# Identification of non-isomorphic $2$-groups with
dihedral central quotient and
isomorphic modular group algebras

Leo Margolis  0000-0002-9413-0775 Universidad Autónoma de Madrid, Departamento
de Matemáticas, C/ Francisco Tomás y Valiente, 7 Facultad de Ciencias, módulo
17, 28049 Madrid, Spain<EMAIL_ADDRESS>and  Taro Sakurai
0000-0003-0608-1852 Department of Mathematics and Informatics, Graduate School
of Science, Chiba University, 1-33, Yayoi-cho, Inage-ku, Chiba-shi, Chiba,
263-8522, Japan<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

The question whether non-isomorphic finite $p$-groups can have isomorphic
modular group algebras was recently answered in the negative by García-Lucas,
Margolis and del Río [J. Reine Angew. Math. 783 (2022), pp. 269–274]. We embed
these negative solutions in the class of two-generated finite $2$-groups with
dihedral central quotient, and solve the original question for all groups
within this class. As a result, we discover new negative solutions and simple
algebra isomorphisms. At the same time, the positive solutions for most of the
groups in this class give some insights what makes the negative solutions
special.

###### Key words and phrases:

Modular isomorphism problem, group algebra, counterexamples, finite
$2$-groups, classification.

###### 2020 Mathematics Subject Classification:

Primary 20C05; Secondary 16S34, 20D15, 16U60

## Introduction

The _modular isomorphism problem_ asks whether non-isomorphic finite
$p$-groups can have isomorphic group algebras over a field of positive
characteristic $p$. It seems to have first been raised in the 1950s and
appears in an influential survey by Brauer [3, Problem 2 and Supplement §9].
Over the following decades, positive results for many classes of finite
$p$-groups were published, cf. [17] for an overview. However, a few years ago,
García-Lucas, Margolis and del Río [10] discovered the first non-isomorphic
finite $2$-groups that have isomorphic group algebras over an arbitrary field
of characteristic $2$.

Nevertheless, the problem remains poorly understood in general and open for
odd primes, for instance. Our goal in this article is to understand better how
the negative solutions presented in [10] come to be. These groups were
discovered during an effort to solve the modular isomorphism problem for two-
generated finite $p$-groups with cyclic derived subgroup, after these had been
classified by Broche, García-Lucas and del Río [4]. Although considerable
effort has been devoted to solving the problem in this wide class, it has not
been achieved completely so far [9, 8]. We consider the negative solutions
from another viewpoint and embed them in a natural class of groups for which
we can completely solve the modular isomorphism problem.

We call a non-abelian group that is generated by two elements of order $2$
_dihedral_. For a group $G$, we call the quotient by its center the _central
quotient_ of $G$. In this article, we study the modular isomorphism problem
for two-generated finite $2$-groups with dihedral central quotient. This class
of groups includes the negative solutions presented in [10], and we are able
to identify all negative solutions to the problem—including new
counterexamples—within this class of groups.

######  Theorem A.

Let $G$ and $H$ be finite $2$-groups and $\mathbb{F}$ a field of
characteristic $2$. Suppose that $G$ is two-generated and the central quotient
of $G$ is dihedral. Then $\mathbb{F}G\cong\mathbb{F}H$ but $G\not\cong H$ if
and only if $G$ and $H$ are isomorphic to the groups

$\displaystyle\langle\,x,y,z\mid x^{2^{n}}$ $\displaystyle=1,$ $\displaystyle
y^{2^{m}}$ $\displaystyle=1,$ $\displaystyle z^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[y,x]$ $\displaystyle=z,$ $\displaystyle[z,x]$
$\displaystyle=z^{-2},$ $\displaystyle[z,y]$ $\displaystyle=z^{-2}\,\rangle,$
$\displaystyle\langle\,a,b,c\mid a^{2^{n}}$ $\displaystyle=1,$ $\displaystyle
b^{2^{m}}$ $\displaystyle=a^{2^{m}},$ $\displaystyle c^{2^{\ell}}$
$\displaystyle=1,$ $\displaystyle[b,a]$ $\displaystyle=c,$
$\displaystyle[c,a]$ $\displaystyle=c^{-2},$ $\displaystyle[c,b]$
$\displaystyle=c^{-2}\,\rangle$

for some integers $n>m>\ell\geq 2$.

The known counterexamples to the modular isomorphism problem presented in [10]
correspond to the case $\ell=2$. It should be emphasized, however, that our
algebra isomorphisms are simpler than the original ones and it makes the proof
cleaner. This is largely due to the change of presentations of the groups,
which is natural in our class.

For the proof of Theorem A, we first give presentations of the groups in our
target class. We then establish an isomorphism between the algebras of the
above groups in Theorem 3.1. The proof of this isomorphism can be read
independently of the rest of this article, as it only needs the presentations
of the groups. We remark that the proof can not be carried over to odd primes,
or at least not in the most naive way (cf. Remark 3.5). Finally, we prove that
no more negative solutions exist within the class of two-generated finite
$2$-groups with dihedral central quotient. This is the most challenging part
of the article, and what can be considered standard methods to attack the
problem turn out to be insufficient to prove this. We hence introduce a
procedure that could be called a “group base approximation”. It allows us,
after considerable effort, to solve the modular isomorphism problem completely
within our class. This procedure could also be used for other classes of
groups with some adjustments.

As a byproduct of identifying negative solutions, we obtain the classification
of two-generated finite $2$-groups with dihedral central quotient. Since such
a classification of groups might be also of independent interest, we provide a
direct proof in Appendix A.

##  1\. Preliminaries

We use standard group-theoretical notation. We always write $G$ and $H$ for
finite $p$-groups for some prime $p$. In most cases, $p=2$ holds. The center
of $G$ is denoted by $\bm{\mathrm{Z}}(G)$, the derived subgroup by
$G^{\prime}$ and the Frattini subgroup by $\operatorname{\Phi}(G)$. By
$\Omega(G)$ we mean the subgroup of $G$ generated by the elements of order
$p$. For a non-negative integer $r$ we write $\mho_{r}(G)$ for the subgroup of
$G$ generated by $g^{p^{r}}$, $g\in G$. For $g,h\in G$ we let
$[g,h]=g^{-1}h^{-1}gh$ denote the commutator. Accordingly we conjugate as
$g^{h}=h^{-1}gh$. We write a cyclic group of order $n$ as $C_{n}$.

Let $\mathbb{F}$ denote a field of characteristic $p$ and $\mathbb{F}G$ the
group algebra of $G$ over $\mathbb{F}$. We write $\Delta(G)$ for the
augmentation ideal in the group algebra $\mathbb{F}G$. Note that $\Delta(G)$
is nilpotent and equal to the Jacobson radical of $\mathbb{F}G$. In
particular, the complement of $\Delta(G)$ equals the set of units in
$\mathbb{F}G$. The center of an algebra $\Lambda$ over $\mathbb{F}$ is denoted
by $\bm{\mathrm{Z}}(\Lambda)$, and the linear subspace spanned by the Lie
commutators is denoted by $[\Lambda,\Lambda]$. We note that
$\Delta(G^{\prime})\mathbb{F}G$ is the relative augmentation ideal of the
derived subgroup and equal to the smallest ideal
$[\mathbb{F}G,\mathbb{F}G]\mathbb{F}G$ of $\mathbb{F}G$ with commutative
quotient. Moreover, we let $\mho_{r}(\Lambda)$ be the subalgebra of $\Lambda$
generated by $T^{p^{r}}$, $T\in\Lambda$.

The center of $\mathbb{F}G$ has a special decomposition which we will use
frequently. Namely
$\bm{\mathrm{Z}}(\mathbb{F}G)=\mathbb{F}\bm{\mathrm{Z}}(G)\oplus(\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G])$.
Moreover, $\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G]$
coincides with the linear subspace of $\bm{\mathrm{Z}}(\mathbb{F}G)$ spanned
by the class sums of non-central elements of $G$. We refer to [24, Section
III.6] for more details and proofs. We will allude to these facts in Section 4
without further mention.

A property of $G$ is an _invariant_ if $\mathbb{F}G\cong\mathbb{F}H$ implies
the same property for $H$. A subset of $\mathbb{F}G$ is called _canonical_ if
it is stable under all algebra automorphisms.

##  2\. Presentations of groups

In this section, we give presentations of two-generated finite $2$-groups with
dihedral central quotient. Compared to many other classes of finite
$p$-groups, this class turns out to be rather accessible.

######  Theorem 2.1.

Let $G$ be a two-generated finite $2$-group with dihedral central quotient.
Then $G/G^{\prime}\cong C_{2^{n}}\times C_{2^{m}}$ and $G^{\prime}\cong
C_{2^{\ell}}$ for some positive integers $n$, $m$ and $\ell$ with $n\geq m$
and $\ell\geq 2$, and $G$ is isomorphic to a group generated by $x$, $y$ and
$z$ with one of the following defining relations.

$\displaystyle\dice{1}:$ $\displaystyle x^{2^{n}}$ $\displaystyle=1,\quad
y^{2^{m}}=1,$ $\displaystyle z^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[y,x]$ $\displaystyle=z,$ $\displaystyle[z,x]$
$\displaystyle=z^{-2},$ $\displaystyle[z,y]$ $\displaystyle=z^{-2};$
$\displaystyle\dice{2}:$ $\displaystyle x^{2^{n}}$ $\displaystyle=1,\quad
y^{2^{m}}=x^{2^{m}},$ $\displaystyle z^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[y,x]$ $\displaystyle=z,$ $\displaystyle[z,x]$
$\displaystyle=z^{-2},$ $\displaystyle[z,y]$ $\displaystyle=z^{-2};$
$\displaystyle\dice{3}:$ $\displaystyle x^{2^{n}}$ $\displaystyle=1,\quad
y^{2^{m}}=z^{2^{\ell-1}},$ $\displaystyle z^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[y,x]$ $\displaystyle=z,$ $\displaystyle[z,x]$
$\displaystyle=z^{-2},$ $\displaystyle[z,y]$ $\displaystyle=z^{-2};$
$\displaystyle\dice{4}:$ $\displaystyle x^{2^{n}}$ $\displaystyle=1,\quad
y^{2^{m}}=x^{2^{m}}z^{2^{\ell-1}},$ $\displaystyle z^{2^{\ell}}$
$\displaystyle=1,$ $\displaystyle[y,x]$ $\displaystyle=z,$
$\displaystyle[z,x]$ $\displaystyle=z^{-2},$ $\displaystyle[z,y]$
$\displaystyle=z^{-2};$ $\displaystyle\dice{5}:$ $\displaystyle x^{2^{n}}$
$\displaystyle=z^{2^{\ell-1}},\quad y^{2^{m}}=1,$ $\displaystyle z^{2^{\ell}}$
$\displaystyle=1,$ $\displaystyle[y,x]$ $\displaystyle=z,$
$\displaystyle[z,x]$ $\displaystyle=z^{-2},$ $\displaystyle[z,y]$
$\displaystyle=z^{-2};$ $\displaystyle\dice{6}:$ $\displaystyle x^{2^{n}}$
$\displaystyle=z^{2^{\ell-1}},\quad y^{2^{m}}=x^{2^{m}},$ $\displaystyle
z^{2^{\ell}}$ $\displaystyle=1,$ $\displaystyle[y,x]$ $\displaystyle=z,$
$\displaystyle[z,x]$ $\displaystyle=z^{-2},$ $\displaystyle[z,y]$
$\displaystyle=z^{-2}.$

We write

(2.1)
$G_{\scriptsizedice{1}},G_{\scriptsizedice{2}},\dotsc,G_{\scriptsizedice{6}}$

for the groups of order $2^{n+m+\ell}$ in Theorem 2.1 and always use the
generators satisfying the above relations, which are written $x,y,z$ or
$a,b,c$.

Evidently, $G_{\scriptsizedice{1}}\cong G_{\scriptsizedice{2}}$ and
$G_{\scriptsizedice{3}}\cong G_{\scriptsizedice{4}}\cong
G_{\scriptsizedice{5}}$ if $n=m$. We will see in the end that all other pairs
of groups are non-isomorphic through investigation of their modular group
algebras. Even though such indirect arguments suffice to prove Theorem A, a
more direct proof is also of interest. Such a proof is presented in Appendix
A.

We also note that the groups (2.1) include finite $2$-groups of maximal class:
the dihedral, semidihedral and generalized quaternion groups (Proposition
A.5).

######  Lemma 2.2.

Let $G$ be a group with dihedral central quotient of order $2^{\ell+1}$. Then
the derived subgroup $G^{\prime}$ is cyclic of order $2^{\ell}$.

###### Proof.

First note that $\ell\geq 2$ as $G/\bm{\mathrm{Z}}(G)$ is non-abelian. Since
the central quotient $G/\bm{\mathrm{Z}}(G)$ is isomorphic to the dihedral
group of order $2^{\ell+1}$, the quotient
$G^{\prime}\bm{\mathrm{Z}}(G)/\bm{\mathrm{Z}}(G)$, which is the derived
subgroup of $G/\bm{\mathrm{Z}}(G)$, has order $2^{\ell-1}$. There are two
elements $x$ and $y$ of $G$ such that $x^{2}$, $y^{2}$ and $(yx)^{2^{\ell}}$
are trivial modulo the center $\bm{\mathrm{Z}}(G)$. Moreover, $x$ and $y$
together with the center of $G$ generate the whole group $G$. We write $z$ for
the commutator $[y,x]$. Since $y^{2}$ is central, we have

$1=[y^{2},x]=[y,x]^{y}[y,x]=z^{y}z$

and the action of $y$ on $z$ by conjugation equals the inversion. The same is
true for $x$. Then a direct calculation of commutators shows that the derived
subgroup $G^{\prime}$ is a cyclic group generated by $z$. Since $z$ is
congruent to $(yx)^{2}$ modulo the center $\bm{\mathrm{Z}}(G)$, the power
$z^{2^{\ell-1}}$ is central. Then $z^{2^{\ell-1}}$ is equal to its inverse by
conjugation. This shows that $G^{\prime}\cap\bm{\mathrm{Z}}(G)$ has order $2$
and hence $G^{\prime}$ has order $2^{\ell}$. ∎

######  Lemma 2.3.

Let $G$ be a two-generated finite $2$-group with dihedral central quotient.
Then $G/G^{\prime}\cong C_{2^{n}}\times C_{2^{m}}$ and $G^{\prime}\cong
C_{2^{\ell}}$ for some positive integers $n$, $m$ and $\ell$ with $n\geq m$
and $\ell\geq 2$, and $G$ is isomorphic to the group

$G_{\theta}=\Biggl{\langle}\,x,y,z\Biggm{|}\begin{gathered}x^{2^{n}}=z^{r2^{\ell-1}},\
y^{2^{m}}=x^{s2^{m}}z^{t2^{\ell-1}},\ z^{2^{\ell}}=1,\\\ [y,x]=z,\
[z,x]=z^{-2},\ [z,y]=z^{-2}\end{gathered}\,\Biggr{\rangle}$

for some $\theta=(r,s,t)$ belonging to

$\Theta=\\{\,(r,s,t)\mid 0\leq r\leq 1,\ \ 0\leq s\leq 2^{n-m}-1,\ \ 0\leq
t\leq 1\,\\}.$

###### Proof.

By Lemma 2.2, the derived subgroup $G^{\prime}$ must be cyclic and the central
quotient $G/\bm{\mathrm{Z}}(G)$ must have order $2^{\ell+1}$. From the proof
of Lemma 2.2, there are three elements $x$, $y$ and $z$ of $G$ such that $x$
and $y$ generate $G$ and

$[y,x]=z,\ z^{2^{\ell}}=1,\ [z,x]=z^{-2},\ [z,y]=z^{-2}.$

We may assume that $x^{2^{n}}$ belongs to $G^{\prime}$, while $x^{2^{n-1}}$
does not. (Otherwise replace $x$, $y$ and $z$ by $y$, $x$ and $z^{-1}$). Since
$x^{2}$ is central, we see that $x^{2^{n}}$ belongs to
$G^{\prime}\cap\bm{\mathrm{Z}}(G)$ which is a cyclic group of order $2$
generated by $z^{2^{\ell-1}}$. Hence

$x^{2^{n}}=z^{r2^{\ell-1}}$

for some $0\leq r\leq 1$.

Let $\bar{G}$ denote the abelianization $G/G^{\prime}$. From [13, Theorem
7.12] there is an element $w$ of $G$ with
$\bar{G}=\langle\bar{x}\rangle\times\langle\bar{w}\rangle$, in particular
$\bar{w}$ has order $2^{m}$. Then $\bar{y}^{2^{m}}=\bar{x}^{s2^{m}}$ for some
$0\leq s\leq 2^{n-m}-1$. Since $y^{-2^{m}}x^{s2^{m}}$ belongs to
$G^{\prime}\cap\bm{\mathrm{Z}}(G)$, we obtain

$y^{2^{m}}=x^{s2^{m}}z^{t2^{\ell-1}}$

for some $0\leq t\leq 1$.

The six relations above define a finite group of order $2^{n+m+\ell}$ and
yield a presentation of $G$. ∎

We consider a partition of the parameter space introduced in Lemma 2.3 as
$\Theta=\Theta_{\scriptsizedice{1}}\cup\dotsb\cup\Theta_{\scriptsizedice{6}}$
defined by the following.

$\displaystyle\Theta_{\scriptsizedice{1}}$
$\displaystyle=\\{\,(r,s,t)\in\Theta\mid r=0,\ t=0,\ s\equiv 0\mod 2\,\\};$
$\displaystyle\Theta_{\scriptsizedice{2}}$
$\displaystyle=\\{\,(r,s,t)\in\Theta\mid r=0,\ t=0,\ s\equiv 1\mod 2\,\\};$
$\displaystyle\Theta_{\scriptsizedice{3}}$
$\displaystyle=\\{\,(r,s,t)\in\Theta\mid r=0,\ t=1,\ s\equiv 0\mod 2\,\\};$
$\displaystyle\Theta_{\scriptsizedice{4}}$
$\displaystyle=\\{\,(r,s,t)\in\Theta\mid r=0,\ t=1,\ s\equiv 1\mod 2\,\\};$
$\displaystyle\Theta_{\scriptsizedice{5}}$
$\displaystyle=\begin{cases}\\{\,(r,s,t)\in\Theta\mid r=1,\ s\equiv 0\mod
2\,\\}&(n>m)\\\ \\{\,(r,s,t)\in\Theta\mid r=1,\ s\equiv 0\mod
2\,\\}\smallsetminus\\{(1,0,1)\\}&(n=m);\end{cases}$
$\displaystyle\Theta_{\scriptsizedice{6}}$
$\displaystyle=\begin{cases}\\{\,(r,s,t)\in\Theta\mid r=1,\ s\equiv 1\mod
2\,\\}&(n>m)\\\ \\{\,(r,s,t)\in\Theta\mid r=1,\ s\equiv 1\mod
2\,\\}\cup\\{(1,0,1)\\}&(n=m).\end{cases}$

######  Lemma 2.4.

Let $n$, $m$ and $\ell$ be positive integers with $n\geq m$ and $\ell\geq 2$,
and let $\theta=(r,s,t)\in\Theta$. Let $s_{0}\equiv s\mod 2$ with
$s_{0}\in\\{0,1\\}$. Then

$G_{\theta}\cong G_{(r,s_{0},(1-r)t)}$

except the case $n=m$ and $\theta=(1,0,1)$. In the exceptional case,

$G_{\theta}\cong G_{(1,1,0)}.$

In particular, we have $G_{\scriptsizetextdice{\textnormal{?}}}\cong
G_{\theta}$ if $\theta\in\Theta_{\scriptsizetextdice{\textnormal{?}}}$ for
$\textdice{\textnormal{?}}\in\\{\dice{1},\dotsc,\dice{6}\\}$.

###### Proof.

Assume that $n>m$ or $\theta\neq(1,0,1)$ so that $r(2-t)2^{n-m}$ is even. Note
here that $n=m$ implies $s=0$.

Take an even number $u$ such that $s+u=s_{0}+r(2-t)2^{n-m}$. Then $x^{u}$ is
central and

$\displaystyle(x^{u}y)^{2^{m}}$
$\displaystyle=x^{u2^{m}}y^{2^{m}}=x^{u2^{m}}x^{s2^{m}}z^{t2^{\ell-1}}=x^{(s+u)2^{m}}z^{t2^{\ell-1}}$
$\displaystyle=x^{s_{0}2^{m}}x^{r(2-t)2^{n}}z^{t2^{\ell-1}}=x^{s_{0}2^{m}}z^{r^{2}(2-t)2^{\ell-1}}z^{t2^{\ell-1}}$
$\displaystyle=x^{s_{0}2^{m}}z^{r(2-t)2^{\ell-1}}z^{t2^{\ell-1}}=x^{s_{0}2^{m}}z^{(1-r)t2^{\ell-1}}.$

Hence $G_{\theta}=\langle x,x^{u}y,z\rangle$ enjoys the defining relations of
$G_{(r,s_{0},(1-r)t)}$.

If $n=m$ and $\theta=(1,0,1)$, then
$y^{2^{m}}=z^{2^{\ell-1}}=x^{2^{n}}=x^{2^{m}}$ and $G_{\theta}=\langle
x,y,z\rangle$ enjoys the defining relations of $G_{(1,1,0)}$. ∎

###### Proof of Theorem 2.1.

It follows from Lemmas 2.2, 2.3 and 2.4. ∎

##  3\. Isomorphic modular group algebras

This section is devoted to proving that groups $G_{\scriptsizedice{1}}$ and
$G_{\scriptsizedice{2}}$ with $n>m>\ell\geq 2$ have isomorphic group algebras
over an arbitrary field of characteristic $2$, but are non-isomorphic groups.
In the next section, we will see that these are the only counterexamples to
the modular isomorphism problem within the class of two-generated finite
$2$-groups with dihedral central quotient.

Arguably the most interesting part of this article is the following, which
generalizes the negative solutions to the modular isomorphism problem
presented in [10].

######  Theorem 3.1.

Let $n\geq m>\ell\geq 2$ and $\mathbb{F}$ a field of characteristic $2$. Then
$\mathbb{F}G_{\scriptsizedice{1}}\cong\mathbb{F}G_{\scriptsizedice{2}}$.

###### Proof.

Let $G=G_{\scriptsizedice{1}}$ and $H=G_{\scriptsizedice{2}}=\langle
a,b,c\rangle$. We construct an algebra homomorphism from $\mathbb{F}G$ to
$\mathbb{F}H$ and then show that it is bijective. Given a group homomorphism
from $G$ to the unit group of $\mathbb{F}H$, we can extend it and obtain an
algebra homomorphism from $\mathbb{F}G$ to $\mathbb{F}H$ by [22, Lemma 1.1.7].
To obtain such a group homomorphism, it suffices to show that the units

$x=a,\ y=b+a+1,\ z=[y,x]\in\mathbb{F}H\smallsetminus\Delta(H)$

satisfy the defining relations for $G$ by [15, Proposition 4.3].

We first verify the commutator relations. Evidently $[y,x]=z$. As $x^{2}$ is
central in $\mathbb{F}H$, the basic commutator identity
$[y,x^{2}]=[y,x][y,x]^{x}$ yields $[z,x]=z^{-2}$. Now use $ba=abc$ to see

$y^{2}=b^{2}+a^{2}+ab+abc+1.$

Observe that $(ab)^{a}=abc=(ab)^{b}$ and $(abc)^{a}=ab=(abc)^{b}$. Hence
$\\{ab,abc\\}$ is a conjugacy class of $H$ and the class sum $ab+abc$ is
central in $\mathbb{F}H$. Thus so is $y^{2}$ which yields $[z,y]=z^{-2}$ as
before.

We proceed to the power relations. Clearly $x^{2^{n}}=1$. As $ab$ commutes
with $c$, it is also easy to raise $y$ to a power:

$\displaystyle y^{2^{m}}$ $\displaystyle=(b^{2}+a^{2}+ab+abc+1)^{2^{m-1}}$
$\displaystyle=b^{2^{m}}+a^{2^{m}}+(ab)^{2^{m-1}}(1+c^{2^{m-1}})+1.$

The first two terms vanish as $b^{2^{m}}=a^{2^{m}}$, and the third term
vanishes as $c^{2^{\ell}}=1$ and $m>\ell$. Thus $y^{2^{m}}=1$. Finally, use
$yx=xyz$ and $ba=abc$ to see

$xy(1+z)=xy+yx=ab+ba=ab(1+c).$

We raise both sides to the power of $2^{\ell}$. Since $ab$ commutes with $c$,
the right hand side becomes $(ab)^{2^{\ell}}(1+c^{2^{\ell}})=0$. On the other
hand, $xy$ commutes with $z$ as

$[z,xy]=[z,y][z,x]^{y}=z^{-2}z^{2}=1.$

Hence the left hand side becomes $(xy)^{2^{\ell}}(1+z^{2^{\ell}})$. As $xy$ is
a unit, we obtain $z^{2^{\ell}}=1$. Therefore all relations are satisfied and
we obtain a suitable algebra homomorphism.

Since $a=x$ and $b=y+x+1$ belong to the image of our algebra homomorphism, it
is surjective. As $G$ and $H$ have the same order, it is bijective. ∎

######  Lemma 3.2.

Let $n>m>\ell\geq 2$. Then $G_{\scriptsizedice{1}}\not\cong
G_{\scriptsizedice{2}}$.

###### Proof.

Let $G=G_{\scriptsizedice{1}}=\langle x,y,z\rangle$ and
$H=G_{\scriptsizedice{2}}=\langle a,b,c\rangle$. The centralizers of the
derived subgroups can distinguish these groups. First we calculate the
exponent of $\bm{\mathrm{{C}}}_{H}(H^{\prime})$.

Recall that the Frattini subgroup $\operatorname{\Phi}(H)$ is generated by
$a^{2}$, $b^{2}$ and $c$. As $a^{2}$ and $b^{2}$ are central we have
$\operatorname{\Phi}(H)\leq\bm{\mathrm{{C}}}_{H}(H^{\prime})$. Moreover, as
$ab\in\bm{\mathrm{{C}}}_{H}(H^{\prime})$ but $ab\notin\operatorname{\Phi}(H)$
we have that $\langle a^{2},ab,b^{2},c\rangle$ is an abelian maximal subgroup
of $H$ and hence equals $\bm{\mathrm{{C}}}_{H}(H^{\prime})$. The orders of
$a^{2}$, $b^{2}$ and $c$ are $2^{n-1}$, $2^{n-1}$ and $2^{\ell}$ respectively.
From $(ab)^{2}=a^{2}bcb=a^{2}b^{2}c^{-1}$ and $b^{2^{m}}=a^{2^{m}}$ we obtain

$(ab)^{2^{n-1}}=a^{2^{n-1}}b^{2^{n-1}}c^{-2^{n-2}}=c^{-2^{n-2}}.$

Since $n>m>\ell\geq 2$, we have $(ab)^{2^{n-1}}=1$ and the exponent of
$\bm{\mathrm{{C}}}_{H}(H^{\prime})$ is $2^{n-1}$.

Similar arguments show that $\bm{\mathrm{{C}}}_{G}(G^{\prime})$ is abelian and
generated by $x^{2}$, $xy$, $y^{2}$ and $z$. Then the exponent of
$\bm{\mathrm{{C}}}_{G}(G^{\prime})$ is equal to $2^{n}$ which is the order of
$xy$. Hence these groups are not isomorphic. ∎

The negative solutions to the modular isomorphism problem presented in [10]
correspond to the groups $G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$
for $n>m>\ell=2$ in our notation. In fact, the isomorphisms from
$G_{\scriptsizedice{1}}=\langle x,y,z\rangle$ and
$G_{\scriptsizedice{2}}=\langle a,b,c\rangle$ to the original groups

$\displaystyle\mathsf{G}$ $\displaystyle=$ $\displaystyle\
\langle\,\mathsf{x},\mathsf{y},\mathsf{z}\mid\mathsf{x}^{2^{n}}$
$\displaystyle=1,$ $\displaystyle\quad\mathsf{y}^{2^{m}}$ $\displaystyle=1,$
$\displaystyle\quad\mathsf{z}^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle\quad[\mathsf{y},\mathsf{x}]$ $\displaystyle=\mathsf{z},$
$\displaystyle\quad\mathsf{z}^{\mathsf{x}}$ $\displaystyle=\mathsf{z}^{-1},$
$\displaystyle\quad\mathsf{z}^{\mathsf{y}}$
$\displaystyle=\mathsf{z}^{-1}\,\rangle,$ $\displaystyle\mathsf{H}$
$\displaystyle=$ $\displaystyle\
\langle\,\mathsf{a},\mathsf{b},\mathsf{c}\mid\mathsf{a}^{2^{n}}$
$\displaystyle=1,$ $\displaystyle\quad\mathsf{b}^{2^{m}}$ $\displaystyle=1,$
$\displaystyle\quad\mathsf{c}^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[\mathsf{b},\mathsf{a}]$ $\displaystyle=\mathsf{c},$
$\displaystyle\quad\mathsf{c}^{\mathsf{a}}$ $\displaystyle=\mathsf{c}^{-1},$
$\displaystyle\quad\mathsf{c}^{\mathsf{b}}$
$\displaystyle=\mathsf{c}\,\rangle$

are given by

$\displaystyle G_{\scriptsizedice{1}}$ $\displaystyle\to\mathsf{G},$
$\displaystyle\quad x$ $\displaystyle\mapsto\mathsf{x},$ $\displaystyle\quad
y$ $\displaystyle\mapsto\mathsf{y},$ $\displaystyle\quad z$
$\displaystyle\mapsto\mathsf{z},$ $\displaystyle G_{\scriptsizedice{2}}$
$\displaystyle\to\mathsf{H},$ $\displaystyle\quad a$
$\displaystyle\mapsto\mathsf{a},$ $\displaystyle\quad b$
$\displaystyle\mapsto\mathsf{a}\mathsf{b},$ $\displaystyle\quad c$
$\displaystyle\mapsto\mathsf{c}.$

######  Remark 3.3.

For a finite $p$-group $G$ with cyclic derived subgroup $G^{\prime}$, García-
Lucas, del Río and Stanojkovski [9, Theorem A] prove that the exponent of the
centralizer $\bm{\mathrm{{C}}}_{G}(G^{\prime})$ is an invariant of its modular
group algebra, provided that the prime $p$ is odd. This shows that the above
pairs of groups, including the original ones presented in [10], are
distinctive for $p=2$.

The groups $G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$ are in fact
isomorphic if $n=m$. We summarize explicitly the negative solutions to the
modular isomorphism problem that we obtain by the previous theorem and lemma.

######  Corollary 3.4.

Let $n>m>\ell\geq 2$. Then the groups

$\displaystyle\langle\,x,y,z\mid x^{2^{n}}$ $\displaystyle=1,$ $\displaystyle
y^{2^{m}}$ $\displaystyle=1,$ $\displaystyle z^{2^{\ell}}$ $\displaystyle=1,$
$\displaystyle[y,x]$ $\displaystyle=z,$ $\displaystyle[z,x]$
$\displaystyle=z^{-2},$ $\displaystyle[z,y]$ $\displaystyle=z^{-2}\,\rangle,$
$\displaystyle\langle\,a,b,c\mid a^{2^{n}}$ $\displaystyle=1,$ $\displaystyle
b^{2^{m}}$ $\displaystyle=a^{2^{m}},$ $\displaystyle c^{2^{\ell}}$
$\displaystyle=1,$ $\displaystyle[b,a]$ $\displaystyle=c,$
$\displaystyle[c,a]$ $\displaystyle=c^{-2},$ $\displaystyle[c,b]$
$\displaystyle=c^{-2}\,\rangle$

are not isomorphic, but have isomorphic group algebras over an arbitrary field
of characteristic $2$.

######  Remark 3.5.

Given the simple proof of the negative solutions to the modular isomorphism
problem for $p=2$, it is natural to ask whether this strategy might be
imitated for groups of odd order. While we currently have no clue, at least
the isomorphism we exhibit and the proof we use can not be imitated directly,
as we will quickly show. Analyzing what facilitates the proof so much, one
important thing is that no actual commutator between any non-trivial units in
the group algebra must be computed: the action of the generators of the group
on the derived subgroup follows from the fact that their squares are central,
and the correct structure of the whole derived subgroup essentially from its
cyclicity. To imitate the proof, it would be necessary to find a finite non-
abelian $p$-group $G$ generated by two elements, say $x$ and $y$, such that
$x^{p}$ and $y^{p}$ are central and $G^{\prime}$ is cyclic. We show that in
this case $G/\bm{\mathrm{Z}}(G)$ is elementary abelian of rank two. For such
groups, a positive answer to the modular isomorphism problem over an arbitrary
field is obtained by Drensky [7].

So assume $p$ is odd, $G=\langle x,y\rangle$ is a finite non-abelian $p$-group
such that $x^{p},y^{p}$ are central in $G$ and $G^{\prime}=\langle z\rangle$
is cyclic. We assume $z=[y,x]$ and let $p^{\ell}$ be the order of $z$. Then
$G$ is regular by [12, III, Satz 10.2 (c)]. Hence $1=[z,x^{p}]=[z,x]^{p}$ by
[12, III, Satz 10.6 (b)]. As $\Omega(G^{\prime})=\langle
z^{p^{\ell-1}}\rangle$, there exists an $0\leq r\leq p-1$ such that
$[z,x]=z^{rp^{\ell-1}}$. This implies $z^{x^{i}}=z^{1+irp^{\ell-1}}$. We
conclude by the standard commutator identity that
$[y,x^{p}]=[y,x][y,x]^{x}\dotsb[y,x]^{x^{p-1}}=z^{k}$, where

$k=\sum_{i=0}^{p-1}(1+irp^{\ell-1})=p+\tfrac{1}{2}r(p-1)p^{\ell}\equiv p\mod
p^{\ell}.$

As $[y,x^{p}]=1$, we conclude that $\ell=1$. Hence $x$ centralizes $z$, and so
does $y$. This implies that $z$ is central. As $x^{p}$ and $y^{p}$ are also
central, we get $\operatorname{\Phi}(G)=\bm{\mathrm{Z}}(G)$, which shows that
$G/\bm{\mathrm{Z}}(G)$ is indeed elementary abelian of rank two.

We add one bibliography remark for readers interested in generalizing the
groups from Theorem 3.1 to odd primes: in several generalizations we have
tried and which were proposed to us the invariants from [2] turned out to
solve the problem in the positive.

For the choice of our isomorphisms in Theorem 3.1, see also Remark 4.21.

##  4\. Non-isomorphic modular group algebras

The rest of this article is devoted to showing that the groups
$G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$ with $n>m>\ell\geq 2$
are indeed the only negative solutions within our class, thereby completing
the proof of Theorem A. Throughout this section, we fix a field $\mathbb{F}$
of characteristic $p$ and assume $p=2$ unless otherwise stated.

We first apply known group-theoretical invariants and then utilize a well-
known argument on power maps to distinguish algebras. However, this turns out
to be insufficient, and in the last part we apply a procedure that could be
called a “group base approximation”. This could also be used for other classes
of groups with some adjustments.

For the proof, we examine all the pairs of groups with $n\geq m$ and $\ell\geq
2$. Basically, we divide it into five cases.

  1. (1)

$n>m>\ell$.

  2. (2)

$n>m\geq 2$ and $m\leq\ell$.

  3. (3)

$n>m=1$.

  4. (4)

$n=m\geq 2$.

  5. (5)

$n=m=1$.

We observe first that the coclass of the groups is $n+m-1$, and the last case
$n=m=1$ corresponds to the finite $2$-groups of maximal class for which a
positive answer to the modular isomorphism problem has been known for decades
[5, 1]. Therefore we may assume $n\geq 2$.

The case $n=m\geq 2$ can be dealt with using a classic invariant, the center
of a group, except the pair that corresponds to $\dice{5}\dice{6}$. We will
use a group base approximation for this case (Lemmas 4.17, 4.18 and 4.19).
What will be used to distinguish modular group algebras for the rest of the
cases is summarized in Table 1.

|  $n>m>\ell$ |  $n>m\geq 2$ and $m\leq\ell$ |  $n>m=1$
---|---|---|---
$\dice{1}\dice{2}$ |  X |  P |  P
$\dice{1}\dice{3}$ |  C |  C |  Q
$\dice{1}\dice{4}$ |  C |  C |  P
$\dice{1}\dice{5}$ |  C |  C |  C
$\dice{1}\dice{6}$ |  C |  C |  C
$\dice{2}\dice{3}$ |  C |  C |  P
$\dice{2}\dice{4}$ |  C |  C |  K
$\dice{2}\dice{5}$ |  C |  C |  C
$\dice{2}\dice{6}$ |  C |  C |  C
$\dice{3}\dice{4}$ |  A |  P |  P
$\dice{3}\dice{5}$ |  C |  C |  C
$\dice{3}\dice{6}$ |  C |  C |  C
$\dice{4}\dice{5}$ |  C |  C |  C
$\dice{4}\dice{6}$ |  C |  C |  C
$\dice{5}\dice{6}$ |  A |  P |  P
Table 1.  Summary of how modular group algebras will be distinguished for
$n>m$. (X) counterexample [Theorem 3.1], (C) center of group [Lemma 4.1], (P)
kernel size of power map [Lemma 4.8], (Q) Quillen’s theorem [Proposition 4.4],
(K) Külshammer’s theorem [Proposition 4.3], (A) group base approximation
[Section 4.3].

###  4.1. Group-theoretical invariants

The first invariant we will use is the isomorphism type of the center of a
group. This is a well-known invariant, and a proof can be found in [24, III,
Theorem 6.6].

######  Lemma 4.1.

Let $G=\langle x,y,z\rangle$ be one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$. Then the center of $G$ is generated by $x^{2}$,
$y^{2}$ and $z^{2^{\ell-1}}$. Consequently:

$\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{1}})$ $\displaystyle=\langle
x^{2}\rangle\times\langle y^{2}\rangle\times\langle z^{2^{\ell-1}}\rangle$
$\displaystyle\cong C_{2^{n-1}}\times C_{2^{m-1}}\times C_{2};$
$\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{2}})$ $\displaystyle=\langle
x^{2}\rangle\times\langle x^{-2}y^{2}\rangle\times\langle
z^{2^{\ell-1}}\rangle$ $\displaystyle\cong C_{2^{n-1}}\times C_{2^{m-1}}\times
C_{2};$ $\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{3}})$
$\displaystyle=\langle x^{2}\rangle\times\langle y^{2}\rangle$
$\displaystyle\cong C_{2^{n-1}}\times C_{2^{m}};$
$\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{4}})$ $\displaystyle=\langle
x^{2}\rangle\times\langle x^{-2}y^{2}\rangle$ $\displaystyle\cong
C_{2^{n-1}}\times C_{2^{m}};$
$\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{5}})$ $\displaystyle=\langle
x^{2}\rangle\times\langle y^{2}\rangle$ $\displaystyle\cong C_{2^{n}}\times
C_{2^{m-1}};$ $\displaystyle\bm{\mathrm{Z}}(G_{\scriptsizedice{6}})$
$\displaystyle=\langle x^{2}\rangle\times\langle x^{-2}y^{2}\rangle$
$\displaystyle\cong C_{2^{n}}\times C_{2^{m-1}}.$

###### Proof.

From the relations, we directly get $x^{2},y^{2}\in\bm{\mathrm{Z}}(G)$. As
$G/\langle x^{2},y^{2}\rangle$ is dihedral with the center generated by the
image of $z^{2^{\ell-1}}$, it follows that $\bm{\mathrm{Z}}(G)=\langle
x^{2},y^{2},z^{2^{\ell-1}}\rangle$. The concrete descriptions of the centers
are then easy to read off from the relations. ∎

######  Proposition 4.2.

Let $n=m$. If $G$ and $H$ are one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ with $\mathbb{F}G\cong\mathbb{F}H$ but $G\not\cong
H$, then $G$ and $H$ are isomorphic to the groups $G_{\scriptsizedice{5}}$ and
$G_{\scriptsizedice{6}}$ with $n=m\geq 2$.

###### Proof.

We observe first that the coclass of the groups is $n+m-1$ in general. Hence
the case $n=m=1$ corresponds to the finite $2$-groups of maximal class for
which a positive answer to the modular isomorphism problem is obtained by
Carlson [5, p. 434] and Bagínski [1].

Thus we may assume that $n=m\geq 2$. Recall that $G_{\scriptsizedice{1}}\cong
G_{\scriptsizedice{2}}$ and $G_{\scriptsizedice{3}}\cong
G_{\scriptsizedice{4}}\cong G_{\scriptsizedice{5}}$ for $n=m$. Since
$\bm{\mathrm{Z}}(G_{\scriptsizedice{1}})\cong C_{2^{n-1}}\times
C_{2^{m-1}}\times C_{2}$ and
$\bm{\mathrm{Z}}(G_{\scriptsizedice{5}})\cong\bm{\mathrm{Z}}(G_{\scriptsizedice{6}})\cong
C_{2^{n}}\times C_{2^{m-1}}$ from Lemma 4.1, the assertion follows. ∎

We will see in Lemmas 4.17, 4.18 and 4.19 that the remaining cases cannot
happen and hence $n>m$.

######  Proposition 4.3.

Let $n>m=1$. Then
$\mathbb{F}G_{\scriptsizedice{2}}\not\cong\mathbb{F}G_{\scriptsizedice{4}}$.

###### Proof.

We will calculate the number of conjugacy classes that consist of squares for
each group. This is an invariant of the group algebra due to the work of
Külshammer [16, Section 1], cf. [11, Section 2.2] for a short proof. To see
that these numbers are different for $G_{\scriptsizedice{2}}$ and
$G_{\scriptsizedice{4}}$, consider a generic element $x^{r}y^{s}z^{t}$ with
$0\leq r\leq 2^{n}-1$, $0\leq s\leq 1$ and $0\leq t\leq 2^{\ell}-1$. Then,
independently of the group,

$(x^{r}y^{s}z^{t})^{2}=\begin{cases}x^{2r}z^{2t}&(r\equiv 0\mod 2,\ s=0)\\\
x^{2r}&(r\equiv 1\mod 2,\ s=0)\\\ x^{2r}y^{2}z^{2t-1}&(r\equiv 1\mod 2,\
s=1)\\\ x^{2r}y^{2}&(r\equiv 0\mod 2,\ s=1).\end{cases}$

Note that if a square is not central, then its centralizer is given by the
maximal subgroup $\bm{\mathrm{{C}}}_{G}(G^{\prime})=\langle
x^{2},y^{2},z,xy\rangle$, the centralizer of $G^{\prime}$. So each conjugacy
class containing squares has either one or two elements; this can be seen from
the generic form of an element. The case $r\equiv 0\mod 2$ and $s=0$ covers
all elements of shape $x^{u}z^{v}$ with $u\equiv 0\mod 4$ and $v\equiv 0\mod
2$. These elements are formally reading the same in $G_{\scriptsizedice{2}}$
and $G_{\scriptsizedice{4}}$ and they also give the same number of classes, as
the centrality of an element of this shape does not depend on which group we
choose. In the case $r\equiv 1\mod 2$ and $s=0$, we get elements of shape
$x^{u}$ with $u\equiv 2\mod 4$. Again this is the same for
$G_{\scriptsizedice{2}}$ and $G_{\scriptsizedice{4}}$. Next we consider the
case $r\equiv 1\mod 2$ and $s=1$. In $G_{\scriptsizedice{2}}$, we have
$x^{2r}y^{2}z^{2t-1}=x^{2r+2}z^{2t-1}$, so these are the elements of shape
$x^{u}z^{v}$ with $u\equiv 0\mod 4$ and $v\equiv 1\mod 2$. In
$G_{\scriptsizedice{4}}$, we have
$x^{2r}y^{2}z^{2t-1}=x^{2r+2}z^{2t-1+2^{\ell-1}}$, which gives the same kind
of elements.

Finally, we consider the case where there is a difference: $r\equiv 0\mod 2$
and $s=1$. In $G_{\scriptsizedice{2}}$, we have $x^{2r}y^{2}=x^{2r+2}$, which
are elements of shape $x^{u}$ with $u\equiv 2\mod 4$. These elements are not
new as they already appeared in the second case (i.e. $r\equiv 1\mod 2$ and
$s=0$). While in $G_{\scriptsizedice{4}}$, we have
$x^{2r}y^{2}=x^{2r+2}z^{2^{\ell-1}}$, which are elements of shape
$x^{u}z^{2^{\ell-1}}$ for $u\equiv 2\mod 4$. These elements are new, so in
particular there are $2^{n-2}$ more conjugacy classes of squares in
$G_{\scriptsizedice{4}}$ than in $G_{\scriptsizedice{2}}$. ∎

######  Proposition 4.4.

Let $n>m=1$. Then
$\mathbb{F}G_{\scriptsizedice{1}}\not\cong\mathbb{F}G_{\scriptsizedice{3}}$.

###### Proof.

Note that $G_{\scriptsizedice{1}}$ contains an elementary abelian subgroup of
rank three, namely $\langle x^{2^{n-1}},y,z^{2^{\ell-1}}\rangle$. By contrast,
$G_{\scriptsizedice{3}}$ does not since the only involutions are
$x^{2^{n-1}},z^{2^{\ell-1}}$ and $x^{2^{n-1}}z^{2^{\ell-1}}$. As the maximal
rank of an elementary abelian subgroup is known to be an invariant of the
group algebra due to the work of Quillen [23, Theorem 6.28], we conclude that
$\mathbb{F}G_{\scriptsizedice{1}}\not\cong\mathbb{F}G_{\scriptsizedice{3}}$. ∎

######  Proposition 4.5.

Let $n>m$. If $G$ and $H$ are one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ with $\mathbb{F}G\cong\mathbb{F}H$ but $G\not\cong
H$, then $G$ and $H$ are isomorphic to one of the following groups.

  1. (1)

$G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$.

  2. (2)

$G_{\scriptsizedice{3}}$ and $G_{\scriptsizedice{4}}$.

  3. (3)

$G_{\scriptsizedice{5}}$ and $G_{\scriptsizedice{6}}$.

###### Proof.

This follows from Lemmas 4.1, 4.3 and 4.4. ∎

We note that computer experiments for the remaining cases in Propositions 4.2
and 4.5 show that the group-theoretical invariants contained in a GAP package
ModIsomExt [18, 19] do not give new information on whether the algebras in
question are isomorphic.

###  4.2. Kernel sizes

We will write $\Delta$ for the augmentation ideal $\Delta(G)$ of $\mathbb{F}G$
for brevity. The next argument we will apply is well known and consists in
computing the sizes of the kernels of certain maps. It can be traced back to
an idea of Brauer [3, Supplement §9] and was applied in practice for the first
time by Passman [21]. We define the standard $2^{m}$-power map

(4.1) $\varphi\colon\Delta/\Delta^{2}\to\Delta^{2^{m}}/\Delta^{1+2^{m}}.$

The kernel of a power map $\varphi$ is defined to be

(4.2)
$K(G)=\\{\,R+\Delta^{2}\in\Delta/\Delta^{2}\mid\varphi(R+\Delta^{2})=0+\Delta^{1+2^{m}}\,\\}.$

Note that this kernel is not an ideal of $\Delta$ in general, but still a
well-defined set stable under automorphisms of the algebra.

A basis for $\Delta$ makes concrete calculations of the kernels feasible, and
we use what is called a Jennings basis, which behaves well with a filtration
$\Delta\supseteq\Delta^{2}\supseteq\dotsb$. We will work with a basis that
looks formally the same in each case. We assume some familiarity with
Jennings’ theory for the proof of the next lemma. If the reader is willing to
accept it, then one can safely skip its proof because we will not use
Jennings’ theory explicitly elsewhere.

We write capital letters for elements of the augmentation ideal corresponding
to group elements, e.g. $X=x+1\in\Delta$ for $x\in G$.

######  Lemma 4.6.

Let $G=\langle x,y,z\rangle$ be one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ and $k$ a non-negative integer. Set
$w=z^{2^{\ell-1}}$ and

$q=\begin{cases}2^{\ell}&(\text{$G=G_{\scriptsizedice{1}}$ or
$G=G_{\scriptsizedice{2}}$})\\\
2^{\max\\{m,\ell\\}}&(\text{$G=G_{\scriptsizedice{3}}$ or
$G=G_{\scriptsizedice{4}}$})\\\
2^{\max\\{n,\ell\\}}&(\text{$G=G_{\scriptsizedice{5}}$ or
$G=G_{\scriptsizedice{6}}$}).\end{cases}$

Then

(4.3)
$\mathfrak{D}_{k}=\left\\{\,X^{r}Y^{s}Z^{t}W^{u}\mathrel{}\middle|\mathrel{}\begin{gathered}0\leq
r\leq 2^{n}-1,\ 0\leq s\leq 2^{m}-1,\\\ 0\leq t\leq 2^{\ell-1}-1,\ 0\leq u\leq
1,\\\ r+s+2t+qu\geq k\end{gathered}\,\right\\}$

is a basis of $\Delta^{k}$. In particular, the image of
$\mathfrak{D}_{k}\smallsetminus\mathfrak{D}_{k+1}$ under the natural
projection $\Delta^{k}\to\Delta^{k}/\Delta^{k+1}$ is a basis of
$\Delta^{k}/\Delta^{k+1}$.

###### Proof.

Let $M_{k}=G\cap(1+\Delta^{k})$, the $k$th dimension subgroup of $G$, for
$k\geq 1$. Observe first that $G^{\prime}=\langle z\rangle$ and
$\operatorname{\Phi}(G)=\langle x^{2},y^{2},z\rangle$ are abelian. It follows
from [6, Theorems 11.2 and 12.9] that $M_{1}=\langle x,y\rangle$ and

(4.4) $2^{e-1}<k\leq 2^{e}\implies
M_{k}=\mho_{e}(G)\mho_{e-1}(G^{\prime})=\langle
x^{2^{e}},y^{2^{e}},z^{2^{e-1}}\rangle$

for $e\geq 1$. In particular, we have

$\displaystyle 0$ $\displaystyle\leq e\leq n-1$ $\displaystyle\implies$
$\displaystyle\quad x^{2^{e}}$ $\displaystyle\in M_{2^{e}}\smallsetminus
M_{1+2^{e}},$ $\displaystyle 0$ $\displaystyle\leq e\leq m-1$
$\displaystyle\implies$ $\displaystyle\quad y^{2^{e}}$ $\displaystyle\in
M_{2^{e}}\smallsetminus M_{1+2^{e}},$ $\displaystyle 1$ $\displaystyle\leq
e\leq\ell-1$ $\displaystyle\implies$ $\displaystyle\quad z^{2^{e-1}}$
$\displaystyle\in M_{2^{e}}\smallsetminus M_{1+2^{e}}$

and by the relations of $G$,

$w\in M_{q}\smallsetminus M_{1+q}.$

Since $x^{2}$ and $y^{2}$ are central, an element of the Jennings basis can be
written in the form

$\displaystyle(x+1)^{r_{0}}(y+1)^{s_{0}}(x^{2}+1)^{r_{1}}(y^{2}+1)^{s_{1}}(z+1)^{t_{0}}\dotsm(w+1)^{u}\dotsm$
$\displaystyle\qquad=(x+1)^{r_{0}}(x^{2}+1)^{r_{1}}\dotsm(y+1)^{s_{0}}(y^{2}+1)^{s_{1}}\dotsm(z+1)^{t_{0}}\dotsm(w+1)^{u}$
$\displaystyle\qquad=(x+1)^{r_{0}}(x+1)^{r_{1}2}\dotsm(y+1)^{s_{0}}(y+1)^{s_{1}2}\dotsm(z+1)^{t_{0}}\dotsm(w+1)^{u}$
$\displaystyle\qquad=(x+1)^{r_{0}+\dotsm+r_{n-1}2^{n-1}}(y+1)^{s_{0}+\dotsm+s_{m-1}2^{m-1}}(z+1)^{t_{0}+\dotsm+t_{\ell-2}2^{\ell-2}}(w+1)^{u}$
$\displaystyle\qquad=(x+1)^{r}(y+1)^{s}(z+1)^{t}(w+1)^{u}$
$\displaystyle\qquad=X^{r}Y^{s}Z^{t}W^{u}$

for

$0\leq
r_{0},r_{1},\dotsc,r_{n-1},s_{0},s_{1},\dotsc,s_{m-1},t_{0},\dotsc,t_{\ell-2},u\leq
1$

where $r=r_{0}+\dotsb+r_{n-1}2^{n-1}$, $s=s_{0}+\dotsb+s_{m-1}2^{m-1}$ and
$t=t_{0}+\dotsb+t_{\ell-2}2^{\ell-2}$. Hence the assertion follows from
Jennings’ theorem [14, Theorem 3.2]. ∎

The _weight_ of an element of
$\mathfrak{D}_{k}\smallsetminus\mathfrak{D}_{k+1}$ is defined to be $k$.

To aid the reader in the following calculations, we explicitly state the bases
of the first few layers of the subsequent quotients of the augmentation ideal
power series when $n\geq m\geq 2$:

$\displaystyle\Delta^{1}/\Delta^{2}$ $\displaystyle:$
$\displaystyle\mathfrak{D}_{1}\smallsetminus\mathfrak{D}_{2}$
$\displaystyle=\\{X,Y\\},$ $\displaystyle\Delta^{2}/\Delta^{3}$
$\displaystyle:$ $\displaystyle\mathfrak{D}_{2}\smallsetminus\mathfrak{D}_{3}$
$\displaystyle=\\{X^{2},XY,Y^{2},Z\\},$ $\displaystyle\Delta^{3}/\Delta^{4}$
$\displaystyle:$ $\displaystyle\mathfrak{D}_{3}\smallsetminus\mathfrak{D}_{4}$
$\displaystyle=\\{X^{3},X^{2}Y,XY^{2},XZ,Y^{3},YZ\\}.$

We will collect some elementary relations between the elements of the basis,
which we will use without further mention:

$\displaystyle XY+YX$ $\displaystyle=Z+XZ+YZ+XYZ,$ $\displaystyle XY+YX$
$\displaystyle\equiv Z\mod\Delta^{3}.$

Now a more concrete expression of the power map (4.1) can be obtained.

######  Lemma 4.7.

Let $G=\langle x,y,z\rangle$ be one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ and $\alpha,\beta\in\mathbb{F}$. Then

$\varphi(\alpha X+\beta
Y+\Delta^{2})=\alpha^{2^{m}}X^{2^{m}}+\beta^{2^{m}}Y^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}}.$

###### Proof.

In general we have

$\displaystyle(\alpha X+\beta Y)^{2}$
$\displaystyle=\alpha^{2}X^{2}+\beta^{2}Y^{2}+(\alpha\beta)(XY+YX)$
$\displaystyle\equiv\alpha^{2}X^{2}+\beta^{2}Y^{2}+(\alpha\beta)Z\mod\Delta^{3}.$

Since $X^{2}$ and $Y^{2}$ are central in $\mathbb{F}G$, we obtain

$\varphi(\alpha X+\beta
Y+\Delta^{2})=\alpha^{2^{m}}X^{2^{m}}+\beta^{2^{m}}Y^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}}.\qed$

######  Lemma 4.8.

Assume $n>m$ and let $G$ be one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$. Then the kernel $K(G)$ of the power map $\varphi$
has the following cardinality as shown in Table 2.

| $m>\ell$ | $m=\ell$ | $m<\ell$
---|---|---|---
1 | $|\mathbb{F}|$ | $|\mathbb{F}|$ | $|\mathbb{F}|$
2 | $|\mathbb{F}|$ | $1$ | $1$
3 | $1$ | $1$ | $|\mathbb{F}|$
4 | $1$ | $|\mathbb{F}|$ | $1$
5 | $|\mathbb{F}|$ | $|\mathbb{F}|$ | $|\mathbb{F}|$
6 | $|\mathbb{F}|$ | $1$ | $1$
Table 2.  Kernel sizes $|K(G)|$ for $n>m$.

###### Proof.

Let $G=\langle x,y,z\rangle$. Every element of $\Delta/\Delta^{2}$ can be
written in the form $\alpha X+\beta Y+\Delta^{2}$ and the scalars
$\alpha,\beta\in\mathbb{F}$ are unique by Lemma 4.6. To measure the size of
the kernel, we write $\varphi(\alpha X+\beta Y+\Delta^{2})$ in Lemma 4.7 as a
linear combination of the images of
$\mathfrak{D}_{2^{m}}\smallsetminus\mathfrak{D}_{1+2^{m}}$ in
$\Delta^{2^{m}}/\Delta^{1+2^{m}}$ for each case. Note that as $n>m$, the
differences between $G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{5}}$ as
well as between $G_{\scriptsizedice{2}}$ and $G_{\scriptsizedice{6}}$ cannot
enter into the expression of $\varphi(\alpha X+\beta Y+\Delta^{2})$, so we
handle these two pairs of groups simultaneously. We also use the equation
$Y^{2^{m}}=X^{2^{m}}+W+X^{2^{m}}W$ for $G=G_{\scriptsizedice{4}}$.

For $m>\ell$, as $Z^{2^{m-1}}$ vanishes, $\varphi(\alpha X+\beta
Y+\Delta^{2})$ is written as follows.

$\displaystyle\dice{1}\dice{5}$ $\displaystyle:$
$\displaystyle\alpha^{2^{m}}X^{2^{m}}+\Delta^{1+2^{m}};$
$\displaystyle\dice{2}\dice{6}$ $\displaystyle:$
$\displaystyle(\alpha^{2^{m}}+\beta^{2^{m}})X^{2^{m}}+\Delta^{1+2^{m}};$
$\displaystyle\dice{3}$ $\displaystyle:$
$\displaystyle\alpha^{2^{m}}X^{2^{m}}+\beta^{2^{m}}W+\Delta^{1+2^{m}};$
$\displaystyle\dice{4}$ $\displaystyle:$
$\displaystyle(\alpha^{2^{m}}+\beta^{2^{m}})X^{2^{m}}+\beta^{2^{m}}W+\Delta^{1+2^{m}}.$

Similarly, for $m<\ell$, the following is obtained.

$\displaystyle\dice{1}\dice{5}$ $\displaystyle:$
$\displaystyle\alpha^{2^{m}}X^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}};$
$\displaystyle\dice{2}\dice{6}$ $\displaystyle:$
$\displaystyle(\alpha^{2^{m}}+\beta^{2^{m}})X^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}};$
$\displaystyle\dice{3}$ $\displaystyle:$
$\displaystyle\alpha^{2^{m}}X^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}};$
$\displaystyle\dice{4}$ $\displaystyle:$
$\displaystyle(\alpha^{2^{m}}+\beta^{2^{m}})X^{2^{m}}+(\alpha\beta)^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}}.$

Finally, for $m=\ell$, the expressions for $G_{\scriptsizedice{1}}$,
$G_{\scriptsizedice{2}}$, $G_{\scriptsizedice{5}}$ and
$G_{\scriptsizedice{6}}$ are as in the case $m<\ell$ and for the other two we
have the following.

$\displaystyle\dice{3}$ $\displaystyle:$
$\displaystyle\alpha^{2^{m}}X^{2^{m}}+(\beta^{2^{m}}+(\alpha\beta)^{2^{m-1}})W+\Delta^{1+2^{m}};$
$\displaystyle\dice{4}$ $\displaystyle:$
$\displaystyle(\alpha^{2^{m}}+\beta^{2^{m}})X^{2^{m}}+(\beta^{2^{m}}+(\alpha\beta)^{2^{m-1}})W+\Delta^{1+2^{m}}.$

All of the terms present are linearly independent by Lemma 4.6. Hence $\alpha
X+\beta Y+\Delta^{2}$ belongs to the kernel $K(G)$ if and only if all of the
coefficients present are equal to zero. Since the Frobenius map on
$\mathbb{F}$ is injective, the cardinalities for $K(G)$ now follow easily. ∎

We remark that in the case $n=m$ the kernel size can also give some useful
information, but only if one restricts the possibilities for $\mathbb{F}$. As
we prefer to work independently of the base field, we do not include the
corresponding calculations.

We summarize the progress we have made on the proof of Theorem A in this
section.

######  Proposition 4.9.

If $G$ and $H$ are one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ with $\mathbb{F}G\cong\mathbb{F}H$ but $G\not\cong
H$, then $G$ and $H$ are isomorphic to one of the following groups.

  1. (1)

$G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$ with $n>m>\ell$.

  2. (2)

$G_{\scriptsizedice{3}}$ and $G_{\scriptsizedice{4}}$ with $n>m>\ell$.

  3. (3)

$G_{\scriptsizedice{5}}$ and $G_{\scriptsizedice{6}}$ with $n>m>\ell$.

  4. (4)

$G_{\scriptsizedice{5}}$ and $G_{\scriptsizedice{6}}$ with $n=m\geq 2$.

###### Proof.

By Proposition 4.2, we may assume that $n>m$. It follows from Proposition 4.5
that $G$ and $H$ are isomorphic to one of the followings:
$G_{\scriptsizedice{1}}$ and $G_{\scriptsizedice{2}}$,
$G_{\scriptsizedice{3}}$ and $G_{\scriptsizedice{4}}$,
$G_{\scriptsizedice{5}}$ and $G_{\scriptsizedice{6}}$. If $m=\ell$ or
$m<\ell$, then the groups in each pair have different kernel sizes by Lemma
4.8, which leads to a contradiction. Hence we have $m>\ell$. ∎

Some more information can be obtained from canonical ideals. The following
proposition could be used to further reduce the cases, though we will not use
it in this sense, but rather differently later on. It seems to have the
potential to also be applicable to other classes of groups.

######  Proposition 4.10.

Let $G$ be a finite $p$-group and $\mathbb{F}$ a field of characteristic $p$.
Assume that $G^{\prime}$ is abelian of exponent $p^{\ell}$. Then

$\mathbb{F}(\mho_{r}(\bm{\mathrm{Z}}(G)))=\mho_{r}(\bm{\mathrm{Z}}(\mathbb{F}G))$

for every integer $r\geq\ell$ and, in particular,
$\Delta(\mho_{r}(\bm{\mathrm{Z}}(G)))$ is canonical in $\mathbb{F}G$.
Consequently, if $\mathbb{F}G\cong\mathbb{F}H$ then
$\mathbb{F}[G/\mho_{r}(\bm{\mathrm{Z}}(G))]\cong\mathbb{F}[H/\mho_{r}(\bm{\mathrm{Z}}(H))]$.

###### Proof.

Recall that $\bm{\mathrm{Z}}(\mathbb{F}G)$ has a basis that is given by class
sums and a decomposition
$\bm{\mathrm{Z}}(\mathbb{F}G)=\mathbb{F}\bm{\mathrm{Z}}(G)\oplus(\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G])$.
We will show that class sums of non-central elements vanish under a
$p^{r}$-power. So let $g\in G$ be a non-central element with conjugacy class
$\\{ga_{1},\dotsc,ga_{s}\\}$ for some $a_{1},\dotsc,a_{s}\in G^{\prime}$. Note
that $p$ divides $s$. The class sum of $g$ equals $g(a_{1}+\dotsb+a_{s})$. As
this element is central in $\mathbb{F}G$, it commutes in particular with $g$,
hence $(a_{1}+\dotsb+a_{s})$ commutes with $g$. As $G^{\prime}$ is abelian of
exponent $p^{\ell}\leq p^{r}$, we conclude that

$(g(a_{1}+\dotsb+a_{s}))^{p^{r}}=g^{p^{r}}(a_{1}^{p^{r}}+\dotsb+a_{s}^{p^{r}})=g^{p^{r}}s=0.\qed$

The next lemma is an easy observation which will be helpful in some
calculations.

######  Lemma 4.11.

Let $G=\langle x,y,z\rangle$ be one of the groups $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$. Then $[X,Y]$ is central in $\mathbb{F}G$.

###### Proof.

We have $[X,Y]=xy+yx=xy+xyz$. As $(xy)^{x}=xyz=(xy)^{y}$ and
$(xyz)^{x}=xy=(xyz)^{y}$, we conclude that $[X,Y]$ is a class sum and hence
central in $\mathbb{F}G$. ∎

###  4.3. Group base approximation

We will handle the rest of the cases by a procedure that could be called a
“group base approximation”. We will first make a few general observations on
group bases satisfying the relations of the normal form given in Theorem 2.1,
and then see in each single case why this leads to a contradiction.

####  4.3.1. General group bases

Now we fix integers $n$, $m$ and $\ell$ with $n\geq m\geq 2$ and $\ell\geq 2$.
Fix a group $G$ from $G_{\scriptsizedice{1}}$, …, $G_{\scriptsizedice{6}}$
generated by $x$, $y$ and $z$ which satisfy the defining relations. Also fix
another group base $H$ in $\mathbb{F}G$ from $G_{\scriptsizedice{1}}$, …,
$G_{\scriptsizedice{6}}$ generated by $a$, $b$ and $c$ which satisfy the
defining relations. Thus, throughout this section, we assume

(4.5) $\begin{gathered}n\geq m\geq 2,\ \ell\geq 2,\
G=G_{\scriptsizetextdice{\textnormal{?}}},\
H=G_{\scriptsizetextdice{\textnormal{?`}}},\
\textdice{\textnormal{?}},\textdice{\textnormal{?`}}\in\\{\dice{1},\dotsc,\dice{6}\\},\\\
G=\langle x,y,z\rangle,\ a,b,c\in 1+\Delta(G),\ H=\langle a,b,c\rangle,\
\mathbb{F}G=\mathbb{F}H.\end{gathered}$

As before, we write $\Delta$ for $\Delta(G)$ and use the basis
$\mathfrak{D}_{k}$ of $\Delta^{k}$ introduced in Lemma 4.6. Recall that
$w=z^{2^{\ell-1}}$, so in this basis $W=Z^{2^{\ell-1}}$, and the weight of $W$
is denoted by $q$; the exact value of $q$ will be irrelevant, except the fact
that $q\geq 4$ as $\ell\geq 2$.

Since $n\geq m\geq 2$, the squares $x^{2}$ and $y^{2}$ are not trivial and we
have

$\Delta=\mathbb{F}X\oplus\mathbb{F}Y\oplus\mathbb{F}XY\oplus\mathbb{F}Z\oplus\mathbb{F}X^{2}\oplus\mathbb{F}Y^{2}\oplus\Delta^{3}.$

From now on square brackets denote Lie commutators in $\Delta$, i.e.
$[U,V]=UV+VU$ for $U,V\in\Delta$. We will frequently use without comment the
commutator formula

$[S,T]=(1+S+T+ST)(1+t^{-1}s^{-1}ts)$

which holds for all $s,t\in G$. In particular, $[X,Y]=(1+X+Y+XY)Z$. Moreover,
as $z^{-1}z^{x},z^{-1}z^{y}\in\langle z^{2}\rangle$, we have $[X,Z],[Y,Z]\in
Z^{2}\mathbb{F}G\subseteq\Delta^{4}$.

We first consider $A$ and $B$ modulo $\Delta^{3}$. Write

(4.6) $\displaystyle A$
$\displaystyle=\alpha_{A}X+\beta_{A}Y+\gamma_{A}XY+\delta_{A}Z+\xi_{A}X^{2}+\eta_{A}Y^{2}+U_{A},$
$\displaystyle B$
$\displaystyle=\alpha_{B}X+\beta_{B}Y+\gamma_{B}XY+\delta_{B}Z+\xi_{B}X^{2}+\eta_{B}Y^{2}+U_{B}$

where $\alpha_{A},\alpha_{B},\dotsc,\eta_{A},\eta_{B}\in\mathbb{F}$ and
$U_{A},U_{B}\in\Delta^{3}$. Note that the scalars are uniquely determined.
This notation will be fixed throughout this section. Observe that
$A+\Delta^{2}$ and $B+\Delta^{2}$ span the two-dimensional vector space
$\Delta/\Delta^{2}$, as $a$ and $b$ generate $H$. Hence they must be linearly
independent, and we have

(4.7) $\displaystyle\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A}\neq 0.$

We first show that $XY$ does not appear as a summand in $A$ and $B$.

######  Lemma 4.12.

We have $\gamma_{A}=\gamma_{B}=0$.

###### Proof.

The proof is identical for $A$ and $B$, so we only show $\gamma_{A}=0$. The
idea is to use the fact that $A^{2}$ is central in $\mathbb{F}G$, as $a^{2}$
is central in $H$.

Let $\Xi$ be the linear subspace of $\Delta$ defined by

$\Xi=\\{\,R\in\Delta^{2}\mid\text{$[R,S]\in\Delta^{5}$ for all
$S\in\Delta$}\,\\}.$

Note that $\Delta^{4}\subseteq\Xi$ and $[X,Y]\in\Xi$ by Lemma 4.11. We will
consider $A^{2}$ modulo $\Xi$. We will use that

$\displaystyle[X,Z],\ $ $\displaystyle[Y,Z]\in
Z^{2}\mathbb{F}G\subseteq\Delta^{4}\subseteq\Xi,$ $\displaystyle[X,XY]$
$\displaystyle=X[X,Y]\equiv XZ\mod\Xi,$ $\displaystyle[Y,XY]$
$\displaystyle=[X,Y]Y\equiv YZ\mod\Xi.$

From (4.6), we get

(4.8) $\displaystyle A^{2}$
$\displaystyle\equiv\alpha_{A}\beta_{A}[X,Y]+\alpha_{A}\gamma_{A}[X,XY]+\beta_{A}\gamma_{A}[Y,XY]$
$\displaystyle\equiv\alpha_{A}\gamma_{A}X[X,Y]+\beta_{A}\gamma_{A}[Y,X]Y$
$\displaystyle\equiv\alpha_{A}\gamma_{A}XZ+\beta_{A}\gamma_{A}YZ\mod\Xi$

where $[X,Y]\in\Xi$ is used to go from the first to the second line. Computing
commutators modulo $\Delta^{5}$ gives

$\displaystyle[XZ,Y]$ $\displaystyle\equiv[YZ,X]\equiv Z^{2}\mod\Delta^{5},$
$\displaystyle[XZ,X]$ $\displaystyle\equiv[YZ,Y]\equiv 0\mod\Delta^{5}.$

So modulo $\Delta^{5}$ computing commutators with $A^{2}$ using (4.8) and the
definition of $\Xi$ we obtain

$\displaystyle[A^{2},X]$
$\displaystyle\equiv\beta_{A}\gamma_{A}Z^{2}\mod\Delta^{5},$
$\displaystyle[A^{2},Y]$
$\displaystyle\equiv\alpha_{A}\gamma_{A}Z^{2}\mod\Delta^{5}.$

As $A^{2}$ is central in $\mathbb{F}G$, we conclude that
$\alpha_{A}\gamma_{A}=\beta_{A}\gamma_{A}=0$. But as $\alpha_{A}\neq 0$ or
$\beta_{A}\neq 0$ by $A\not\in\Delta^{2}$, this implies $\gamma_{A}=0$. ∎

The next lemma allows us to make conclusions about scalars from a relation
between $A^{2^{n}}$ and $B^{2^{n}}$.

######  Lemma 4.13.

We have

$\displaystyle A^{2^{n}}$
$\displaystyle\equiv\alpha_{A}^{2^{n}}X^{2^{n}}+\beta_{A}^{2^{n}}Y^{2^{n}}\mod\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G],$
$\displaystyle B^{2^{n}}$
$\displaystyle\equiv\alpha_{B}^{2^{n}}X^{2^{n}}+\beta_{B}^{2^{n}}Y^{2^{n}}\mod\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G].$

###### Proof.

We first observe that $\bm{\mathrm{Z}}(G)$ is generated by $x^{2}$, $y^{2}$
and $w$. Thus

$\mathbb{F}\bm{\mathrm{Z}}(G)\cap\Delta=\sum_{\begin{subarray}{c}r,s,t\geq
0\\\ 2r+2s+qt\geq
1\end{subarray}}\mathbb{F}X^{2r}Y^{2s}W^{t}\subseteq\mathbb{F}X^{2}+\mathbb{F}Y^{2}+\Delta^{4}.$

We will prove the claims only for $A$ as the arguments for $B$ are identical.
Recall that (4.6) and Lemma 4.12 imply

$A=\alpha_{A}X+\beta_{A}Y+\delta_{A}Z+\xi_{A}X^{2}+\eta_{A}Y^{2}+U_{A}.$

So, by $[X,Y]\equiv Z\mod\Delta^{3}$, it yields

(4.9)
$A^{2}\equiv\alpha_{A}^{2}X^{2}+\beta_{A}^{2}Y^{2}+\alpha_{A}\beta_{A}Z\mod\Delta^{3}.$

Using the decomposition
$\bm{\mathrm{Z}}(\mathbb{F}G)=\mathbb{F}\bm{\mathrm{Z}}(G)\oplus(\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G])$,
we get

(4.10) $A^{2}=\alpha_{A}^{2}X^{2}+\beta_{A}^{2}Y^{2}+R_{A}+S_{A}$

for some $R_{A}\in\mathbb{F}\bm{\mathrm{Z}}(G)\cap\Delta$ and
$S_{A}\in\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G]$.

Now $S_{A}$ can be written as a linear combination of class sums of non-
central elements. It follows from $G^{\prime}=\langle z\rangle$ that

$S_{A}\in\mathbb{F}Z+\Delta^{3}.$

Indeed, let $g=x^{i}y^{j}z^{k}$ be a representative of a non-central conjugacy
class of $G$ with $i,j,k\geq 0$. Since the conjugacy class of $g$ can be
written as

$\\{gz^{e_{1}},\dotsc,gz^{e_{t}}\\}$

for some non-trivial power $t$ of 2 and some $e_{1},\dotsc,e_{t}\geq 0$, the
class sum of the conjugacy class can be written as follows.

$\displaystyle gz^{e_{1}}+\dotsb+gz^{e_{t}}$
$\displaystyle=x^{i}y^{j}z^{k}(z^{e_{1}}+\dotsb+z^{e_{t}})$
$\displaystyle=(1+X)^{i}(1+Y)^{j}(1+Z)^{k}((1+Z)^{e_{1}}+\dotsb+(1+Z)^{e_{t}})$
$\displaystyle\equiv(1+X)^{i}(1+Y)^{j}(1+Z)^{k}(e_{1}+\dotsb+e_{t})Z$
$\displaystyle\equiv(e_{1}+\dotsb+e_{t})Z\mod\Delta^{3}.$

Hence $S_{A}$ belongs to the desired vector space.

Since $X^{2}+\Delta^{3}$, $Y^{2}+\Delta^{3}$, $Z+\Delta^{3}$ are linearly
independent in $\Delta^{2}/\Delta^{3}$, it follows, reformulating (4.9) and
(4.10) as

$R_{A}+S_{A}\equiv\alpha_{A}\beta_{A}Z\mod\Delta^{3},$

that $R_{A}\in\Delta^{4}$. Since $X^{2^{n+1}}=0$, $X^{2^{n}}Y^{2^{n}}=0$,
$Y^{2^{n+1}}=0$ and $W^{2}=0$ by the relations of $G$, we get
$R_{A}^{2^{n-1}}=0$. Raising (4.10) to a power hence yields

$A^{2^{n}}=\alpha_{A}^{2^{n}}X^{2^{n}}+\beta_{A}^{2^{n}}Y^{2^{n}}+S_{A}^{2^{n-1}},$

and the claim follows. ∎

The element $C=1+c$ will play an important role and we first approximate it
modulo $\Delta^{4}$.

######  Lemma 4.14.

The following congruence holds:

$C\equiv\lambda Z+\mu XZ+\nu YZ\mod\Delta^{4},$

where $\lambda=\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A}$,
$\mu=\lambda(1+\alpha_{A}+\alpha_{B})$ and
$\nu=\lambda(1+\beta_{A}+\beta_{B})$.

###### Proof.

From the definitions, we have

$C=1+c=1+b^{-1}a^{-1}ba=1+(1+B)^{-1}(1+A)^{-1}(1+B)(1+A).$

For $R\in\Delta$, one has $(1+R)^{-1}=1+R+R^{2}+\dotsb+R^{k}$ for some $k$
such that $R^{k}=0$. Hence

$\displaystyle C$ $\displaystyle\equiv
1+(1+B+B^{2}+B^{3})(1+A+A^{2}+A^{3})(1+B)(1+A)$
$\displaystyle\equiv[A,B]+A[A,B]+B[A,B]$
$\displaystyle=(1+A+B)[A,B]\mod\Delta^{4}$

by an explicit multiplication, with cancellation afterwards as we may in
characteristic $2$. Then, by $[X,Z],[Y,Z]\in\Delta^{4}$,
$U_{A},U_{B}\in\Delta^{3}$ and $\gamma_{A}=\gamma_{B}=0$ (i.e. Lemma 4.12),
from (4.6) we obtain

$C\equiv(1+\alpha_{A}X+\beta_{A}Y+\alpha_{B}X+\beta_{B}Y)[\alpha_{A}X+\beta_{A}Y,\alpha_{B}X+\beta_{B}Y]\mod\Delta^{4}.$

Since $[X,Y]=(1+X+Y+XY)Z$, straightforward calculations yield that

$\displaystyle C$
$\displaystyle\equiv(1+(\alpha_{A}+\alpha_{B})X+(\beta_{A}+\beta_{B})Y)(\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A})(Z+XZ+YZ)$
$\displaystyle\equiv(\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A})(Z+(1+\alpha_{A}+\alpha_{B})XZ+(1+\beta_{A}+\beta_{B})YZ)\mod\Delta^{4}.\qed$

We now define a variation of a power map that will turn out to be very useful
to us. First set
$\Gamma=\Delta(G^{\prime})\mathbb{F}G=[\mathbb{F}G,\mathbb{F}G]\mathbb{F}G$.
Note that $\Gamma=Z\mathbb{F}G$, and thus the subset

(4.11)
$\mathfrak{C}=\left\\{\,X^{r}Y^{s}Z^{t}W^{u}\mathrel{}\middle|\mathrel{}\begin{gathered}0\leq
r\leq 2^{n}-1,\ 0\leq s\leq 2^{m}-1,\\\ 0\leq t\leq 2^{\ell-1}-1,\ 0\leq u\leq
1,\\\ 2t+qu\geq 1\end{gathered}\,\right\\}$

of the basis of $\mathbb{F}G$ defined in Lemma 4.6 is a basis of $\Gamma$.

Recall that $q$ denotes the weight of $W$. Set $d=1+2^{\ell-1}+q$, and define
the $2^{\ell-1}$-power map

(4.12)
$\psi\colon\Gamma/\Gamma^{2}\to\Gamma^{2^{\ell-1}}/\left(\Gamma^{1+2^{\ell-1}}+\Delta^{d}\right).$

As will become clear, the reason for the definition of $d$ is that $W$,
$X^{2^{\ell-1}}W$ and $Y^{2^{\ell-1}}W$ are non-zero modulo
$\Gamma^{1+2^{\ell-1}}+\Delta^{d}$, while other elements in $\mathfrak{C}$
that span the image of $\psi$ are zero.

######  Lemma 4.15.

For $T\in\Delta^{2}$ and $\lambda,\mu,\nu\in\mathbb{F}$, we have

$\displaystyle\psi\left(\lambda Z+\mu XZ+\nu YZ+TZ+\Gamma^{2}\right)$
$\displaystyle\qquad=\lambda^{2^{\ell-1}}W+\mu^{2^{\ell-1}}X^{2^{\ell-1}}W+\nu^{2^{\ell-1}}Y^{2^{\ell-1}}W+\left(\Gamma^{1+2^{\ell-1}}+\Delta^{d}\right).$

###### Proof.

To prove the lemma, we factor $\psi$ as a composition of maps. Define the
square-map

$\psi_{i}\colon\Gamma^{2^{i}}/\Gamma^{1+2^{i}}\to\Gamma^{2^{i+1}}/\Gamma^{1+2^{i+1}}$

for $i\geq 0$ and the natural projection

$\pi\colon\Gamma^{2^{\ell-1}}/\Gamma^{1+2^{\ell-1}}\to\Gamma^{2^{\ell-1}}/\left(\Gamma^{1+2^{\ell-1}}+\Delta^{d}\right).$

Then $\psi=\pi\circ\psi_{\ell-2}\circ\dotsb\circ\psi_{1}\circ\psi_{0}$. Note
that this expression is well defined as $\ell\geq 2$.

Claim 1: For $g\in G$ and $i\geq 0$, one has
$[g,Z^{2^{i}}]\in\Gamma^{2^{i+1}}$.

It suffices to show that $1+g^{-1}Z^{2^{i}}g\equiv
1+Z^{2^{i}}\mod\Gamma^{2^{i+1}}$. From the general formula
$(1+R)^{-1}=1+R+R^{2}+\dotsb$ for $R\in\Delta$, we get

$1+Z^{-2^{i}}=(1+Z^{2^{i}})^{-1}=1+Z^{2^{i}}+(Z^{2^{i}})^{2}+\dotsb\equiv
1+Z^{2^{i}}\mod\Gamma^{2^{i+1}}.$

Note that $\\{z^{2^{i}},z^{-2^{i}}\\}$ is a conjugacy class of $G$. Thus
$1+g^{-1}Z^{2^{i}}g$ is equal to $1+Z^{2^{i}}$ or $1+Z^{-2^{i}}$ and the claim
holds.

Claim 2: For $R,S\in\mathbb{F}G$, one has

$\psi_{i}\left(RZ^{2^{i}}+SZ^{2^{i}}+\Gamma^{1+2^{i}}\right)=R^{2}Z^{2^{i+1}}+S^{2}Z^{2^{i+1}}+\Gamma^{1+2^{i+1}}.$

From the previous claim we know
$[R,Z^{2^{i}}],[S,Z^{2^{i}}]\in\Gamma^{2^{i+1}}$, so

$\displaystyle\psi_{i}$
$\displaystyle\left(RZ^{2^{i}}+SZ^{2^{i}}+\Gamma^{1+2^{i}}\right)$
$\displaystyle=(RZ^{2^{i}})^{2}+(SZ^{2^{i}})^{2}+[RZ^{2^{i}},SZ^{2^{i}}]+\Gamma^{1+2^{i+1}}$
$\displaystyle=R^{2}Z^{2^{i+1}}+S^{2}Z^{2^{i+1}}+[R,S]Z^{2^{i+1}}+\Gamma^{1+2^{i+1}}.$

As $[R,S]\in\Gamma$, the claim follows.

Now applying the last claim stepwise to $\psi_{0}$, $\psi_{1}$, …,
$\psi_{\ell-2}$ gives

$\displaystyle(\psi_{\ell-2}\circ\dotsb\circ\psi_{1}\circ\psi_{0})(\lambda
Z+\mu XZ+\nu YZ+TZ+\Gamma^{2})$
$\displaystyle\qquad=\lambda^{2^{\ell-1}}W+\mu^{2^{\ell-1}}X^{2^{\ell-1}}W+\nu^{2^{\ell-1}}Y^{2^{\ell-1}}W+T^{2^{\ell-1}}W+\Gamma^{1+2^{\ell-1}}.$

As $T^{2^{\ell-1}}\in\Delta^{2^{\ell}}$, this means
$T^{2^{\ell-1}}W\in\Delta^{d}$. Thus applying $\pi$ eliminates the last term,
and the lemma follows. ∎

####  4.3.2. Specific cases

Next we consider the coefficients $\alpha_{A}$ and $\alpha_{B}$. As the
calculations are no longer uniform for all groups, we introduce case
distinctions.

######  Lemma 4.16.

If $G\cong G_{\scriptsizedice{3}}$, $H\cong G_{\scriptsizedice{4}}$ and
$n>m>\ell$, then $\alpha_{A}=\alpha_{B}$.

If $G\cong G_{\scriptsizedice{5}}$ and $H\cong G_{\scriptsizedice{6}}$, then
we have the following.

  1. (1)

$m>\ell$ implies $\alpha_{A}=\alpha_{B}$.

  2. (2)

$n=m=\ell$ implies
$\alpha_{A}(\alpha_{A}+\beta_{A})=\alpha_{B}(\alpha_{B}+\beta_{B})$.

  3. (3)

$n=m<\ell$ implies $\alpha_{A}\beta_{A}=\alpha_{B}\beta_{B}=0$.

###### Proof.

Recall that $\varphi$ denotes the $2^{m}$-power map
$\Delta/\Delta^{2}\to\Delta^{2^{m}}/\Delta^{1+2^{m}}$. In this proof, the fact
that the Frobenius map on $\mathbb{F}$ is injective will be used without
further comment.

First consider $G\cong G_{\scriptsizedice{5}}$ and $H\cong
G_{\scriptsizedice{6}}$. By Lemma 4.7 and $Y^{2^{m}}=0$, we have

$\varphi(A+\Delta^{2})=\alpha_{A}^{2^{m}}X^{2^{m}}+(\alpha_{A}\beta_{A})^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}}$

and, if additionally $n=m$ holds,

$\varphi(A+\Delta^{2})=\alpha_{A}^{2^{m}}W+(\alpha_{A}\beta_{A})^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}}.$

Similar expressions hold for $\varphi(B+\Delta^{2})$. Note that the relation
$a^{2^{m}}=b^{2^{m}}$, which holds in $H$, implies
$\varphi(A+\Delta^{2})=\varphi(B+\Delta^{2})$.

Now, consider $m>\ell$. Then $Z^{2^{m-1}}=0$, and so
$\varphi(A+\Delta^{2})=\varphi(B+\Delta^{2})$ implies
$\alpha_{A}^{2^{m}}X^{2^{m}}+\Delta^{1+2^{m}}=\alpha_{B}^{2^{m}}X^{2^{m}}+\Delta^{1+2^{m}}$.
As $X^{2^{m}}\in\mathfrak{D}_{2^{m}}\smallsetminus\mathfrak{D}_{1+2^{m}}$, we
have $\alpha_{A}=\alpha_{B}$. Note that if additionally $n=m$ holds, then we
have $X^{2^{m}}=W$, but this does not change the conclusion, as then
$W\in\mathfrak{D}_{2^{m}}\smallsetminus\mathfrak{D}_{1+2^{m}}$.

Next, consider $n=m=\ell$. Then $\varphi(A+\Delta^{2})=\varphi(B+\Delta^{2})$
implies

$(\alpha_{A}^{2^{m}}+(\alpha_{A}\beta_{A})^{2^{m-1}})W+\Delta^{1+2^{m}}=(\alpha_{B}^{2^{m}}+(\alpha_{B}\beta_{B})^{2^{m-1}})W+\Delta^{1+2^{m}}.$

As $W\in\mathfrak{D}_{2^{m}}\smallsetminus\mathfrak{D}_{1+2^{m}}$, this
implies
$\alpha_{A}^{2}+\alpha_{A}\beta_{A}=\alpha_{B}^{2}+\alpha_{B}\beta_{B}$ and
hence the claim.

Finally, consider $n=m<\ell$. Then $W\in\Delta^{1+2^{m}}$ and
$C^{2^{\ell-1}}\in\Delta^{1+2^{m}}$. So because of
$a^{2^{m}}=b^{2^{m}}=c^{2^{\ell-1}}$, we have

$\varphi(A+\Delta^{2})=\varphi(B+\Delta^{2})=0+\Delta^{1+2^{m}}.$

As

$\displaystyle\varphi(A+\Delta^{2})$
$\displaystyle=(\alpha_{A}\beta_{A})^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}},$
$\displaystyle\varphi(B+\Delta^{2})$
$\displaystyle=(\alpha_{B}\beta_{B})^{2^{m-1}}Z^{2^{m-1}}+\Delta^{1+2^{m}},$

this together with
$Z^{2^{m-1}}\in\mathfrak{D}_{2^{m}}\smallsetminus\mathfrak{D}_{1+2^{m}}$
implies $\alpha_{A}\beta_{A}=\alpha_{B}\beta_{B}=0$.

Finally, consider $G\cong G_{\scriptsizedice{3}}$, $H\cong
G_{\scriptsizedice{4}}$ and $n>m>\ell$. Then by Lemma 4.7 and using
$Y^{2^{m}}=W$, $Z^{2^{\ell}}=0$ and $m>\ell$ we get from (4.6):

(4.13) $\displaystyle\varphi(A+\Delta^{2})$
$\displaystyle=\alpha_{A}^{2^{m}}X^{2^{m}}+\beta_{A}^{2^{m}}W+\Delta^{1+2^{m}},$
$\displaystyle\varphi(B+\Delta^{2})$
$\displaystyle=\alpha_{B}^{2^{m}}X^{2^{m}}+\beta_{B}^{2^{m}}W+\Delta^{1+2^{m}}.$

Now $H\cong G_{\scriptsizedice{4}}$ means $b^{2^{m}}=a^{2^{m}}c^{2^{\ell-1}}$,
so

(4.14) $\displaystyle\varphi(B+\Delta^{2})$
$\displaystyle=B^{2^{m}}+\Delta^{1+2^{m}}=1+a^{2^{m}}c^{2^{\ell-1}}+\Delta^{1+2^{m}}$
$\displaystyle=A^{2^{m}}+C^{2^{\ell-1}}+A^{2^{m}}C^{2^{\ell-1}}+\Delta^{1+2^{m}}.$

We next consider $A^{2^{m}}$ and $B^{2^{m}}$ modulo $\Gamma+\Delta^{1+2^{m}}$
which is done by “deleting the $Z$’s and $C$’s”, roughly speaking. Formally,
(4.13) and (4.14) imply

$\displaystyle\alpha_{A}^{2^{m}}X^{2^{m}}\equiv A^{2^{m}}\equiv
B^{2^{m}}\equiv\alpha_{B}^{2^{m}}X^{2^{m}}\mod\Gamma+\Delta^{1+2^{m}}.$

Now as $X^{2^{m}}$ does not lie in $\Gamma+\Delta^{1+2^{m}}$ by (4.11) and
(4.3), we get $\alpha_{A}=\alpha_{B}$. ∎

With these preparations, we are finally ready to answer the modular
isomorphism problem for all the remaining cases. In the first two cases,
$m\leq\ell$ holds, and we use the explicit approximations of $A^{2^{n}}$ and
$B^{2^{n}}$ given by Lemma 4.13. In the last two cases, $m>\ell$ holds, and we
use the explicit approximation of $C^{2^{\ell-1}}$ given by Lemmas 4.14 and
4.15.

######  Lemma 4.17.

If $n=m<\ell$, then
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$.

###### Proof.

Assume the notation (4.6), $n=m<\ell$, $G\cong G_{\scriptsizedice{5}}$ and
$H\cong G_{\scriptsizedice{6}}$. By the relations $X^{2^{n}}=W$ and
$Y^{2^{n}}=0$, we have

$A^{2^{n}}\equiv\alpha_{A}^{2^{n}}W\mod\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G]$

by Lemma 4.13. Similarly, we have

$B^{2^{n}}\equiv\alpha_{B}^{2^{n}}W\mod\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G].$

As in $H$ the relation $a^{2^{n}}=b^{2^{n}}$ holds by $n=m$, from the fact
that $W$ is a non-zero element in $\mathbb{F}\bm{\mathrm{Z}}(G)$ and the
decomposition
$\bm{\mathrm{Z}}(\mathbb{F}G)=\mathbb{F}\bm{\mathrm{Z}}(G)\oplus(\bm{\mathrm{Z}}(\mathbb{F}G)\cap[\mathbb{F}G,\mathbb{F}G])$,
we conclude that $\alpha_{A}=\alpha_{B}$.

As $n=m<\ell$, Lemma 4.16 implies $\alpha_{A}\beta_{A}=\alpha_{B}\beta_{B}=0$.
Hence, as $\alpha_{A}=\alpha_{B}$, we get
$\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A}=0$, which contradicts (4.7). ∎

The proof of the next case parallels that of the first case.

######  Lemma 4.18.

If $n=m=\ell$, then
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$.

###### Proof.

Assume the notation (4.6), $n=m=\ell$, $G\cong G_{\scriptsizedice{5}}$ and
$H\cong G_{\scriptsizedice{6}}$. We conclude that $\alpha_{A}=\alpha_{B}$ by
the same arguments as in the proof of Lemma 4.17.

As $n=m=\ell$, Lemma 4.16 and $\alpha_{A}=\alpha_{B}$ imply
$\alpha_{A}\beta_{A}=\alpha_{B}\beta_{B}$. Hence
$\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A}=0$, which contradicts (4.7). ∎

######  Lemma 4.19.

If $m>\ell$, then
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$.

###### Proof.

Assume the notation (4.6), $m>\ell$, $G\cong G_{\scriptsizedice{5}}$ and
$H\cong G_{\scriptsizedice{6}}$. As $n>\ell$ we obtain
$\mho_{n-1}(\bm{\mathrm{Z}}(\mathbb{F}G))=\mathbb{F}(\mho_{n-1}(\bm{\mathrm{Z}}(G)))$
by Proposition 4.10. This is generated by $W$ as an algebra since Lemma 4.1
shows that $\mho_{n-1}(\bm{\mathrm{Z}}(G))$ is generated by $w$. Thus
$\mho_{n-1}(\bm{\mathrm{Z}}(\mathbb{F}G))\cap\Gamma=\mathbb{F}W$. As $a^{2}$
is central and $c^{2^{\ell-1}}=a^{2^{n}}$ by the defining relations of $H$, we
see that $C^{2^{\ell-1}}=(A^{2})^{2^{n-1}}$ lies in
$\mho_{n-1}(\bm{\mathrm{Z}}(\mathbb{F}G))$. Hence
$C^{2^{\ell-1}}\in\mathbb{F}W$.

Recall that $\psi$ is defined as a $2^{\ell-1}$-power map on
$\Gamma/\Gamma^{2}$. Now by Lemmas 4.14 and 4.15 we know

$\psi(C+\Gamma^{2})=\lambda^{2^{\ell-1}}W+\mu^{2^{\ell-1}}X^{2^{\ell-1}}W+\nu^{2^{\ell-1}}Y^{2^{\ell-1}}W+\left(\Gamma^{1+2^{\ell-1}}+\Delta^{d}\right),$

where $\lambda=\alpha_{A}\beta_{B}+\alpha_{B}\beta_{A}$,
$\mu=\lambda(1+\alpha_{A}+\alpha_{B})$ and
$\nu=\lambda(1+\beta_{A}+\beta_{B})$.

As $X^{2^{\ell-1}}W$ does not lie in $\mathbb{F}W$ and neither in
$\Gamma^{1+2^{\ell-1}}+\Delta^{d}$ by (4.11) and (4.3), the coefficient of it
in the expression of $\psi(C+\Gamma^{2})$ is $0$ by
$C^{2^{\ell-1}}\in\mathbb{F}W$. This coefficient equals a power of $\mu$ and
by Lemma 4.16, this implies $\lambda=0$, which contradicts (4.7). ∎

The proof of the last case parallels that of the previous case.

######  Lemma 4.20.

If $n>m>\ell$, then
$\mathbb{F}G_{\scriptsizedice{3}}\not\cong\mathbb{F}G_{\scriptsizedice{4}}$.

###### Proof.

Assume the notation (4.6), $n>m>\ell$, $G\cong G_{\scriptsizedice{3}}$ and
$H\cong G_{\scriptsizedice{4}}$. As $m>\ell$ we obtain
$\mho_{m-1}(\bm{\mathrm{Z}}(\mathbb{F}G))=\mathbb{F}(\mho_{m-1}(\bm{\mathrm{Z}}(G)))$
by Proposition 4.10. This is generated by $X^{2^{m}}$ and $W$ as an algebra
since Lemma 4.1 shows that $\mho_{m-1}(\bm{\mathrm{Z}}(G))$ is generated by
$x^{2^{m}}$ and $w$. Recall that $d=1+2^{\ell-1}+q$. Now, the weight of
$X^{2^{m}}W$ is $2^{m}+q$ which is bigger than $d$ as $m>\ell$. Thus
$\mho_{m-1}(\bm{\mathrm{Z}}(\mathbb{F}G))\cap\Gamma\subseteq\mathbb{F}W+\Delta^{d}$.
As $a^{-2}b^{2}$ is central and
$(a^{-2}b^{2})^{2^{m-1}}=a^{-2^{m}}b^{2^{m}}=a^{-2^{m}}a^{2^{m}}c^{2^{\ell-1}}=c^{2^{\ell-1}}$
by the defining relations of $H$, we see that
$C^{2^{\ell-1}}=(1+a^{-2}b^{2})^{2^{m-1}}$ lies in
$\mho_{m-1}(\bm{\mathrm{Z}}(\mathbb{F}G))$. Hence
$C^{2^{\ell-1}}\in\mathbb{F}W+\Delta^{d}$.

The rest of the argument is essentially the same as at the end of the proof of
the previous lemma: by the previous paragraph we known that the coefficient of
$X^{2^{\ell-1}}W$ in the expression of $\psi(C+\Gamma^{2})$ is $0$. On the
other hand, by Lemmas 4.14 and 4.15, this coefficient equals a power of $\mu$.
By Lemma 4.16 we conclude that $\lambda=0$, which contradicts (4.7). ∎

Now we give a proof of Theorem A using the results obtained so far.

###### Proof of Theorem A.

It remains to show that the groups in 3.4 are the only counterexamples to the
modular isomorphism problem in our class.

First we show that $H$ is two-generated and the central quotient of $H$ is
dihedral. The minimal number of generators is a well-known invariant and a
proof can be found in [22, Lemma 14.2.7]. Note that the central quotient of
$H$ is dihedral if and only if $|H:\bm{\mathrm{Z}}(H)|\geq 8$ and
$|H:H^{\prime}\bm{\mathrm{Z}}(H)|=4$; thus it is also an invariant by a result
of Margolis, Sakurai and Stanojkovski [20, Corollary 2.8, Lemma 5.9]. Since
the order and the isomorphism type of the abelianization are invariant, the
groups have common parameters $n$, $m$ and $\ell$.

By Theorems 2.1, 3.1 and 4.9, it remains to show that
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$
when either $n=m\geq 2$ or $n>m>\ell$ and that
$\mathbb{F}G_{\scriptsizedice{3}}\not\cong G_{\scriptsizedice{4}}$ when
$n>m>\ell$. If $n=m\geq 2$, then we conclude that
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$
from Lemmas 4.17, 4.18 and 4.19. If $n>m>\ell$, then we conclude that
$\mathbb{F}G_{\scriptsizedice{5}}\not\cong\mathbb{F}G_{\scriptsizedice{6}}$
from Lemma 4.19, and
$\mathbb{F}G_{\scriptsizedice{3}}\not\cong\mathbb{F}G_{\scriptsizedice{4}}$
from Lemma 4.20. ∎

######  Remark 4.21.

The arguments involved in the group base approximation partly explain the
choice we make for the isomorphism in Theorem 3.1. There are of course many
more choices for this isomorphism, but Lemma 4.12 imposes some restrictions,
for example. A similar argument, as in the first case proved in Lemma 4.16,
can be used to show that $\alpha_{A}=\alpha_{B}$ needs to hold in the setting
of Theorem 3.1. Eventually the choice we make for the isomorphism seems to be
the easiest.

Moreover, the arguments in Section 4 suggest that the straightforward
calculations in the proof of Theorem 3.1 not only make this proof easy, but
also make it possible. Any relation in a group $H$ that requires to compute an
actual commutator in the group algebra of another group $G$ for some given
elements is not only difficult to verify by hand or even by computer; it
imposes conditions that are difficult to meet in the first place.

##  Appendix A Distinction of groups

In the class of two-generated finite $2$-groups with dihedral central
quotient, each group has a presentation of the form in Theorem 2.1. The goal
of this appendix is to establish by group-theoretical arguments when these
groups are non-isomorphic and to obtain the classification of groups within
this class. Let $n$, $m$ and $\ell$ denote positive integers. Recall that a
group is _homocyclic_ if it is isomorphic to a direct product of copies of a
cyclic group.

######  Theorem A.1.

The six groups $G_{\scriptsizedice{1}}$, …, $G_{\scriptsizedice{6}}$ are
pairwise non-isomorphic if they do not have homocyclic abelianizations (i.e.
$n>m$).

######  Theorem A.2.

The three groups $G_{\scriptsizedice{1}}$, $G_{\scriptsizedice{5}}$,
$G_{\scriptsizedice{6}}$ are pairwise non-isomorphic if they have homocyclic
abelianizations (i.e. $n=m$). Moreover, in this case
$G_{\scriptsizedice{1}}\cong G_{\scriptsizedice{2}}$ and
$G_{\scriptsizedice{3}}\cong G_{\scriptsizedice{4}}\cong
G_{\scriptsizedice{5}}$.

These theorems follow immediately from Theorems A and 3.2; nevertheless, we
will provide a direct group-theoretical proof. We do this by considering the
centers and maximal quotients of the groups. We will write

$G_{\scriptsizedice{1}}(n,m,\ell),\dotsc,G_{\scriptsizedice{6}}(n,m,\ell)$

to make the parameters $n$, $m$ and $\ell$ explicit when necessary.

###  A.1. Maximal quotients

To describe the maximal quotients, we first describe the socle in each case,
as the central involutions are exactly the generators of minimal normal
subgroups. By the socle of a finite $p$-group, we mean the subgroup generated
by central elements of order $p$. From Lemma 4.1, we directly get the socles
in all cases.

######  Lemma A.3.

Assume $n\geq 2$. Then the following holds.

$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{1}}))$
$\displaystyle=\begin{cases}\langle x^{2^{n-1}}\rangle\times\langle
y^{2^{m-1}}\rangle\times\langle z^{2^{\ell-1}}\rangle\cong C_{2}\times
C_{2}\times C_{2}&(m\geq 2)\\\ \langle x^{2^{n-1}}\rangle\times\langle
z^{2^{\ell-1}}\rangle\phantom{\langle y^{2^{m-1}}\rangle\times{}}\cong
C_{2}\times C_{2}&(m=1);\end{cases}$
$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{2}}))$
$\displaystyle=\begin{cases}\langle x^{2^{n-1}}\rangle\times\langle
x^{-2^{m-1}}y^{2^{m-1}}\rangle\times\langle z^{2^{\ell-1}}\rangle\cong
C_{2}\times C_{2}\times C_{2}&(m\geq 2)\\\ \langle
x^{2^{n-1}}\rangle\times\langle z^{2^{\ell-1}}\rangle\phantom{\langle
x^{-2^{m-1}}y^{2^{m-1}}\rangle\times{}}\cong C_{2}\times
C_{2}&(m=1);\end{cases}$
$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{3}}))$
$\displaystyle=\langle x^{2^{n-1}}\rangle\times\langle
z^{2^{\ell-1}}\rangle\cong C_{2}\times C_{2};$
$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{4}}))$
$\displaystyle=\langle x^{2^{n-1}}\rangle\times\langle
z^{2^{\ell-1}}\rangle\cong C_{2}\times C_{2};$
$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{5}}))$
$\displaystyle=\begin{cases}\langle y^{2^{m-1}}\rangle\times\langle
z^{2^{\ell-1}}\rangle\cong C_{2}\times C_{2}&(m\geq 2)\\\ \langle
z^{2^{\ell-1}}\rangle\phantom{\langle y^{2^{m-1}}\rangle\times{}}\cong
C_{2}&(m=1);\end{cases}$
$\displaystyle\Omega(\bm{\mathrm{Z}}(G_{\scriptsizedice{6}}))$
$\displaystyle=\begin{cases}\langle
x^{-2^{m-1}}y^{2^{m-1}}\rangle\times\langle z^{2^{\ell-1}}\rangle\cong
C_{2}\times C_{2}&(m\geq 2)\\\ \langle z^{2^{\ell-1}}\rangle\phantom{\langle
x^{-2^{m-1}}y^{2^{m-1}}\rangle\times{}}\cong C_{2}&(m=1).\end{cases}$

The next lemma describes all the maximal quotients of the six groups in
Theorem 2.1.

######  Lemma A.4.

Assume $n\geq 2$. Let $G$ be one of the groups
$G_{\scriptsizedice{1}}(n,m,\ell)$, …, $G_{\scriptsizedice{6}}(n,m,\ell)$ and
$Q$ a maximal quotient of $G$. Then $Q\cong
G_{\scriptsizetextdice{\textnormal{?}}}(n^{\prime},m^{\prime},\ell^{\prime})$
for some $\textdice{\textnormal{?}}\in\\{\dice{1},\dotsc,\dice{6}\\}$ and
$n^{\prime},m^{\prime},\ell^{\prime}$ satisfying
$n^{\prime}+m^{\prime}+\ell^{\prime}=n+m+\ell-1$; we need to consider five
cases

  1. (1)

$n>m+1$ and $m\geq 2$ (Table 7)

  2. (2)

$n=m+1$ and $m\geq 2$ (Table 7)

  3. (3)

$n=m$ and $m\geq 2$ (Table 7)

  4. (4)

$n>m+1$ and $m=1$ (Table 7)

  5. (5)

$n=m+1$ and $m=1$ (Table 7)

for all the possible values of the parameters for $Q$, and those are
summarized in Tables 7, 7, 7, 7 and 7.

| $(n-1,m,\ell)$ | $(n,m-1,\ell)$ | $(n,m,\ell-1)$
---|---|---|---
$G_{\scriptsizedice{1}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{5}}$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{3}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{2}}(n,m,\ell)$ |  $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{6}}$ |  $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{4}}$ | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{3}}(n,m,\ell)$ |  $G_{\scriptsizedice{3}}$ $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{4}}(n,m,\ell)$ |  $G_{\scriptsizedice{4}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{5}}(n,m,\ell)$ |  | $G_{\scriptsizedice{5}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{6}}(n,m,\ell)$ |  | $G_{\scriptsizedice{6}}$ | $G_{\scriptsizedice{2}}$

Table 3. Maximal quotients when $n>m+1$, $m\geq 2$.

| $(n-1,m,\ell)$ | $(n,m-1,\ell)$ | $(n,m,\ell-1)$
---|---|---|---
$G_{\scriptsizedice{1}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{5}}$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{3}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{2}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{6}}$ |  $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{4}}$ | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{3}}(n,m,\ell)$ |  $G_{\scriptsizedice{5}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{4}}(n,m,\ell)$ | $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{5}}(n,m,\ell)$ |  | $G_{\scriptsizedice{5}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{6}}(n,m,\ell)$ |  | $G_{\scriptsizedice{6}}$ | $G_{\scriptsizedice{2}}$

Table 4. Maximal quotients when $n=m+1$, $m\geq 2$.

| $(n-1,m,\ell)$ | $(n,m-1,\ell)$ | $(n,m,\ell-1)$
---|---|---|---
$G_{\scriptsizedice{1}}(n,m,\ell)$ |  |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{3}}$ $G_{\scriptsizedice{4}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{2}}(n,m,\ell)$ |  |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{3}}$ $G_{\scriptsizedice{4}}$ | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{3}}(n,m,\ell)$ |  | $G_{\scriptsizedice{5}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{4}}(n,m,\ell)$ |  | $G_{\scriptsizedice{5}}$ | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{5}}(n,m,\ell)$ |  | $G_{\scriptsizedice{5}}$ | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{6}}(n,m,\ell)$ |  | $G_{\scriptsizedice{6}}$ | $G_{\scriptsizedice{2}}$

Table 5. Maximal quotients when $n=m$, $m\geq 2$.

| $(n-1,m,\ell)$ | $(n,m-1,\ell)$ | $(n,m,\ell-1)$
---|---|---|---
$G_{\scriptsizedice{1}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{2}}(n,m,\ell)$ |  $G_{\scriptsizedice{2}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{3}}(n,m,\ell)$ |  $G_{\scriptsizedice{3}}$ $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{4}}(n,m,\ell)$ |  $G_{\scriptsizedice{4}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{5}}(n,m,\ell)$ |  |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{6}}(n,m,\ell)$ |  |  | $G_{\scriptsizedice{2}}$

Table 6. Maximal quotients when $n>m+1$, $m=1$.

| $(n-1,m,\ell)$ | $(n,m-1,\ell)$ | $(n,m,\ell-1)$
---|---|---|---
$G_{\scriptsizedice{1}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{2}}(n,m,\ell)$ |  $G_{\scriptsizedice{1}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{3}}(n,m,\ell)$ |  $G_{\scriptsizedice{5}}$ $G_{\scriptsizedice{6}}$ |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{4}}(n,m,\ell)$ | $G_{\scriptsizedice{5}}$ |  | $G_{\scriptsizedice{2}}$
$G_{\scriptsizedice{5}}(n,m,\ell)$ |  |  | $G_{\scriptsizedice{1}}$
$G_{\scriptsizedice{6}}(n,m,\ell)$ |  |  | $G_{\scriptsizedice{2}}$

Table 7. Maximal quotients when $n=m+1$, $m=1$.

###### Proof.

In all cases, the proof is obtained by applying Lemma 2.4 to each maximal
quotient in each case, which corresponds to a minimal normal subgroup
generated by a non-trivial element of the socle in Lemma A.3. ∎

We finish this subsection with an illustration of the studied groups by a
graph, which can be derived from Lemma A.4 in Figure 1.

Figure 1.  Finite $2$-groups with dihedral central quotient whose
abelianization is isomorphic to $C_{2^{n}}\times C_{2^{m}}$ and not
homocyclic. An arrow from a group $G$ to $H$ is drawn if $H$ is isomorphic to
a maximal quotient of $G$, that is, $G$ is an immediate descendant of $H$ .

###  A.2. Inductive argument

We will now proceed to prove Theorems A.1 and A.2. This will be achieved by a
double induction. The outer induction runs on the first parameter, i.e. $n$,
and the inner on the second parameter, i.e. $m$. We will need the following
two cases for the base of the inductions.

Recall the standard presentations of finite $2$-groups of maximal class: the
dihedral, semidihedral and generalized quaternion groups.

$\displaystyle D_{2^{n+1}}=\langle\,a,b\mid a^{2}$ $\displaystyle=1,$
$\displaystyle\quad b^{2^{n}}$ $\displaystyle=1,$ $\displaystyle\quad b^{a}$
$\displaystyle=b^{-1}\,\rangle;$ $\displaystyle S_{2^{n+1}}=\langle\,a,b\mid
a^{2}$ $\displaystyle=1,$ $\displaystyle\quad b^{2^{n}}$ $\displaystyle=1,$
$\displaystyle\quad b^{a}$ $\displaystyle=b^{2^{n-1}-1}\,\rangle;$
$\displaystyle Q_{2^{n+1}}=\langle\,a,b\mid a^{2}$
$\displaystyle=b^{2^{n-1}},$ $\displaystyle\quad b^{2^{n}}$ $\displaystyle=1,$
$\displaystyle\quad b^{a}$ $\displaystyle=b^{-1}\,\rangle.$

######  Proposition A.5.

Assume $\ell\geq 2$. Then

$\displaystyle D_{2^{\ell+2}}\cong G_{\scriptsizedice{1}}(1,1,\ell);\quad
S_{2^{\ell+2}}\cong G_{\scriptsizedice{5}}(1,1,\ell);\quad Q_{2^{\ell+2}}\cong
G_{\scriptsizedice{6}}(1,1,\ell).$

In particular, $G_{\scriptsizedice{1}}(1,1,\ell)$,
$G_{\scriptsizedice{5}}(1,1,\ell)$ and $G_{\scriptsizedice{6}}(1,1,\ell)$ are
pairwise non-isomorphic.

###### Proof.

Since the three cases have the same kind of isomorphism $a\mapsto y$ and
$b\mapsto xy$, we only prove that $Q_{2^{\ell+2}}\cong
G_{\scriptsizedice{6}}(1,1,\ell)$ here.

Let $G_{\scriptsizedice{6}}(1,1,\ell)=\langle x,y,z\rangle$ and define $a=y$
and $b=xy$. Since

$\displaystyle b^{2}$ $\displaystyle=xyxy=x^{2}y[y,x]y=x^{2}yzy$
$\displaystyle=z^{2^{\ell-1}}yzy=z^{2^{\ell-1}}y^{2}z[z,y]=z^{2^{\ell}}z[z,y]=z^{-1},$

we have, using that $y^{2}$ is central in the last equation,

$\displaystyle a^{2}$
$\displaystyle=y^{2}=z^{2^{\ell-1}}=z^{-2^{\ell-1}}=b^{2^{\ell}};$
$\displaystyle b^{2^{\ell+1}}$ $\displaystyle=z^{-2^{\ell}}=1;$ $\displaystyle
b^{a}$ $\displaystyle=(xy)^{y}=y^{-1}xy^{2}=yx=xy[y,x]=xyz=bb^{-2}=b^{-1}.$

Hence there is an epimorphism from $Q_{2^{\ell+2}}$ to
$G_{\scriptsizedice{6}}(1,1,\ell)$. Since both groups have the same order,
they must be isomorphic. ∎

######  Lemma A.6.

Assume $n\geq 2$. Then $G_{\scriptsizedice{1}}(n,1,1)\not\cong
G_{\scriptsizedice{2}}(n,1,1)$.

###### Proof.

Observe that $G_{\scriptsizedice{1}}(n,1,1)$ has a maximal subgroup that is
isomorphic to $C_{2^{n-1}}\times C_{2}\times C_{2}$, namely $\langle
x^{2},y,z\rangle$, while $G_{\scriptsizedice{2}}(n,1,1)$ does not. Indeed, we
have $\operatorname{\Phi}(G_{\scriptsizedice{2}}(n,1,1))=\langle
x^{2},z\rangle$, so that the maximal subgroups of
$G_{\scriptsizedice{2}}(n,1,1)$ are $\langle x,z\rangle$, $\langle y,z\rangle$
and $\langle xy,x^{2}\rangle$. Note here that $(xy)^{2}=x^{2}y^{2}z=x^{4}z$,
so that $z\in\langle xy,x^{2}\rangle$. Hence all the maximal subgroups of
$G_{\scriptsizedice{2}}(n,1,1)$ are two-generated. ∎

We are ready for the inductions.

###### Proof of Theorems A.1 and A.2.

We first note that when $n=m$, then the isomorphisms
$G_{\scriptsizedice{1}}(n,m,\ell)\cong G_{\scriptsizedice{2}}(n,m,\ell)$ and
$G_{\scriptsizedice{3}}(n,m,\ell)\cong G_{\scriptsizedice{4}}(n,m,\ell)\cong
G_{\scriptsizedice{5}}(n,m,\ell)$ are clear from the presentations.

We proceed by induction on $n$. If $n=1$, then also $m=1$ and the groups
$G_{\scriptsizedice{1}}(n,m,\ell)$, $G_{\scriptsizedice{5}}(n,m,\ell)$ and
$G_{\scriptsizedice{6}}(n,m,\ell)$ are pairwise non-isomorphic by Proposition
A.5.

So assume that $n\geq 2$ and that Theorems A.1 and A.2 hold for smaller values
of $n$. We proceed by induction on $m$. When $m=1$, then by Lemma 4.1 the
centers of $G_{\scriptsizedice{1}}(n,m,\ell)$,
$G_{\scriptsizedice{2}}(n,m,\ell)$, $G_{\scriptsizedice{3}}(n,m,\ell)$ and
$G_{\scriptsizedice{4}}(n,m,\ell)$ are not isomorphic to the centers of
$G_{\scriptsizedice{5}}(n,m,\ell)$ and $G_{\scriptsizedice{6}}(n,m,\ell)$. By
induction and Tables 7 and 7, only $G_{\scriptsizedice{1}}(n,m,\ell)$ or
$G_{\scriptsizedice{2}}(n,m,\ell)$ can map onto one of the groups
$G_{\scriptsizedice{1}}(n-1,m,\ell)$ or $G_{\scriptsizedice{2}}(n-1,m,\ell)$,
so these two groups are not isomorphic to either of
$G_{\scriptsizedice{3}}(n,m,\ell)$ or $G_{\scriptsizedice{4}}(n,m,\ell)$. To
see that $G_{\scriptsizedice{1}}(n,m,\ell)$ is not isomorphic to
$G_{\scriptsizedice{2}}(n,m,\ell)$ note that, again by induction and Tables 7
and 7, $G_{\scriptsizedice{2}}(n,m,\ell)$ maps onto
$G_{\scriptsizedice{6}}(n-1,m,\ell)$, while $G_{\scriptsizedice{1}}(n,m,\ell)$
does not. To see that $G_{\scriptsizedice{3}}(n,m,\ell)$ is not isomorphic to
$G_{\scriptsizedice{4}}(n,m,\ell)$ note that, by Tables 7 and 7, exactly one
of them maps onto $G_{\scriptsizedice{6}}(n-1,m,\ell)$ by induction. To
distinguish between $G_{\scriptsizedice{5}}(n,m,\ell)$ and
$G_{\scriptsizedice{6}}(n,m,\ell)$ we observe that by Tables 7 and 7 the only
maximal quotient of $G_{\scriptsizedice{5}}(n,m,\ell)$ is
$G_{\scriptsizedice{1}}(n,m,\ell-1)$, while the only maximal quotient of
$G_{\scriptsizedice{6}}(n,m,\ell)$ is $G_{\scriptsizedice{2}}(n,m,\ell-1)$. If
$\ell=2$, then these quotients are not isomorphic and hence neither are the
groups $G_{\scriptsizedice{5}}(n,m,\ell)$ and
$G_{\scriptsizedice{6}}(n,m,\ell)$ by Lemma A.6. If $\ell>2$, then the
quotients $G_{\scriptsizedice{1}}(n,m,\ell-1)$ and
$G_{\scriptsizedice{2}}(n,m,\ell-1)$ are also not isomorphic, as we showed
earlier in this paragraph. This finishes the case $m=1$.

Consider next $n>m\geq 2$ and assume that Theorems A.1 and A.2 hold for
smaller values of $n$ or $m$. Looking on the centers of the groups in Lemma
4.1, we see that we only need to show that
$G_{\scriptsizedice{1}}(n,m,\ell)\not\cong G_{\scriptsizedice{2}}(n,m,\ell)$,
$G_{\scriptsizedice{3}}(n,m,\ell)\not\cong G_{\scriptsizedice{4}}(n,m,\ell)$
and $G_{\scriptsizedice{5}}(n,m,\ell)\not\cong
G_{\scriptsizedice{6}}(n,m,\ell)$. To distinguish between
$G_{\scriptsizedice{1}}(n,m,\ell)$ and $G_{\scriptsizedice{2}}(n,m,\ell)$ note
that by induction and Tables 7 and 7 the group
$G_{\scriptsizedice{1}}(n,m,\ell)$ maps onto
$G_{\scriptsizedice{1}}(n,m-1,\ell)$, while $G_{\scriptsizedice{2}}(n,m,\ell)$
does not. To see that $G_{\scriptsizedice{3}}(n,m,\ell)\not\cong
G_{\scriptsizedice{4}}(n,m,\ell)$ we note that, again by induction and Tables
7 and 7, exactly one of them maps onto $G_{\scriptsizedice{6}}(n-1,m,\ell)$.
Finally, to observe that $G_{\scriptsizedice{5}}(n,m,\ell)$ and
$G_{\scriptsizedice{6}}(n,m,\ell)$ are not isomorphic we also use induction
and Tables 7 and 7 and see that $G_{\scriptsizedice{5}}(n,m,\ell)$ maps onto
$G_{\scriptsizedice{5}}(n,m-1,\ell)$, while $G_{\scriptsizedice{6}}(n,m,\ell)$
does not.

The last case to consider is $n=m$. Recall that then
$G_{\scriptsizedice{1}}(n,m,\ell)\cong G_{\scriptsizedice{2}}(n,m,\ell)$ and
$G_{\scriptsizedice{3}}(n,m,\ell)\cong G_{\scriptsizedice{4}}(n,m,\ell)\cong
G_{\scriptsizedice{5}}(n,m,\ell)$. So after looking at the centers in Lemma
4.1, it only remains to show that $G_{\scriptsizedice{5}}(n,m,\ell)$ and
$G_{\scriptsizedice{6}}(n,m,\ell)$ are not isomorphic. By induction and Table
7, we observe that $G_{\scriptsizedice{6}}(n,m,\ell)$ maps onto
$G_{\scriptsizedice{6}}(n,m-1,\ell)$ which is not the case for
$G_{\scriptsizedice{5}}(n,m,\ell)$. This finishes the inner induction and
hence also the induction step of the outer induction. ∎

Acknowledgments: This work has been supported by the Madrid Government
(Comunidad de Madrid - Spain) under the multiannual Agreement with UAM in the
line for the Excellence of the University Research Staff in the context of the
V PRICIT (Regional Programme of Research and Technological Innovation). The
authors acknowledges financial support from the Spanish Ministry of Science
and Innovation, through the Severo Ochoa Programme for Centers of Excellence
in R& D (CEX2019-000904-S) and through the Ramon y Cajal grant program.

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