac{q}{2}m_{8}}c_{9}^{\frac{q}{4}m_{9}}\ldots
c_{41}^{\frac{q}{4}m_{41}}c_{42}^{\frac{q}{8}m_{42}}\ldots
c_{1377}^{\frac{q}{8}m_{1377}},$ (6)

where $m_{3},m_{4},\ldots,m_{1377}$ are integers. (The number of basic
commutators of weight at most 4 is 8, the number of weight at most 7 is 41,
and the number of weight at most 13 is 1377.) First let

$K_{2}=\gamma_{3}(F)^{\frac{q}{2}}\gamma_{5}(F)^{\frac{q}{4}}\gamma_{8}(F)^{\frac{q}{8}}\gamma_{14}(F).$

We show that every element in $\gamma_{2}(F)^{q}$ can be written as
$[b,a]^{qm_{3}}$ modulo $K_{2}$. The elements of $\gamma_{2}(F)^{q}$ are
products of $q^{th}$ powers of elements in $\gamma_{2}(F)$. Let
$x,y\in\gamma_{2}(F)$. Then from equation (2) in Section 3 we see that

$(xy)^{q}=x^{q}y^{q}[y,x]^{\binom{q}{2}}[y,x,x]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{1377}\alpha)^{f_{1377}(q)}$

where $\alpha:F\rightarrow F$ is the endomorphism mapping $a,b$ to $x,y$. Now
$[y,x]$, and $[y,x,x]\in\gamma_{4}(F)$ and $\binom{q}{2}$ and $\binom{q}{3}$
are divisible by $\frac{q}{2}$. So $[y,x]^{\binom{q}{2}}$ and
$[y,x,x]^{\binom{q}{3}}\in K_{2}$. Similarly all the terms
$(c_{5}\alpha)^{f_{5}(q)}$, $\ldots$, $(c_{1377}\alpha)^{f_{1377}(q)}$ lie in
$K_{2}$. So $(xy)^{q}=x^{q}y^{q}$ modulo $K_{2}$. This means that every
product of $q^{th}$ powers of elements in $\gamma_{2}(F)$ can be written as
the $q^{th}$ power of a single element in $\gamma_{2}(F)$ modulo $K_{2}$. So
consider $x^{q}$ when $x\in\gamma_{2}(F)$. We can write $x=[b,a]^{m_{3}}g$ for
some $g\in\gamma_{3}(F)$. Then, as we have just seen,
$x^{q}=[b,a]^{qm_{3}}g^{q}$ modulo $K_{2}$, and since $g^{q}\in K_{2}$ this
means that $x^{q}=[b,a]^{qm_{3}}$ modulo $K_{2}$.

Now let
$K_{3}=\gamma_{5}(F)^{\frac{q}{4}}\gamma_{8}(F)^{\frac{q}{8}}\gamma_{14}(F)$.
Then $K_{2}$ is generated modulo $K_{3}$ by $(\frac{q}{2})^{th}$ powers of
elements in $\gamma_{3}(F)$. Using the same argument as above we see that if
$x,y\in\gamma_{3}(F)$ then $(xy)^{\frac{q}{2}}=x^{\frac{q}{2}}y^{\frac{q}{2}}$
modulo $K_{3}$. So every product of $(\frac{q}{2})^{th}$ powers in
$\gamma_{3}(F)$ can be expressed modulo $K_{3}$ as $x^{\frac{q}{2}}$ with
$x\in\gamma_{3}(F)$. Let $x=c_{4}^{m_{4}}c_{5}^{m_{5}}\ldots c_{8}^{m_{8}}g$,
with $g\in\gamma_{5}(F)$. Then

$\displaystyle x^{\frac{q}{2}}$
$\displaystyle=c_{4}^{\frac{q}{2}m_{4}}(c_{5}^{m_{5}}\ldots
c_{8}^{m_{8}}g)^{\frac{q}{2}}\text{ modulo }K_{3}$
$\displaystyle=c_{4}^{\frac{q}{2}m_{4}}c_{5}^{\frac{q}{2}m_{5}}(c_{6}^{m_{6}}\ldots
c_{8}^{m_{8}}g)^{\frac{q}{2}}\text{ modulo }K_{3}$ $\displaystyle=\ldots$
$\displaystyle=c_{4}^{\frac{q}{2}m_{4}}\ldots
c_{8}^{\frac{q}{2}m_{8}}g^{\frac{q}{2}}\text{ modulo }K_{3}$
$\displaystyle=c_{4}^{\frac{q}{2}m_{4}}\ldots c_{8}^{\frac{q}{2}m_{8}}\text{
modulo }K_{3}.$

So every element of $\gamma_{2}(F)^{q}\gamma_{3}(F)^{\frac{q}{2}}$ can be
expressed modulo $K_{3}$ in the form

$[b,a]^{qm_{3}}c_{4}^{\frac{q}{2}m_{4}}\ldots c_{8}^{\frac{q}{2}m_{8}}.$

Continuing in this way we see that every element $k\in K$ can be written in
the form (6) modulo $\gamma_{14}(F)$. Since every element of
$F/\gamma_{14}(F)$ can be uniquely expressed in the form

$c_{1}^{n_{1}}c_{2}^{n_{2}}\ldots c_{1377}^{n_{1377}}\gamma_{14}(F)$

for some integers $n_{1},n_{2},\ldots,n_{1377}$, the expression (6) for $k$
modulo $\gamma_{14}(F)$ is unique.

Similarly every element of $N$ can be expressed uniquely modulo
$\gamma_{14}(F)$ in the form

$[b,a]^{8qn_{3}}c_{4}^{4qn_{4}}\ldots c_{8}^{4qn_{8}}c_{9}^{2qn_{9}}\ldots
c_{41}^{2qn_{41}}c_{42}^{qn_{42}}\ldots c_{1377}^{qn_{1377}},$

and if $k\in K$ is given by (6) then $k\in N$ if and only if $8|m_{i}$ for
$3\leq i\leq 1377$. So $K$ is generated by

$C=\\{[b,a]^{q},c_{4}^{\frac{q}{2}},\ldots,c_{8}^{\frac{q}{2}},c_{9}^{\frac{q}{4}},\ldots,c_{41}^{\frac{q}{4}},c_{42}^{\frac{q}{8}},\ldots,c_{1377}^{\frac{q}{8}}\\}$

modulo $\gamma_{14}(F)$, and all these generators have order 8 modulo $N$. We
show that provided $q\geq 32$ these generators commute with each other modulo
$N$, so that $K/N$ is abelian. It follows that $K/N$ is a direct sum of 1375
copies of the cyclic group of order 8. If $k\in K$ has the form (6) then we
let $[m_{3},m_{4},\ldots,m_{1377}]$ be the _representative vector_ of $kN$,
and we think of this vector as an element in $C_{8}^{1375}$. Multiplying two
elements of $K/N$ corresponds to adding their representative vectors, and
$k\in N$ if and only if the representative vector of $kN$ is zero.

To show that the elements in $C$ commute with each other we first let
$c,d\in\gamma_{3}(F)$ and consider the commutator $[c^{r},d^{s}]$ for general
$r,s>0$. The subgroup $\langle c,d\rangle$ has class at most 4 modulo $N$, and
we expand $[c^{r},d^{s}]$ modulo $\gamma_{5}(\langle c,d\rangle)$. Taking
$x=d^{s}$ we see that

$[c^{r},x]=[c,x]^{r}[c,x,c]^{\binom{r}{2}}[c,x,c,c]^{\binom{r}{3}}.$

Also

$[c,x]=[c,d^{s}]=[c,d]^{s}[c,d,d]^{\binom{s}{2}}[c,d,d,d]^{\binom{s}{3}}.$

So, modulo $\gamma_{5}(\langle c,d\rangle)$,

$\displaystyle[c^{r},d^{s}]$
$\displaystyle=([c,d]^{s}[c,d,d]^{\binom{s}{2}}[c,d,d,d]^{\binom{s}{3}})^{r}([c,d]^{s}[c,d,d]^{\binom{s}{2}},c])^{\binom{r}{2}}[[c,d]^{s},c,c]^{\binom{r}{3}}$
$\displaystyle=[c,d]^{rs}[c,d,d]^{r\binom{s}{2}}[c,d,d,d]^{r\binom{s}{3}}[c,d,c]^{\binom{r}{2}s}[c,d,d,c]^{\binom{r}{2}\binom{s}{2}}[c,d,c,c]^{\binom{r}{3}s}.$

Now assume that $q\geq 32$ and let $r=s=\frac{q}{2}$ then $2q$ divides $rs$
and all the other exponents in the product above are divisible by $q$ so that
$[c^{\frac{q}{2}},d^{\frac{q}{2}}]\in N$. So the elements in
$\\{c_{i}^{\frac{q}{2}}:\,$wt$\,c_{i}=3,4\\}$ commute with each other modulo
$N$. Similarly if $c\in\gamma_{3}(F)$ and $d\in\gamma_{5}(F)$ then $\langle
c,d\rangle$ is nilpotent of class at most 3 modulo $N$, and in the expansion
of $[c^{\frac{q}{2}},d^{\frac{q}{4}}]$ the exponent of $[c,d]$ is divisible by
$2q$, and all the exponents of terms of weight 3 are divisible by $q$, so that
$[c^{\frac{q}{2}},d^{\frac{q}{4}}]\in N$. So elements in
$\\{c_{i}^{\frac{q}{2}}:\,$wt$\,c_{i}=3,4\\}$ commute with elements in
$\\{c_{i}^{\frac{q}{4}}\,:\,5\leq\,$wt$\,c_{i}\leq 7\\}$ modulo $N$. If
$c\in\gamma_{3}(F)$ and $d\in\gamma_{8}(F)$ then $\langle c,d\rangle$ is
nilpotent of class at most 2 modulo $N$, so that

$[c^{\frac{q}{2}},d^{\frac{q}{8}}]=[c,d]^{\frac{q^{2}}{16}}\in N.$

In the same way, if $c,d\in\gamma_{5}(F)$ then $\langle c,d\rangle$ is
nilpotent of class at most 2 modulo $N$, and

$\displaystyle[c^{\frac{q}{4}},d^{\frac{q}{4}}]$
$\displaystyle=[c,d]^{\frac{q^{2}}{16}}\in N,$
$\displaystyle[c^{\frac{q}{4}},d^{\frac{q}{8}}]$
$\displaystyle=[c,d]^{\frac{q^{2}}{32}}\in N.$

Finally, if $c,d\in\gamma_{7}(F)$ then $[c,d]\in N$, so $[c^{r},c^{s}]\in N$
for all $r,s$. So all the elements in $C\backslash\\{[b,a]^{q}\\}$ commute
with each other modulo $N$ (provided $q\geq 32$).

It remains to show that $[b,a]^{q}$ commutes with elements in $C$. Let
$d\in\gamma_{3}(F)$ and let $c=[b,a]$. We obtain an expression for
$[d^{r},c^{s}]$ modulo $N$ similar to the expression we obtained above for
$[c^{r},d^{s}]$ modulo $N$ when $c,d\in\gamma_{3}(F)$. In this case we have
$\gamma_{7}(\langle c,d\rangle)\leq N$, so we obtain an expression involving
basic commutators in $c,d$ of weight at most 6. A complete list of basic
commutators up to weight 5 is

$\displaystyle
c,d,[d,c],[d,c,c],[d,c,d],[d,c,c,c],[d,c,c,d],[d,c,d,d],[d,c,c,c,c],$
$\displaystyle[d,c,c,c,d],[d,c,c,d,d],[d,c,d,d,d],[d,c,c,[d,c]],[d,c,d,[d,c]].$

We also need the first basic commutator of weight 6: $[d,c,c,c,c,c]$. All the
other basic commutators of weight 6 in $c,d$ lie in $N$. We call these basic
commutators (including $[d,c,c,c,c,c]$) $d_{1},d_{2},\ldots,d_{15}$. Then

$[d^{r},c^{s}]=d_{3}^{n_{3}}d_{4}^{n_{4}}\ldots d_{15}^{n_{15}}\text{ modulo
}N$

where $n_{3},n_{4},\ldots,n_{15}$ equal

$\displaystyle
rs,r\binom{s}{2},\binom{r}{2}s,r\binom{s}{3},\binom{r}{2}\binom{s}{2},\binom{r}{3}s,r\binom{s}{4},\binom{r}{2}\binom{s}{3},\binom{r}{3}\binom{s}{2},\binom{r}{4}s,$
$\displaystyle
3\binom{r}{2}\binom{s}{3}+2\binom{r}{2}\binom{s}{2}+r\binom{s}{3},4\binom{r}{3}\binom{s}{2}+2\binom{r}{3}s+3\binom{r}{2}\binom{s}{2}+\binom{r}{2}s,r\binom{s}{5}.$

The derivation of these exponents is straightforward, but tedious, so I will
omit it. It is easy to use a computer to check that they are correct for any
number of $r,s$. Using this expression for $[d^{r},c^{s}]$ with
$c^{s}=[b,a]^{q}$ it is straightforward to check that $[b,a]^{q}$ commutes
with all the elements in $C$ modulo $N$.

This completes our proof that $K/N$ is a direct product of 1375 copies of the
cyclic group of order 8.

We have shown that every element $k\in K$ can be expressed uniquely modulo
$\gamma_{14}(F)$ in the form (6) above. But there is a problem in that the
coefficients $m_{3},m_{4},\ldots,m_{1377}$ which appear in (6) can depend on
$q$. To illustrate this, consider the following example. Let
$c_{i},c_{j},c_{k}$ be the basic commutators

$[b,a,a,a,a],\,[b,a,a,a,b],\,[[b,a,a,a,b],[b,a,a,a,a]].$

Then working modulo $\gamma_{14}(F)$ we have

$(c_{i}c_{j})^{\frac{q}{4}}=c_{i}^{\frac{q}{4}}c_{j}^{\frac{q}{4}}c_{k}^{\frac{q}{8}(\frac{q}{4}-1)}.$

However $8|\frac{q}{4}$ provided $q\geq 32$, and so the representative vector
of $(c_{i}c_{j})^{\frac{q}{4}}N$, thought of as an element in $C_{8}^{1375}$,
is

$[0,\ldots,0,1,0,\ldots,0,1,0,\ldots,0,-1,0,\ldots,0]$

which does not depend on $q$.

To solve this problem in generality we need to investigate the binomial
coefficients $\binom{q}{d}$ for $1\leq d\leq 13$. These are all divisible by
$\frac{q}{8}$, so we can write $\binom{q}{d}=\frac{q}{8}n$ for some integer
$n$. We show that $n\operatorname{mod}8$ only depends on $d$, and not on $q$
(provided $q\geq 32$).

Consider $\binom{q}{8}$ for example. We want to show that
$\binom{q}{8}=\frac{q}{8}n$ where $n\operatorname{mod}8$ does not depend on
$q$.

$\displaystyle\frac{q}{8}n$ $\displaystyle=\frac{\allowbreak
q^{8}-28q^{7}+322q^{6}-1960q^{5}+6769q^{4}-13\,132\allowbreak
q^{3}+13\,068q^{2}-5040q}{8!}$ $\displaystyle=-\frac{q}{8}+\frac{2^{2}\times
3267\times q^{2}+rq^{3}}{2^{7}\times 315}$

for some integer $r$. The fraction $\frac{2^{2}\times 3267\times
q^{2}+sq^{3}}{2^{7}\times 315}$ is actually an integer, and since $q=2^{k}$
with $k\geq 5$ we can write this integer as $qm$ for some integer $m$.
Dividing through by $\frac{q}{8}$ we have $n=-1+8m$, and so
$n=-1\operatorname{mod}8$.

For another example consider $\binom{q}{12}$, and again write this binomial
coefficient as $\frac{q}{8}n$.

$\frac{q}{8}n=-\frac{q}{3\times 4}+\frac{2^{7}\times 1024785\times
q^{2}+rq^{3}}{12!}$

for some integer $r$. Multiplying both sides of this equation by 3 we have

$\frac{q}{8}3n=-\frac{q}{4}+sq$

for some integer $s$. So $3n\operatorname{mod}8=-2$ and
$n\operatorname{mod}8=2$.

The binomial coefficients $\binom{q}{d}$ where $d$ is odd are all divisible by
$q$, so can all be written in the form $\frac{q}{8}n$ where
$n\operatorname{mod}8=0$. The binomial coefficients $\binom{q}{d}$ for
$d=2,4,6,10$ can be handled in the same way as we dealt with the cases
$d=8,12$.

Similarly the binomial coefficients $\binom{q}{d}$ for $1\leq d\leq 7$ are all
divisible by $\frac{q}{4}$ and can all be written in the form $\frac{q}{4}n$
where $n\operatorname{mod}8$ does not depend on $q$.

And finally the binomial coefficients $\binom{q}{d}$ for $1\leq d\leq 3$ are
all divisible by $\frac{q}{2}$ and can all be written in the form
$\frac{q}{2}n$ where $n\operatorname{mod}8$ does not depend on $q$.

Now we return to the issue of representing an element $[y^{q},x]$ in the form
(6) above. We need to show that provided $q\geq 32$ then the representative
vector of $[y^{q},x]N$ depends only on $x,y$, and not on $q$. We introduce the
notation rv$\,(kN)$ for the representative vector of $kN$ when $k\in K$.

Equation (3) from Section 3 gives

$[y^{q},x]=[y,x]^{q}[y,x,y]^{\binom{q}{2}}[y,x,y,y]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{1377}\alpha)^{f_{1377}(q)}$

modulo $N$, where $\alpha$ is the endomorphism of $F$ mapping $a,b$ to
$y,[y,x]$. And so

$\text{rv}\,([y^{q},x]N)=\text{rv}\,([y,x]^{q}N)+\text{rv}\,([y,x,y]^{\binom{q}{2}}N)+\ldots+\text{rv}\,((c_{1377}\alpha)^{f_{1377}(q)}N).$

We show that all the summands on the right hand side of this equation depend
only on $x,y$, and not on $q$.

First consider a summand rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ where
wt$\,c_{i}\geq 8$. From our analysis of binomial coefficients we know that
$f_{i}(q)=n\frac{q}{8}$ for some integer $n$ where $n\bmod 8$ is independent
of $q$ (provided $q\geq 32$). Furthermore $(c_{i}\alpha)\in\gamma_{8}(F)$ so
that $(c_{i}\alpha)^{q}\in N$. So

$(c_{i}\alpha)^{f_{i}(q)}N=(c_{i}\alpha)^{(n\bmod 8)\frac{q}{8}}N.$

We show that rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ depends only on $x,y$ by
showing that if $g\in\gamma_{8}(F)$ then rv$\,(g^{\frac{q}{8}}N)$ depends only
on $g$ and not on $q$. Let

$g=c_{42}^{\beta_{42}}\ldots c_{1377}^{\beta_{1377}}\text{ modulo }N,$

for some integers $\beta_{i}$. Then since $\gamma_{8}(F)$ is abelian modulo
$N$,

$g^{\frac{q}{8}}=c_{42}^{\frac{q}{8}\beta_{42}}\ldots
c_{1377}^{\frac{q}{8}\beta_{1377}}\text{ modulo }N,$

and rv$\,(g^{\frac{q}{8}}N)$ equals

$[0,\ldots,0,\beta_{42},\ldots,\beta_{1377}]$

which depends on $g$, and not on $q$.

Next consider a summand rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ where
$4\leq\text{wt}\,c_{i}<8$. We can write $f_{i}(q)=n\frac{q}{4}$ for some
integer $n$ where $n\bmod 8$ is independent of $q$. The element
$c_{i}\alpha\in\gamma_{5}(F)$ and so $(c_{i}\alpha)^{2q}\in N$. So

$(c_{i}\alpha)^{f_{i}(q)}N=(c_{i}\alpha)^{(n\bmod 8)\frac{q}{4}}N.$

We show that rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ depends only on $x,y$ by
showing that if $g\in\gamma_{5}(F)$ then rv$\,(g^{\frac{q}{4}}N)$ depends only
on $g$ and not on $q$. Let

$g=c_{9}^{\beta_{9}}\ldots c_{1377}^{\beta_{1377}}\text{ modulo }N,$

for some integers $\beta_{i}$. Then

$g^{\frac{q}{4}}=c_{9}^{\frac{q}{4}\beta_{9}}\ldots
c_{1377}^{\frac{q}{4}\beta_{1377}}h^{\binom{\frac{q}{4}}{2}}\text{ modulo }N,$

where

$h=\prod_{9\leq i<j\leq 1377}[c_{j}^{\beta_{j}},c_{i}^{\beta_{i}}].$

So

$\text{rv}\,(g^{\frac{q}{4}}N)=[0,\ldots,0,\beta_{9},\ldots,\beta_{41},2\beta_{42},\ldots,2\beta_{1377}]+u$

where $u=\,$rv$\,(h^{\binom{\frac{q}{4}}{2}}N)$. As we have seen

$h^{\binom{\frac{q}{4}}{2}}N=h^{-\frac{q}{8}}N$

so $u$ depends only on $h$, which in turn depends only on $g$. So
rv$\,(g^{\frac{q}{4}}N)$ depends only on $g$ and not on $q$.

Now consider a summand rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ where wt$\,c_{i}=2$
or $3$. We can write $f_{i}(q)=n\frac{q}{2}$ for some integer $n$ where
$n\bmod 8$ is independent of $q$. The element $c_{i}\alpha\in\gamma_{3}(F)$
and so $(c_{i}\alpha)^{4q}\in N$. So

$(c_{i}\alpha)^{f_{i}(q)}N=(c_{i}\alpha)^{(n\bmod 8)\frac{q}{2}}N.$

We show that rv$\,((c_{i}\alpha)^{f_{i}(q)}N)$ depends only on $x,y$ by
showing that if $g\in\gamma_{3}(F)$ then rv$\,(g^{\frac{q}{2}}N)$ depends only
on $g$ and not on $q$. Let $g=c_{4}^{\beta_{4}}h$ modulo $N$ where

$h=c_{5}^{\beta_{5}}\ldots c_{1377}^{\beta_{1377}}.$

Then from equation (2) in Section 3 we see that

$g^{\frac{q}{2}}=c_{4}^{\frac{q}{2}\beta_{4}}h^{\frac{q}{2}}[h,c_{4}^{\beta_{4}}]^{\binom{\frac{q}{2}}{2}}(c_{4}\gamma)^{f_{4}(\frac{q}{2})}\ldots(c_{1377}\gamma)^{f_{1377}(\frac{q}{2})}\text{
modulo }N$

where $\gamma$ is the endomorphism of $F$ mapping $a,b$ to
$c_{4}^{\beta_{4}},h$. If wt$\,c_{i}>4$ then $c_{i}\gamma\in\gamma_{14}(F)$
and so

$g^{\frac{q}{2}}=c_{4}^{\frac{q}{2}\beta_{4}}h^{\frac{q}{2}}[h,c_{4}^{\beta_{4}}]^{\binom{\frac{q}{2}}{2}}(c_{4}\gamma)^{f_{4}(\frac{q}{2})}\ldots(c_{8}\gamma)^{f_{8}(\frac{q}{2})}\text{
modulo }N$

and

$\text{rv}\,(g^{\frac{q}{2}}N)=\text{rv}\,(c_{4}^{\frac{q}{2}\alpha_{4}}N)+\text{rv}\,(h^{\frac{q}{2}}N)+\ldots+\text{rv}\,((c_{8}\gamma)^{f_{8}(\frac{q}{2})}N).$

Clearly rv$\,(c_{4}^{\frac{q}{2}\alpha_{4}}N)=[0,\alpha_{4},0,\ldots,0]$
depends only on $g$. And we can assume by induction on the length of the
product $h=c_{5}^{\beta_{5}}\ldots c_{1377}^{\beta_{1377}}$ that
rv$\,(h^{\frac{q}{2}}N)$ depends only on $h$, and hence only on $g$.

So consider the element $[h,c_{4}^{\beta_{4}}]$. This element lies in
$\gamma_{6}(F)$ so that $[h,c_{4}^{\beta_{4}}]^{2q}\in N$. Furthermore

$\binom{\frac{q}{2}}{2}=\frac{q}{4}(\frac{q}{2}-1)=-\frac{q}{4}\text{ modulo
}2q$

provided $q\geq 32$. So

$[h,c_{4}^{\beta_{4}}]^{\binom{\frac{q}{2}}{2}}N=[h,c_{4}^{\beta_{4}}]^{-\frac{q}{4}}N$

and rv$\,([h,c_{4}^{\beta_{4}}]^{\binom{\frac{q}{2}}{2}}N)$ (as we have seen
above) depends only on $[h,c_{4}^{\beta_{4}}]$, which in turn depends only on
$g$.

For $i=4,5,6,7,8$ $c_{i}\gamma\in\gamma_{9}(F)$, so that $(c_{i}\gamma)^{q}\in
N$. It is straightforward to see that $\frac{q}{8}|f_{i}(\frac{q}{2})$ for
$i=4,5,6,7,8$, and it is also straightforward to see that $f_{i}(\frac{q}{2})$
modulo $q$ equals $4\frac{q}{8}$, $6\frac{q}{8}$, $7\frac{q}{8}$,
$5\frac{q}{8}$, $5\frac{q}{8}$ for $i=4,5,6,7,8$ provided $q\geq 32$. It
follows that
$(c_{4}\gamma)^{f_{4}(\frac{q}{2})}N=(c_{4}\gamma)^{4\frac{q}{8}}N$ and
rv$\,((c_{4}\gamma)^{f_{4}(\frac{q}{2})}N)$ depends only on $c_{4}\gamma$ and
hence only on $g$. Similarly rv$\,((c_{i}\gamma)^{f_{i}(\frac{q}{2})}N)$
depends only on $g$ for $i=5,6,7,8$.

So rv$\,(g^{\frac{q}{2}}N)$ is a sum of vectors each of which depends only on
$g$.

Finally consider the summand rv$\,([y,x]^{q}N)$. Let $[y,x]=[b,a]^{\beta}h$
where $h\in\gamma_{3}(F)$. Then from equation (2) in Section 3 we see that

$[y,x]^{q}=[b,a]^{q\beta}h^{q}[h,[b,a]^{\beta}]^{\binom{q}{2}}(c_{4}\gamma)^{f_{4}(q)}\ldots(c_{23}\gamma)^{f_{23}(q)}\text{
modulo }N.$

where $\gamma$ is the endomorphism of $F$ sending $a,b$ to $[b,a]^{\beta},h$.
(For $i>23$ $c_{i}\gamma\in\gamma_{14}(F)$). So rv$\,([y,x]^{q}N)$ equals

$[\beta,0,\ldots,0]+\text{rv}\,(h^{q}N)+\text{rv}\,([h,[b,a]^{\beta}]^{\binom{q}{2}}N)+\ldots+\text{rv}\,((c_{23}\gamma)^{f_{23}(q)}N).$

We have already shown that rv$\,(h^{q}N)$ depends only on $h$, and hence only
on $x,y$, so consider rv$\,([h,[b,a]^{\beta}]^{\binom{q}{2}}N)$. Since
$[h,[b,a]^{\beta}]\in\gamma_{5}(F)$ it follows that $[h,[b,a]^{\beta}]^{2q}\in
N$. So, as we have seen above,
$[h,[b,a]^{\beta}]^{\binom{q}{2}}N=[h,[b,a]^{\beta}]^{-\frac{q}{2}}N$, so that
rv$\,([h,[b,a]^{\beta}]^{\binom{q}{2}}N)$ depends only on $x,y$. Both
$c_{4}\gamma$ and $c_{5}\gamma$ lie in $\gamma_{5}(F)$, and so
$c_{4}\gamma^{2q}\in N$ and $c_{5}\gamma^{2q}\in N$. Now
$f_{4}(q)=\binom{q}{3}$ and $f_{5}(q)=\binom{q}{2}+2\binom{q}{3}$, and so
$(c_{4}\gamma)^{f_{4}(q)}N=(c_{4}\gamma)^{q}N$ and
$(c_{5}\gamma)^{f_{5}(q)}N=(c_{5}\gamma)^{-\frac{q}{2}}N$, and
rv$\,((c_{i}\gamma)^{f_{i}(q)}N)$ depends only on $x,y$ for $i=4,5$.

The elements $c_{i}\gamma$ for $i=6,7,\ldots,23$ lie in $\gamma_{8}(F)$, and
so all have order dividing $q$ modulo $N$. And the exponents $f_{i}(q)$ for
$i=6,7,\ldots,23$ can all be expressed in the form $\frac{q}{8}n_{i}$ modulo
$q$, where $n_{i}\bmod 8$ does not depend on $q$ (provided $q\geq 32$). So
rv$\,((c_{i}\gamma)^{f_{i}(q)}N)$ depends only on $x,y$ for $i=6,7,\ldots,23$.

Putting all this together we see that rv$\,([y^{q},x]N)$ depends only on
$x,y$, and not on $q$.

As we stated earlier in this section, $H=F/M$ where

$M=\langle[y^{q},x]\,:\,x,y\in F\rangle N.$

As $x,y$ range over $F$ the elements rv$\,([y^{q},x]N)$ generate an (additive)
subgroup $S\leq C_{8}^{1375}$. An element $k\in K$ lies in $M$ if and only if
rv$\,(kN)\in S$. The key point is that provided $q\geq 32$ the subgroup $S$
does not depend on $q$.

Now consider the claim that $[b,a]$ has order $8q$ in $H$. As stated above, I
have used the $p$-quotient algorithm to confirm this for $q=8,16,32$. For
$q\geq 32$ this is equivalent to showing that in $F$, $[b,a]^{8q}\in M$ and
$[b,a]^{4q}\notin M$. We have shown above that $[b,a]^{8q}\in N$. On the other
hand, if $q\geq 32$ then $[b,a]^{4q}$ has representative vector
$[4,0,0,\ldots,0]$, and since my computations show that $[b,a]^{4q}\notin M$
when $q=32$ this implies that $[4,0,0,\ldots,0]\notin S$, and hence that
$[b,a]^{4q}\notin M$ for any $q\geq 32$.

Next consider the claim that $b^{-q}a^{-q}(ab)^{q}$ has order $8$ in $H$. This
is equivalent to showing that $b^{-8q}a^{-8q}(ab)^{8q}\in M$ and that
$b^{-4q}a^{-4q}(ab)^{4q}\notin M$. From equation (2) in Section 3 we see that

$b^{-8q}a^{-8q}(ab)^{8q}=[b,a]^{\binom{8q}{2}}[b,a,a]^{\binom{8q}{3}}c_{5}^{f_{5}(8q)}\ldots
c_{1377}^{f_{1377}(8q)}\text{ modulo }N,$

and the properties of the integer-valued polynomials $f_{i}(t)$ imply that

$b^{-8q}a^{-8q}(ab)^{8q}\in K.$

My computer calculations show that $b^{-8q}a^{-8q}(ab)^{8q}\in M$ for $q=32$.
So the representative vector of $b^{-8q}a^{-8q}(ab)^{8q}$ lies in $S$ when
$q=32$. It is not really relevant, but the representative vector of
$b^{-8q}a^{-8q}(ab)^{8q}$ is

$[4,0,0,4,4,4,0,0,\ldots,0].$

So $b^{-8q}a^{-8q}(ab)^{8q}\in M$ for all $q\geq 32$. The element
$b^{-4q}a^{-4q}(ab)^{4q}$ also lies in $K$, and my computer calculations show
that if $q=32$ then $b^{-4q}a^{-4q}(ab)^{4q}\notin M$. So the representative
vector of $b^{-4q}a^{-4q}(ab)^{4q}$ does not lie in $S$ when $q=32$, and this
implies that it does not lie in $S$ for any $q\geq 32$. So
$b^{-4q}a^{-4q}(ab)^{4q}\notin M$ for $q\geq 32$.

The claims in Theorem 5 and Theorem 7 for the orders of $b^{-q}a^{-q}(ab)^{q}$
and $[b,a]$ in Schur covers of $R(2,q;c)$ for $c<12$ follow similarly. We
replace $N$ by $N\gamma_{c+2}(F)$. If $c_{r}$ is the last basic commutator of
weight $c+1$ then any element in $K/N$ has a unique representative vector
$[m_{3},m_{4},\ldots,m_{r}]$ for all $q\geq 32$, and the proof goes through in
the same way as above.

There is a slight problem in showing that $b^{-q}a^{-q}(ab)^{q}$ is non-
trivial in the Schur cover of $R(2,q;c)$ since $b^{-q}a^{-q}(ab)^{q}\notin K$.
But we can directly calculate a Schur cover of $R(2,q;1)$ by hand, and show
that $b^{-q}a^{-q}(ab)^{q}\neq 1$ in this cover. Similarly we can show that
$[b,a]^{\frac{q}{2}}$ is non-trivial in a Schur cover of $R(2,q;1)$, so that
the order of $[b,a]$ in a Schur cover of $R(2,q;c)$ is at least $q$ for any
$c$.

Finally, let $G$ be a finite 2-group with exponent $q>2$ and nilpotency class
at most 3. We show that if $H$ is the covering group of $G$ then $H^{\prime}$
has exponent dividing $q$. Our calculations with the covering group of
$R(2,q;3)$ show that $[y,x]^{q}=[y,x,x]^{\frac{q}{2}}=1$ for all $x,y\in H$.
Groups satisfying the 2-Engel identity $[y,x,x]=1$ are nilpotent of class at
most 3. So if $h\in\gamma_{4}(H)$, $h$ can be expressed as a product of terms
$[y,x,x]$ and their inverses, with $x,y\in H$. So $h^{\frac{q}{2}}$ is a
product of terms $[y,x,x]^{\frac{q}{2}}$ and their inverses, and hence
$h^{\frac{q}{2}}=1$. This implies that $\gamma_{4}(H)$ has exponent dividing
$\frac{q}{2}$. Since $H^{\prime}$ is generated by commutators which have order
dividing $q$, and $\gamma_{4}(H)$ has exponent dividing $\frac{q}{2}$, we see
that that $H^{\prime}$ has exponent dividing $q$.

##  6 Proofs of Theorem 6 and Theorem 8

The proofs of Theorem 6 and Theorem 8 are essentially the same as the proofs
of Theorem 5 and Theorem 7. Let $q=3^{k}$ where $k\geq 2$, let $R(2,q;12)=F/R$
where $F$ is the free group of rank two generated by $a,b$, and let
$H=F/[F,R]$. We can use the $p$-quotient algorithm to show that Theorem 6 and
Theorem 8 hold true when $q=9$ or 27, and we show that the fact that they hold
true for $q=27$ shows that they also hold true for higher powers of 3.

We let $q=3^{k}$ where $k\geq 3$, and let $c_{1},c_{2},\ldots,c_{1377}$ be our
list of basic commutators of weight at most $13$. As in the proof of Theorem 5
and Theorem 7 we let $x,y\in H$ and obtain a relation

$[y,x]^{q}[y,x,y]^{\binom{q}{2}}[y,x,y,y]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{1377}\alpha)^{f_{1377}(q)}=1$
(7)

where $\alpha$ is the homomorphism from $F$ to $H$ mapping $a,b$ to $y,[y,x]$.
If wt$\,c_{i}=w$ then $c_{i}\alpha$ is a commutator in $x$ and $y$, with $w$
entries $y$, and $\deg f_{i}\leq w$. The binomial coefficients $q$ and
$\binom{q}{2}$ are both divisible by $q$, the binomial coefficients
$\binom{q}{d}$ for $d<9$ are divisible by $\frac{q}{3}$, and the binomial
coefficients $\binom{q}{d}$ for $d<27$ are divisible by $\frac{q}{9}$.

If we let $y\in\gamma_{5}(H)$ then all commutators in $H$ with 3 or more
entries $y$ are trivial and we obtain the relation
$[y,x]^{q}[y,x,y]^{\binom{q}{2}}=1$. This implies that $[y,x]^{q}=1$, and so
(since $\gamma_{6}(H)$ is nilpotent of class 2) $\gamma_{6}(H)$ has exponent
$q$.

Now let $y\in\gamma_{2}(H)$, and replace $q$ by $3q$ in equation (7). Using
the fact that $\gamma_{6}(H)$ has exponent $q$ we see that
$[y,x]^{3q}[y,x,y]^{\binom{3q}{2}}=1$, which implies that $[y,x]^{3q}=1$, and
hence that $\gamma_{3}(H)$ has exponent $3q$.

Finally replace $q$ by $9q$ in equation (7) and using the facts that
$\gamma_{3}(H)$ has exponent $3q$ and $\gamma_{6}(H)$ has exponent $q$ we see
that $[y,x]^{9q}=1$ for all $x,y\in H$, and that $H^{\prime}$ has exponent
$9q$.

Let $N$ be the normal subgroup

$\gamma_{2}(F)^{9q}\gamma_{3}(F)^{3q}\gamma_{7}(F)^{q}\gamma_{14}(F)<F.$

Then $H=F/M$ where $M=\langle[y^{q},x]\,:\,x,y\in F\rangle N$.

Now let

$K=\gamma_{2}(F)^{q}\gamma_{3}(F)^{\frac{q}{3}}\gamma_{7}(F)^{\frac{q}{9}}\gamma_{14}(F).$

Just as in Section 5 we can show that every element $k\in K$ can be uniquely
expressed modulo $\gamma_{14}(F)$ in the form

$k=[b,a]^{qm_{3}}[b,a,a]^{\frac{q}{3}m_{4}}\ldots
c_{23}^{\frac{q}{3}m_{23}}c_{24}^{\frac{q}{9}m_{24}}\ldots
c_{1377}^{\frac{q}{9}m_{1377}}$ (8)

(There are 23 basic commutators $c_{i}$ of weight at most 6.) Similarly every
element of $N$ can be uniquely expressed modulo $\gamma_{14}(F)$ in the form

$[b,a]^{9qn_{3}}[b,a,a]^{3qn_{4}}\ldots
c_{23}^{3qn_{23}}c_{24}^{qn_{24}}\ldots c_{1377}^{qn_{1377}}.$

If $k\in K$ is given by equation (8) then $k\in N$ if and only if $9|m_{i}$
for $i=3,4,\ldots,1377$. And just as in Section 5 we can show that $K/N$ is
abelian and is a direct product of 1375 copies of the cyclic group of order 9.
We let $[m_{3},m_{4},\ldots,m_{1377}]$ be the representative vector for $kN$,
and think of this vector as an element in $C_{9}^{1375}$. Multiplying elements
of $K/N$ corresponds to adding their representative vectors, and the element
$k$ lies in $N$ if and only if the representative vector of $kN$ is $0$.

A similar argument to that given in Section 5 for binomial coefficients
$\binom{q}{d}$ $(d\leq 13)$ for $q$ a power of 2 at least as big as 32, shows
that if $q\geq 27$ is a power of 3 then all the binomial coefficients
$\binom{q}{d}$ $d\leq 13$ can be expressed in the form $\frac{q}{9}n$ for some
integer $n$ where $n\operatorname{mod}9$ depends only on $d$, and not on $q$.
Similarly, if $q\geq 27$ is a power of 3 then all the binomial coefficients
$\binom{q}{d}$ $d<9$ can be expressed in the form $\frac{q}{3}n$ for some
integer $n$ where $n\operatorname{mod}9$ depends only on $d$, and not on $q$.
And finally, if $q\geq 27$ is a power of 3 then $\binom{q}{1}=q$ and
$\binom{q}{2}=qn$ where $n\bmod 9=4$. Using the same argument as we used in
Section 5, we see that if $q\geq 27$ and $x,y\in F$, then $[y^{q},x]\in K$,
and the representative vector of $[y^{q},x]N$ depends only on $x,y$, and not
on $q$. As we stated above, $H=F/M$ where $M=\langle[y^{q},x]\,:\,x,y\in
F\rangle N$. The quotient $M/N$ is a subgroup of $K/N$ and the set of
representative vectors of elements in this subgroup is a subgroup $S\leq
C_{9}^{1375}$. If $k\in K$, then $k\in M$ if and only if the representative
vector of $kN$ lies in $S$.

The remainder of the proof of Theorem 6 goes through in the same way as in
Section 5, as does the proof of the claims in Theorem 8 for the order of
$[b,a]$ in Schur covers of $R(2,q;c)$ for various $c$.

Now let $G$ be a finite 3-group with exponent $q>3$ and class at most 8, and
let $H$ be the covering group for $G$. We show that $H^{\prime}$ has exponent
dividing $q$. We have shown that commutators in $H$ have order dividing $q$,
but we need to show that products of commutators have order dividing $q$. Our
calculations with the covering group of $R(2,q;8)$ show that
$[y,x,x,x]^{\frac{q}{3}}=1$ for all $x,y\in H$. It is known that 3-Engel
groups are locally nilpotent, and Werner Nickel’s nilpotent quotient algorithm
in Magma [2] has a facility for computing Engel groups. The free 3-Engel group
of rank 5 has class 9, and it is an easy calculation with the nilpotent
quotient algorithm to show that 3-Engel groups satisfy the identity
$[x_{1},x_{2},x_{3},x_{4},x_{5}]^{20}=1$. This implies that in a free group
$[x_{1},x_{2},x_{3},x_{4},x_{5}]^{20}$ can be expressed as a product of terms
$[y,x,x,x]$ and their inverses. Now $\gamma_{5}(H)$ is abelian and is
generated by elements with order dividing $q$. So $\gamma_{5}(H)$ has exponent
dividing $q$, which is coprime to 20. So if $h\in\gamma_{5}(H)$ then $h$ can
be expressed as a product of terms $[y,x,x,x]$ and their inverses (with
$x,y\in H$). This implies that $h^{\frac{q}{3}}$ is a product of terms
$[y,x,x,x]^{\pm\frac{q}{3}}$ (which are all trivial) and terms

$[[y,x,x,x]^{\pm 1},[z,t,t,t]^{\pm 1}]^{\binom{\frac{q}{3}}{2}}$

which are also trivial. So $\gamma_{5}(H)$ has exponent dividing
$\frac{q}{3}$. This, combined with the fact that $H^{\prime}$ is generated by
elements with order dividing $q$, implies that $H^{\prime}$ has exponent
dividing $q$.

##  7 Proof of Theorem 9 and Theorem 10

To prove Theorem 9 we need to show that if $q$ is a power of $5$ then the
order of $[b,a]$ in a Schur cover of $R(2,q;c)$ is $q$ for $c<9$, and $5q$ for
$c=9$. It is an easy calculation with the $p$-quotient algorithm to show that
this is the case for $q=5,25$. We use the same argument as in Section 5 and
Section 6 to show that this implies that the theorem holds true for all powers
of 5.

Let $q=5^{k}$ where $k\geq 2$, let $R(2,q;9)=F/R$ where $F$ is the free group
of rank two generated by $a,b$, and let $H=F/[F,R]$. Let
$c_{1},c_{2},\ldots,c_{226}$ be our list of basic commutators of weight at
most $10$. As in the last two sections we let $x,y\in H$ and obtain a relation

$[y,x]^{q}[y,x,y]^{\binom{q}{2}}[y,x,y,y]^{\binom{q}{3}}(c_{5}\alpha)^{f_{5}(q)}\ldots(c_{226}\alpha)^{f_{226}(q)}=1$
(9)

where $\alpha$ is the homomorphism from $F$ to $H$ mapping $a,b$ to $y,[y,x]$.
If wt$\,c_{i}=w$ then $c_{i}\alpha$ is a commutator in $x$ and $y$, with $w$
entries $y$, and $\deg f_{i}\leq w$. The binomial coefficients $\binom{q}{d}$
for $d<5$ are divisible by $q$, and the binomial coefficients $\binom{q}{d}$
for $d<25$ are divisible by $\frac{q}{5}$.

If we let $y\in\gamma_{2}(H)$ then all commutators in $H$ with 5 or more
entries $y$ are trivial and we see that $[y,x]^{q}$ is a product of $q^{th}$
powers of commutators of higher weight in $x,y$. So $[y,x]^{q}=1$, and
$\gamma_{3}(H)$ has exponent $q$.

Now replace $q$ by $5q$ in (9) and we obtain $[y,x]^{5q}=1$, which implies
that $\gamma_{2}(H)$ has exponent $5q$.

So we let

$N=\gamma_{2}(F)^{5q}\gamma_{3}(F)^{q}\gamma_{11}(F)$

and we let

$K=\gamma_{2}(F)^{q}\gamma_{3}(F)^{\frac{q}{5}}\gamma_{11}(F).$

Just as in Section 5 and Section 6 we can show that $K/N$ is a direct product
of 224 copies of the cyclic group of order 5. Every element $k\in K$ can be
uniquely expressed modulo $N$ in the form

$k=[b,a]^{qm_{3}}c_{4}^{\frac{q}{5}m_{4}}\ldots c_{226}^{\frac{q}{5}m_{226}}$

where $0\leq m_{i}<5$ for $i=3,4,\ldots,226$. We let
$[m_{3},m_{4},\ldots,m_{226}]$ be the representative vector for $kN$, and we
think of it as an element in $C_{5}^{224}$.

Just as in Section 5 we can show that if $x,y\in F$ then $[x^{q},y]\in K$, and
the representative vector of $[x^{q},y]N$ depends only on $x,y$, and not on
$q$. The rest of the proof that the order of $[b,a]$ in a Schur cover of
$R(2,q;c)$ is $q$ for $c<9$, and $5q$ for $c=9$ goes through just as in
Section 5.

If $G$ is any group of exponent $q=5^{k}$ ($k\geq 1$) with class less than 9,
and if $H$ is the cover of $G$ then commutators in $H$ have order dividing
$q$, and so (since $H^{\prime}$ has class at most 4) $H^{\prime}$ has exponent
dividing $q$, which implies that $M(G)$ has exponent dividing $q$.

The proof of Theorem 10 is almost identical to the proof of Theorem 9.

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