# MULTI-EGS-GROUPS: EXPONENT OF CONGRUENCE QUOTIENTS

Elena Maini  Department of Mathematics, University of Padova, 35121 Padova,
Italy<EMAIL_ADDRESS>
###### Abstract.

Given a multi-EGS-group $K$ acting on the $p$-adic rooted tree, where $p$ is
any prime number, we compute the exponent of the congruence quotient
$K_{n}=K/\operatorname{St}_{K}(n)$ for all $n\geq 1$. The formula that we
obtain for $\exp(K_{n})$ only depends on $p$, $n$ and the periodicity of $K$.

##  1\. Introduction

From the early 20th century, the so-called Burnside Problem has been an object
of great interest, being formulated in different versions which led to various
and elaborate solutions. One of its versions, known as the General Burnside
Problem, asks whether a finitely generated periodic group has to be finite.
Since the 1960s, several examples of infinite finitely generated periodic
groups have been found: among them, some of the easiest turned out to be
groups acting on regular rooted trees. As an evidence of this, we mention the
Grigorchuk groups (defined in [3]) and the Gupta-Sidki group (defined in [4]).
The Gupta-Sidki group belongs to a broader family of subgroups of the group of
automorphisms of the $p$-adic rooted tree $\mathcal{T}$ (where $p$ is a prime
number), which will be the object of our paper: the family of the so-called
multi-EGS-groups.

In order to define this family of groups, we start by recalling some basic
notions from the theory of automorphisms of trees. We use the following
notation: the vertices of $\mathcal{T}$ are the elements of the free monoid
generated by $X=\\{1,\dots,p\\}$, that is, the words of finite length in the
alphabet $X$ (the word of length zero, denoted by $\emptyset$, is said to be
the root of $\mathcal{T}$ and is the only vertex that has no parent). For
$n\in\mathbb{N}$, we denote with $X^{n}$ the set of all the words of length
$n$; the finite tree whose vertices are the words of length $\leq n$ is
$\mathcal{T}_{n}$ and
$\operatorname{Aut}\mathcal{T}_{n}\simeq\operatorname{Aut}\mathcal{T}/\operatorname{St}(n)$,
where $\operatorname{St}(n)$ is the subgroup of
$\operatorname{Aut}\mathcal{T}$ consisting of all the automorphisms that do
not move the elements of $X^{n}$. If $G$ is a subgroup of
$\operatorname{Aut}\mathcal{T}$, we put
$\operatorname{St}_{G}(n)=\operatorname{St}(n)\cap G$ and we call
$G_{n}=G/\operatorname{St}_{G}(n)$ the nth congruence quotient of $G$. The
sequence $\\{\operatorname{St}_{G}(n)\\}_{n\geq 0}$ is a chain of finite-index
normal subgroups of $G$, which makes $G$ into a residually finite group and
gives rise to the sequence of finite groups $\\{G_{n}\\}_{n\geq 0}$.
Understanding properties of $G_{n}$ as $n$ varies is a typical way to
understand the structure of the whole $G$. One can easily prove, for example,
that

(1) $\exp(G)=\lim_{n\rightarrow\infty}\exp(G_{n}).$

It is clear that, if $H\leq G\leq\operatorname{Aut}\mathcal{T}$, $H_{n}$ can
be naturally embedded in $G_{n}$. We will write $\operatorname{St}_{G_{n}}(k)$
for the quotient $\operatorname{St}_{G}(k)/\operatorname{St}_{G}(n)$, whenever
$k$ is an integer less than or equal to $n$.

Now, for any $g\in\operatorname{Aut}\mathcal{T}$ and for any vertex $u$, we
denote with $g_{(u)}$ the permutation of $X$ which satisfies
$g(ux)=g(u)g_{(u)}(x)$ for any $x\in X$. We refer to $g_{(u)}$ as the label of
$g$ at $u$. The set of all labels of $g$ is the portrait of $g$ and the
automorphism $g$ is completely determined by its portrait. Moreover, for any
vertex $u$, we write $g_{u}$ for the automorphism of
$\operatorname{Aut}\mathcal{T}$ that has portrait given by
${(g_{u})}_{(v)}=g_{(uv)}$ for any vertex $v$. We call $g_{u}$ the section of
$g$ at $u$ and we say that a subgroup $G$ of $\operatorname{Aut}\mathcal{T}$
is self-similar if, for every $g\in G$, all the sections of $g$ belong to $G$.
It is well known that the map

$\displaystyle\psi\text{ }:\text{ }$ $\displaystyle\operatorname{St}(1)\text{
}\longrightarrow\text{
}\operatorname{Aut}\mathcal{T}\times\overset{p}{\cdots}\times\operatorname{Aut}\mathcal{T}$
$\displaystyle\hskip 14.22636ptg\hskip 17.07182pt\longmapsto\hskip
19.91684pt(g_{1},\dots,g_{p})$

is a group isomorphism and clearly, if $G$ is self-similar,
$\psi(\operatorname{St}_{G}(1))\subseteq G\times\overset{p}{\cdots}\times G$.
From now onwards, we will denote by the letter $a$ the automorphism of
$\mathcal{T}$ with portrait

$a_{(u)}=\begin{cases}(1\hskip 2.84544pt2\hskip 2.84544pt\dots\hskip
2.84544ptp)\hskip 28.45274pt&\text{ if }u\text{ is the root of }\mathcal{T}\\\
1\hskip 28.45274pt&\text{ otherwise.}\end{cases}$

Notice that $a$ has order $p$ and then we will be allowed to write $a^{e}$
whenever $e$ is an element in $\mathbb{F}_{p}$. It follows from an easy
computation that for any $g\in\operatorname{St}(1)$ with
$\psi(g)=(g_{1},\dots,g_{p-1},g_{p})$ we have

(2) $\psi(g^{a})=(g_{p},g_{1},\dots,g_{p-1})$

and then

(3)
$\displaystyle\begin{split}\psi((ag)^{p})&=\psi(g^{(a^{p-1})})\psi(g^{(a^{p-2})})\dots\psi(g^{(a)})\psi(g)\\\
&=(\hskip 2.84544pt(g_{2}\dots g_{p-1}g_{p}g_{1})\hskip 2.84544pt,\hskip
2.84544pt(g_{3}\dots g_{p}g_{1}g_{2})\hskip 2.84544pt,\hskip
2.84544pt\dots\hskip 2.84544pt,\hskip 2.84544pt(g_{1}\dots
g_{p-2}g_{p-1}g_{p})\hskip 2.84544pt).\end{split}$

The isomorphism $\psi$, besides being typically used to understand the
structure of groups acting on $\mathcal{T}$, is a useful tool to define
recursively automorphisms. As an evidence of this, we give the following
definition. For any vector
$\mathbf{e}=(e_{1},\dots,e_{p-1})\in(\mathbb{F}_{p})^{p-1}$ and for any $j\in
X=\\{1,\dots,p\\}$, we refer to the automorphism $b\in\operatorname{St}(1)$
defined by

$\psi(b)=(a^{e_{p-j+1}},\dots,a^{e_{p-1}},b,a^{e_{1}},\dots,a^{e_{p-j}})$

as the directed automorphism along the path $P_{j}=(\emptyset,j,jj,\dots)$
with defining vector $\mathbf{e}$. The use of this term is justified by the
fact that the label of $b$ at a vertex $u$ can be non trivial only when the
parent of $u$ lies in $P_{j}$. We are then ready to say what we mean with the
term multi-EGS-group.

######  Definition 1.

Let $E^{(1)},\dots,E^{(p)}$ be vector subspaces of $(\mathbb{F}_{p})^{p-1}$
not all equal to the null subspace. For every $j\in\\{1,\dots,p\\}$ we denote
with $B^{(j)}$ the set of all directed automorphisms along $P_{j}$ with
defining vector in $E^{(j)}$. Then we say that $K=\langle
a,B^{(1)},\dots,B^{(p)}\rangle$ is the multi-EGS-group associated to the tuple
$E=(E^{(1)},\dots,E^{(p)})$.

We observe that, with the notation of the above definition, for every
$j\in\\{1,\dots,p\\}$ there is a natural isomorphism from $E^{(j)}$ to
$B^{(j)}$, which makes $B^{(j)}$ into an elementary abelian $p$-subgroup of
$\operatorname{Aut}\mathcal{T}$. Such isomorphism is the map sending a vector
$\mathbf{e}\in E^{(j)}$ to the directed automorphism (along $P_{j}$) defined
by $\mathbf{e}$. Let us remark, in addition, that

(4) $K=\langle a,\mathcal{B}^{(1)},\dots,\mathcal{B}^{(p)}\rangle$

whenever $\mathcal{E}^{(j)}$ is a generating set for $E^{(j)}$ and
$\mathcal{B}^{(j)}$ is the set of all directed automorphisms along $P_{j}$
with defining vector in $\mathcal{E}^{(j)}$. Every multi-EGS-group is then
finitely generated.

Some subfamilies of multi-EGS-groups have played an important role in modern
group theory and deserve a different name.

######  Definition 2.

Let $K$ a multi-EGS-group as in Definition 1.

If $E^{(1)},\dots,E^{(p-1)}$ are all equal to the null subspace, we say that
$K$ is the multi-GGS-group with defining space $E^{(p)}$.

If $E^{(1)},\dots,E^{(p-1)}$ are all equal to the null subspace and $E^{(p)}$
has dimension $1$, we say that $K$ is the GGS-group with defining space
$E^{(p)}$. In this case, taken any non zero vector $\mathbf{e}\in E^{(p)}$,
$K$ is also said to be the GGS-group with defining vector $\mathbf{e}$ (this
is consistent thanks to (4)).

The aforementioned Gupta-Sidki group is the GGS-group obtained by choosing
$p\geq 3$ and $\mathbf{e}=(1,-1,0,\dots,0)$. Theorem 1 of [6] ensures that
every GGS-group is infinite. With the same argument, one can prove that
$\langle a,b\rangle$ is infinite whenever $b$ is a directed automorphism with
non-zero defining vector. It readily follows that every multi-EGS-group is
infinite.

Another version of the problem raised by Burnside, known as the Burnside
Problem, asks whether a finitely generated group with finite exponent has to
be finite. The answer is still negative, but examples of infinite finitely
generated groups with finite exponent cannot be found among the subgroups of
$\operatorname{Aut}\mathcal{T}$. To motivate this, we look at a third version
of the Burnside Problem, which goes by the name of Restricted Burnside Problem
and reads: given a finite group with exponent $m$ and $d$ generators, is it
possible to find a bound for the order of such group which only depends on $m$
and $d$? In the early 1990s, Zelmanov showed, in two remarkable papers ([7]
and [8]), that the answer to the above question is affirmative. As a
consequence, every finitely generated residually finite group with finite
exponent has to be finite. Hence every infinite finitely generated subgroup
$G$ of $\operatorname{Aut}\mathcal{T}$ (and in particular every multi-EGS-
group) must have infinite exponent and, by (1), the sequence $\exp(G_{n})$
goes to infinity when $n$ goes to infinity. The aim of this paper will be to
determine how fast $\exp(G_{n})$ goes to infinity when $G$ is a multi-EGS-
group. This goal will be achieved in Theorem A, which is our main result.

Since any multi-EGS-group is infinite and finitely generated, the multi-EGS-
groups giving a negative answer to the General Burnside Problem are exactly
the periodic ones. Moreover, by Lemma 3.13 in [5], we have that a multi-EGS-
group $K$ is periodic if and only if, with the notation of Definition 1, the
subspaces ${E^{(1)}},\dots,{E^{(p)}}$ are all contained in the hyperplane of
$(\mathbb{F}_{p})^{p-1}$ defined by
$V=\\{(f_{1},\dots,f_{p-1})\in(\mathbb{F}_{p})^{p-1}:f_{1}+\dots+f_{p-1}=0\\}$.
Given a multi-EGS-group $K$, Theorem A provides an explicit formula for
$\exp(K_{n})$, which only depends on the periodicity of $K$.

######  Theorem A.

Let $K$ be a multi-EGS-group. Then for every $n\geq 1$

$\exp(K_{n})=\begin{cases}p^{\lfloor\frac{n+1}{2}\rfloor}&\text{ if }K\text{
is periodic}\\\ p^{n}&\text{ otherwise}\end{cases}$

where $\lfloor\frac{n+1}{2}\rfloor$ is the greatest integer number less than
or equal to $\frac{n+1}{2}$.

We observe that, given $K$ any multi-EGS-group, $\exp(K_{n})$ can be trivially
bounded from above. Indeed, if we set

$\Gamma=\\{f\in\operatorname{Aut}\mathcal{T}:f_{(u)}\text{ is a power of
}\sigma=(12\dots p)\text{ for any vertex }u\\}$

we have $K\leq\Gamma$ and, since $\Gamma_{n}$ is a Sylow $p$-subgroup of
$\operatorname{Sym}(p^{n})$,

(5) $\exp(K_{n})\mid\exp(\Gamma_{n})=p^{n}.$

What is surprising and more delicate to be proved is that, starting from
$\exp(K_{1})=p$, the exponent of $K_{n}$ does grow by $p$ whenever $n$
increases by 1 (if $K$ is non-periodic) or whenever $n$ increases by 2 (if $K$
is periodic). Therefore, not only, as the affirmative answer to the Restricted
Burnside Problem guarantees, the increasing sequence $\exp(K_{n})$ cannot
converge to a finite number, but its growth to infinity is exponential in $n$
(which is, according to (5), the quickest growth we could expect).

Notation. We will use the convention that $x^{y}=y^{-1}xy$ and
$[x,y]=x^{-1}y^{-1}xy$. Moreover, we will write $fg$ for the composition of
two maps $f$ and $g$, where we apply first $f$ and then $g$ (i.e.
$fg(u)=g(f(u))$).

Acknowledgements. I am deeply grateful to G. A. Fernández-Alcober for the
valuable suggestions and the continued support. I thank the Basque Center for
Applied Mathematics (BCAM) and the SEND Mobility Consortium for financial
contributions to the carrying out of this work.

##  2\. Non-periodic multi-EGS-groups

In this section we prove Theorem A for non-periodic multi-EGS-groups. By (5),
it is enough to show that $\exp(K_{n})\geq p^{n}$ for any non-periodic multi-
EGS-group $K$.

It can be easily checked that multi-EGS-groups are self-similar and that will
be an important ingredient for the purpose of proving Theorem A. We highlight
some properties of self-similar subgroups of $\operatorname{Aut}\mathcal{T}$
which will be useful in this and in the following section.

######  Lemma 1.

Let $S$ be a self-similar subgroup of $\operatorname{Aut}\mathcal{T}$. Then,
for any natural numbers $k,n$ such that $k\leq n$, the map

$\displaystyle\hskip 14.22636pt\operatorname{St}_{S_{n}}(k)\hskip 8.5359pt$
$\displaystyle\longrightarrow S_{n-k}\times\overset{p^{k}}{\cdots}\times
S_{n-k}$ $\displaystyle\hskip 11.38092ptg\operatorname{St}_{S}(n)\hskip
5.69046pt$
$\displaystyle\longmapsto\big{(}g_{u}\operatorname{St}_{S}(n-k)\big{)}_{u\in
X^{k}}$

is an injective group homomorphism.

We say that an element $g\in S$ has order $t$ in $S_{n}$ if
$g\operatorname{St}_{S}(n)$ has order $t$ in $S_{n}$. Then, provided that $S$
is self-similar, we have the following consequence of Lemma 1:

(6) $\text{if a component of }\psi(g)\text{ has order }t\text{ in
}S_{n}\implies g\text{ has order }\geq t\text{ in }S_{n+1}.$

Another consequence of Lemma 1 is stated in the following result.

######  Lemma 2.

Let $S\leq\operatorname{Aut}\mathcal{T}$ be a self-similar group. Then

$\exp(S_{n+k})\leq\exp(S_{n})\cdot\exp(S_{k})$

for every $n,k\geq 0$.

###### Proof.

Let us observe that for any $n,k\geq 0$

$S_{n}=\frac{S}{\operatorname{St}_{S}(n)}\simeq\frac{\frac{S}{\operatorname{St}_{S}(n+k)}}{\frac{\operatorname{St}_{S}(n)}{\operatorname{St}_{S}(n+k)}}=\frac{S_{n+k}}{\operatorname{St}_{S_{n+k}}(n)}$

which yields that $\exp(S_{n+k})$ divides
$\exp(S_{n})\cdot\exp(\operatorname{St}_{S_{n+k}}(n))$. Then it suffices to
show that $\exp(\operatorname{St}_{S_{n+k}}(n))$ divides $\exp(S_{k})$. This
follows from the fact that $\operatorname{St}_{S_{n+k}}(n)$ can be embedded in
$S_{k}\times\overset{p^{n}}{\cdots}\times S_{k}$ by Lemma 1. ∎

We first prove Theorem A for non-periodic GGS-groups. From now until the end
of the paper, we will denote with $G$ the GGS-group with defining vector
$\mathbf{e}=(e_{1},\dots,e_{p-1})$ and we will indicate with $c$ the directed
automorphism along $P_{p}$ defined by $\mathbf{e}$. From what we observed in
(4), $G=\langle a,c\rangle$. If we set $c_{i}:=c^{(a^{i})}$ for any integer
$i$, we have $c_{i}=c_{j}$ whenever $i\equiv j(\text{mod }p)$ and we get, by
(2), that

(7)
$\displaystyle\begin{split}\psi(c_{0})&=(a^{e_{1}},a^{e_{2}},\dots,a^{e_{p-1}},c)\\\
\psi(c_{1})&=(c,a^{e_{1}},\dots,a^{e_{p-2}},a^{e_{p-1}})\\\ &\vdots\\\
\psi(c_{p-1})&=(a^{e_{2}},a^{e_{3}},\dots,c,a^{e_{1}}).\end{split}$

We will denote, in addition, by $\pi_{i}$ the projection
$\pi_{i}:G\times\overset{p}{\dots}\times G\longrightarrow G$ on the $i$th
component.

######  Lemma 3.

Assume $G$ to be a non-periodic GGS-group and let $x$ be the only integer in
$\\{0,\dots,p-1\\}$ whose class in $\mathbb{F}_{p}$ is equal to
$e_{1}+\dots+e_{p-1}$. Then, for every $n\geq 1$, any element of form
$a^{x}c_{j}$ with $j\in\\{0,\dots,p-1\\}$ has order $\geq p^{n}$ in $G_{n}$.

Proof. We argue by induction on $n$. For $n=1$, $G_{1}\simeq\langle a\rangle$
is the cyclic group of order $p$. Since $G$ is not periodic, $x\not\equiv
0(\text{mod }p)$ and then, for any $j\in\\{0,\dots,p-1\\}$,
$a^{x}c_{j}\not\in\operatorname{St}_{G}(1)$, which yields that $a^{x}c_{j}$
has order $p$ in $G_{1}$.

Assume now $n\geq 2$ and let $j\in\\{0,\dots,p-1\\}$. We have

$\displaystyle(a^{x}c_{j})^{p}$
$\displaystyle=a^{xp}{c_{j}}^{(a^{x(p-1)})}{c_{j}}^{(a^{x(p-2)})}\dots
c_{j}^{(a^{x})}c_{j}$ $\displaystyle=c_{j+x(p-1)}c_{j+x(p-2)}\dots
c_{j+x}c_{j}$

and, since $x\not\equiv 0(\text{mod }p)$,
$\\{j+x(p-1),j+x(p-2),\dots,j+x,j\\}$ is a set of representatives for the
congruence classes modulo $p$. In other words, $(a^{x}c_{j})^{p}$ is the
product, in some order, of the $p$ factors $c_{0},\dots,c_{p-1}$. Then, by
(7), the first component of $\psi((a^{x}c_{j})^{p})$ is the product, in some
order, of the $p$ factors $a^{e_{1}},\dots,a^{e_{p-1}},c$. This means that

$\displaystyle\psi\pi_{1}((a^{x}c_{j})^{p})$ $\displaystyle=a^{f_{1}}\dots
a^{f_{k}}ca^{f_{k+1}}\dots a^{f_{p-1}}$
$\displaystyle=a^{f_{1}+\dots+f_{p-1}}c^{(a^{f_{k+1}+\dots+f_{p-1}})}$

for some $0\leq k\leq p-1$ and $f_{1}+\dots+f_{p-1}=e_{1}+\dots+e_{p-1}$. Then
$\psi\pi_{1}((a^{x}c_{j})^{p})$ is of the form $a^{x}c_{s}$ for some
$s\in\\{0,\dots,p-1\\}$ and by induction hypothesis it has order $\geq
p^{n-1}$ in $G_{n-1}$. This yields, by (6), that $a^{x}c_{j}$ has order $\geq
p^{n}$ in $G_{n}$. ∎

Now the following result is straightforward.

######  Proposition 4.

Let $G$ be a non-periodic GGS-group. Then

$\exp(G_{n})=p^{n}.$

for every $n\geq 1$.

To extend Proposition 4 to any non-periodic multi-EGS-group, it will suffice
to realise that a non-periodic multi-EGS-group always contains a conjugate of
a non-periodic GGS-group.

######  Theorem 5.

Let $K$ be a non-periodic multi-EGS-group. Then

$\exp(K_{n})=p^{n}$

for every $n\geq 1$.

Proof. Take $K$ a multi-EGS-group as in Definition 1. For every
$j\in\\{1,\dots,p\\}$ and every $\mathbf{e}\in E^{(j)}$, we set
$H_{\mathbf{e}}^{(j)}=\langle a,b\rangle$, where $b$ is the directed
automorphism along $P_{j}$ with defining vector $\mathbf{e}$.

We first show that, fixed $j\in\\{1,\dots,p\\}$ and
$\mathbf{e}=(e_{1},\dots,e_{p-1})\in E^{(j)}\setminus\\{\mathbf{0}\\}$,
$H_{\mathbf{e}}^{(j)}=\langle a,b\rangle$ is conjugate to the GGS-group
$L=\langle a,d\rangle$, where $d$ is the directed automorphism along $P_{p}$
with defining vector $\mathbf{e}$. We denote with $\sigma$ the $p$-cycle
$(12\dots p)\in\mathrm{Sym}(p)$ and we call $f$ the automorphism of
$\mathcal{T}$ that has all the labels equal to $\sigma^{p-j}$. In other words,
$f=xa^{p-j}$, with $x\in\operatorname{St}(1)$ defined by
$\psi(x)=(f,\dots,f)$. By (2) we have $\psi(x^{a})=(f,\dots,f)=\psi(x)$, then
$x$ commutes with $a$ and $f$ commutes with $a$. It follows from this and
again from (2) that

$\psi(b^{x})=(a^{e_{p-j+1}},\dots,a^{e_{p-1}},b^{f},a^{e_{1}},\dots,a^{e_{p-j}})$

and

$\psi(b^{f})=(a^{e_{1}},\dots,a^{e_{p-1}},b^{f}),$

which yields that $b^{f}=d$. All in all, we have
$(H_{\mathbf{e}}^{(j)})^{f}=L$, as desired.

Now, $K$ is not periodic and then there is $\mathbf{e}$ which lies in some
${E^{(j)}}$ but not in $V$. For what we just proved, $H_{\mathbf{e}}^{(j)}\leq
K$ is conjugate to a non-periodic GGS-group. Thus the $n$th congruence
quotient of $H_{\mathbf{e}}^{(j)}$ is isomorphic to the $n$th congruence
quotient of such GGS-group, and it can be embedded in $K_{n}$. It follows that
$\exp(K_{n})\geq p^{n}$ by Proposition 4, and we have the claim thanks to (5).
∎

##  3\. Periodic multi-EGS-groups

In this section we prove Theorem A for a periodic multi-EGS-group $K$. As for
the non-periodic case, it will be easy to show that the expression provided by
Theorem A is an upper bound for $\exp(K_{n})$. More efforts will be needed to
prove that such expression is a lower bound too.

We first compute the exponent of $K_{2}$. For such purpose, we need to fix the
following notation: for every $W\subseteq(\mathbb{F}_{p})^{p-1}$, we indicate
with $\overline{W}$ the subset of $(\mathbb{F}_{p})^{p}$ given by

$\overline{W}=\\{(e_{p-j+1},\dots,e_{p-1},0,e_{1},\dots,e_{p-j}):1\leq j\leq
p\text{ and }(e_{1},\dots,e_{p-1})\in W\\}.$

######  Lemma 6.

Let $K$ be a multi-EGS-group as in Definition 1. Then

$|K_{2}|=p^{t+1},$

where $t$ is the dimension of the $\mathbb{F}_{p}$-vector space
$\langle\overline{E^{(j)}}:j=1,\dots,p\rangle$.

Proof. Observe that $K=\langle a\rangle\ltimes\operatorname{St}_{K}(1)$ and
$K_{2}$ has order equal to $|\langle
a\rangle|\cdot|\frac{\operatorname{St}_{K}(1)}{\operatorname{St}_{K}(2)}|=p\cdot|\frac{\operatorname{St}_{K}(1)}{\operatorname{St}_{K}(2)}|$.
Then it suffices to show that $|\operatorname{St}_{K_{2}}(1)|=p^{t}$, with $t$
as in the statement. Now, $K_{1}\simeq\langle a\rangle\simeq\mathbb{F}_{p}$
and the composition $\varphi$ between the map

$\displaystyle\hskip 14.22636pt\hskip
5.69046pt\operatorname{St}_{K_{2}}(1)\hskip 8.5359pt$
$\displaystyle\longrightarrow K_{1}\times\overset{p}{\cdots}\times K_{1}$
$\displaystyle\hskip 11.38092ptg\operatorname{St}_{K}(2)\hskip 5.69046pt$
$\displaystyle\longmapsto(g_{1}\operatorname{St}_{K}(1),\dots,g_{p}\operatorname{St}_{K}(1))$

and the map

$\displaystyle\hskip 48.36958ptK_{1}\times\overset{p}{\cdots}\times
K_{1}\hskip 42.67912pt$ $\displaystyle\longrightarrow(\mathbb{F}_{p})^{p}$
$\displaystyle(a^{f_{1}}\operatorname{St}_{K}(1),\dots,a^{f_{p}}\operatorname{St}_{K}(1))\hskip
5.69046pt$ $\displaystyle\longmapsto(f_{1},\dots,f_{p}),$

is, by Lemma 1, an injective homomorphism. It can be easily proved, by using
(2), that $\bigcup_{j=1}^{p}\overline{E^{(j)}}=\varphi(\Delta)$ where
$\Delta=\\{{b}^{(a^{l})}:b\in\bigcup_{j=1}^{p}B^{(j)}\text{ and }0\leq l\leq
p-1\\}$.

If we prove that $\operatorname{St}_{K}(1)$ is generated by $\Delta$, we will
get that $\varphi(\operatorname{St}_{K_{2}}(1))$ is the vector subspace of
$(\mathbb{F}_{p})^{p}$ generated by $\bigcup_{j=1}^{p}\overline{E^{(j)}}$, and
this will prove the claim since
$|\varphi(\operatorname{St}_{K_{2}}(1))|=|\operatorname{St}_{K_{2}}(1)|$.

Then we are only left to show that $\operatorname{St}_{K}(1)$ is generated by
$\Delta$. This follows from the fact that
$\langle\Delta\rangle\leq\operatorname{St}_{K}(1)$, $\langle\Delta\rangle$ is
normal in $K$ and the quotients $\frac{K}{\langle\Delta\rangle}$,
$\frac{K}{\operatorname{St}_{K}(1)}$ have the same order (equal to $p$) since
they are both generated by the class of $a$. ∎

######  Lemma 7.

Let $K$ be a periodic multi-EGS-group. Then the exponent of $K_{2}$ is $p$.

Proof. Since $K$ is periodic, keeping as usual the notation of Definition 1,
we have that $\bigcup_{j=1}^{p}\overline{E^{(j)}}$ is contained in
$U=\\{(f_{1},\dots,f_{p})\in(\mathbb{F}_{p})^{p}:f_{1}+\dots+f_{p}=0\\}$.
Since $U$ is a vector space of dimension $p-1$, Lemma 6 yields that
$|K_{2}|\leq p^{p}$. This implies, by Theorem 2.8 of [1], that $K_{2}$ is a
regular $p$-group and, being generated by elements of order $p$, it must have
exponent $p$ by Corollary 2.11 of [1]. ∎

Since $\exp(K_{2})=p$ and $K$ is self-similar, Lemma 2 yields that
$\exp(K_{n})$ can grow at most by $p$ whenever $n$ increases by 2. This allows
us to bound from above $\exp(K_{n})$.

######  Proposition 8.

Let $K$ be a periodic multi-EGS-group. Then

$\exp(K_{n})\leq p^{\lfloor\frac{n+1}{2}\rfloor}$

for every $n\geq 1$.

Analogously to the non-periodic case, our strategy to bound $\exp(K_{n})$ from
below will be to do that when $K$ is a GGS-group and then to look for a
suitable GGS-group inside any multi-EGS-group. The most noteworthy part of our
proof comes with the next lemma.

######  Lemma 9.

Assume $G$ to be a periodic GGS-group and let $n\geq 1$ be an integer. Then

$\exp(G_{n+2})\geq\exp(G_{n})\cdot p$

Proof. We deduce from (5) that $\exp(G_{n})=p^{k}$ for some $k\geq 1$; then
there is $g\in G$ which has order $p^{k}$ in $G_{n}$. In order to prove the
claim, it is enough to find an element of $G$ that has order $\geq p^{k+1}$ in
$G_{n+2}$.

By Proposition 2.2 in [2], the composition
$\psi\pi_{i}:\operatorname{St}_{G}(1)\longrightarrow G$ is a surjective
homomorphism. Hence there is $h\in\operatorname{St}_{G}(1)$ such that
$\psi\pi_{1}(h)=g$ and, by (6), the order of $h$ in $G_{n+1}$ is $\geq p^{k}$.

Since $h\in\operatorname{St}_{G}(1)=\langle c_{0},\dots,c_{p-1}\rangle$ (see
the last part of the proof of Lemma 6) and $c_{0},\dots,c_{p-1}$ have order
$p$, we can write

$h=c_{j_{1}}\dots c_{j_{r}}$

for some $r\geq 1$ and $j_{1},\dots,j_{r}\in\\{0,\dots,p-1\\}$. By Theorem
2.16 in [2], we may assume $e_{1}=1$, up to replacing $G$ with a GGS-group
that is conjugate to $G$ (obviously the periodicity of the group and the
exponents of congruence quotients do not change if we replace $G$ with a
conjugate of $G$). Then we define the element

$f_{1}={c_{1}}^{({c}^{j_{1}})}{c_{1}}^{({c}^{j_{2}})}\dots
c_{1}^{({c}^{j_{r}})}.$

Observe that for any integer $\lambda$

$\displaystyle\psi({c_{1}}^{({c}^{\lambda})})$
$\displaystyle=(c,a,\dots,a^{e_{p-2}},a^{e_{p-1}})^{(a^{\lambda},a^{\lambda
e_{2}},\dots,a^{\lambda e_{p-1}},c^{\lambda})}$
$\displaystyle=(c_{\lambda},a,\dots,a^{e_{p-2}},{(a^{e_{p-1}})}^{(c^{\lambda})})$

and then

$\psi(f_{1})=(h,a^{r},\dots,a^{re_{p-2}},F)$

where

$\displaystyle F$
$\displaystyle=(a^{e_{p-1}})^{(c^{j_{1}})}\dots(a^{e_{p-1}})^{(c^{j_{r}})}$
$\displaystyle=a^{e_{p-1}}[a^{e_{p-1}},c^{j_{1}}]\dots
a^{e_{p-1}}[a^{e_{p-1}},c^{j_{r}}]$
$\displaystyle=a^{re_{p-1}}[a^{e_{p-1}},c^{j_{1}}]^{(a^{(r-1)e_{p-1}})}[a^{e_{p-1}},c^{j_{2}}]^{(a^{(r-2)e_{p-1}})}\dots[a^{e_{p-1}},c^{j_{r}}].$

We set
$y=[a^{e_{p-1}},c^{j_{1}}]^{(a^{(r-1)e_{p-1}})}[a^{e_{p-1}},c^{j_{2}}]^{(a^{(r-2)e_{p-1}})}\dots[a^{e_{p-1}},c^{j_{r}}]$,
so that $F=a^{re_{p-1}}y$.

At this point, it is convenient to consider the case in which $\mathbf{e}$ is
non-symmetric and the case in which $\mathbf{e}$ is symmetric separately. In
both cases $\mathbf{e}\in(\mathbb{F}_{p})^{p-1}\setminus\\{\mathbf{0}\\}$
cannot be constant (because the sum of its components is 0 by the periodicity
of $G$) and in particular $p\neq 2$.

Assume first $\mathbf{e}$ non-symmetric. In this case, Lemma 3.4 of [2] yields
that
$\gamma_{2}(G)\times\overset{p}{\dots}\times\gamma_{2}(G)\subseteq\psi(\operatorname{St}_{G}(1))$.
Then there exists $f_{2}\in\operatorname{St}_{G}(1)$ such that
$\psi(f_{2})=(1,\dots,1,y^{-1})$ and we get

$\psi(f_{1}f_{2})=(h,a^{r},a^{re_{2}},\dots,a^{re_{p-2}},a^{re_{p-1}}).$

Since $G$ is periodic we have $e_{1}+e_{2}+\dots+e_{p-1}=0$ and (3) yields
that

$\psi\pi_{p}((af_{1}f_{2})^{p})=h\cdot a^{r(e_{1}+e_{2}+\dots+e_{p-1})}=h.$

This implies, by (6), that $af_{1}f_{2}$ has order $\geq p^{k+1}$ in
$G_{n+2}$, as desired.

Let us now consider the case in which $\mathbf{e}$ is symmetric. Since we are
assuming $e_{1}=1$, we have $e_{p-1}=1$.

Recall that we set $h=c_{j_{1}}\dots c_{j_{r}}$ and the only assumption we
made on $h$ is $\psi\pi_{1}(h)=g$. We want to show that, thanks to the
symmetry of $\mathbf{e}$, we may assume $r\equiv j_{1}+\dots+j_{r}\text{ (mod
}p)$. Since $2$ is invertible in $\mathbb{F}_{p}$ (remember that $p\neq 2$),
there is an integer $\alpha$ such that $2\alpha\equiv
r-(j_{1}+\dots+j_{r})\text{ (mod }p)$. Then the element

$h^{\prime}=({c_{0}}^{p-1}c_{2})^{\alpha}h$

can be written in the form $h^{\prime}=c_{l_{1}}\dots c_{l_{s}}$, with
$s=p\alpha+r\equiv r$ and $l_{1}+\dots+l_{s}\equiv
2\alpha+j_{1}+\dots+j_{r}\equiv r\text{ (mod }p)$. Moreover, remembering that
$e_{1}=e_{p-1}=1$ and using (7), we have
$\psi\pi_{1}(h^{\prime})=(a^{p-1}a)^{\alpha}\psi\pi_{1}(h)=g$. Then
$h^{\prime}=c_{l_{1}}\dots c_{l_{s}}$ satisfies $s\equiv
l_{1}+\dots+l_{s}\text{ (mod }p)$ and $\psi\pi_{1}(h^{\prime})=g$. This allows
us to suppose, from now on, that $r\equiv j_{1}+\dots+j_{r}\text{ (mod }p)$.

We have

$\displaystyle y$
$\displaystyle=[a,c^{j_{1}}]^{(a^{r-1})}[a,c^{j_{2}}]^{(a^{r-2})}\dots[a,c^{j_{r}}]$
$\displaystyle\equiv[a,c^{j_{1}}][a,c^{j_{2}}]\dots[a,c^{j_{r}}]$
$\displaystyle\equiv[a,c]^{j_{1}+\dots+j_{r}}$
$\displaystyle\equiv[a,c]^{r}\text{ (mod }\gamma_{3}(G)\text{)},$

where the last congruence holds because $[a,c]$ has order $p$ in the quotient
$\gamma_{2}(G)/\gamma_{3}(G)$ (see Theorem 2.1 in [2]). This yields
$y=z\cdot[a,c]^{r}$ for some $z\in\gamma_{3}(G)$. Working again modulo
$\gamma_{3}(G)$, we get

$\displaystyle[a,h]^{(a^{-1})}$ $\displaystyle\equiv[a,c_{j_{1}}\dots
c_{j_{r}}]$ $\displaystyle\equiv[a,c]^{(a^{j_{1}})}\dots[a,c]^{(a^{j_{r}})}$
$\displaystyle\equiv[a,c]^{r}\text{ (mod }\gamma_{3}(G)\text{)}$

and $[a,c]^{r}=[a,h]^{(a^{-1})}\cdot w$ for some $w\in\gamma_{3}(G)$. Now,
since $\mathbf{e}$ is non-constant, Lemma 3.2 of [2] ensures that
$\gamma_{3}(G)\times\overset{p}{\dots}\times\gamma_{3}(G)\subseteq\psi(\operatorname{St}_{G}(1))$.
Then there must be $f_{3}\in\operatorname{St}_{G}(1)$ with
$\psi(f_{3})=(w^{-1},1,\dots,1,z^{-1})$ and

$\displaystyle\psi(f_{1}[c_{0},c_{1}]^{r}f_{3})$
$\displaystyle=(h,a^{r},\dots,a^{re_{p-2}},a^{r}\cdot
y)([a,c]^{r},1,\dots,1,[c,a]^{r})(w^{-1},1,\dots,1,z^{-1})$
$\displaystyle=(h\cdot[a,h]^{(a^{-1})},a^{r},\dots,a^{re_{p-2}},a^{r})$
$\displaystyle=(h^{(a^{-1})},a^{re_{1}},\dots,a^{re_{p-2}},a^{re_{p-1}}),$

where the last equality holds by the commutator identity
$[a,h]^{(a^{-1})}=[h,a^{-1}]$. This implies, analogously to the previous case,
that

$\psi\pi_{p}((af_{1}[c_{0},c_{1}]^{r}f_{3})^{p})=h^{(a^{-1})}$

and that $af_{1}[c_{0},c_{1}]^{r}f_{3}$ has order $\geq p^{k+1}$ in $G_{n+2}$.
∎

Now the next proposition follows immediately by induction on $n$.

######  Proposition 10.

Let $G$ be a periodic GGS-group. Then

$\exp(G_{n})=p^{\lfloor\frac{n+1}{2}\rfloor}$

for every $n\geq 1$.

To prove Theorem A, we are only left to extend Proposition 10 to any periodic
multi-EGS-group.

######  Theorem 11.

Let $K$ be a periodic multi-EGS-group. Then

$\exp(K_{n})=p^{\lfloor\frac{n+1}{2}\rfloor}$

for every $n\geq 1$.

Proof. The proof follows the same line of the proof of Theorem 5. The only
difference is that now $K$ is periodic, hence $E^{(j)}\subseteq V$ for every
$j\in\\{1,\dots,p\\}$ and, since $E^{(1)},\dots,E^{(p)}$ are not all equal to
the null subspace, we can pick
$\mathbf{e}\in\bigcup_{j=1}^{p}E^{(j)}\setminus\\{\mathbf{0}\\}\subseteq V$.
It follows that $H_{\mathbf{e}}^{(j)}$ is conjugate to a periodic GGS-group
and the claim follows by Proposition 10 and Proposition 8. ∎

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