\markboth{A. Gaglione, S. Lipschutz, D. Spellman}{Burnside Groups}

\wstoc{Burnside Groups}{A. Gaglione, S. Lipschutz, D. Spellman}

\title{Groups Universally Equivalent to Free
Burnside Groups of Prime Exponent and a Question of Philip Hall}
%{\large GROUPS UNIVERSALLY EQUIVALENT TO FREE BURNSIDE GROUPS OF 
%PRIME EXPONENT AND A QUESTION OF PHILIP HALL }
\author{A. Gaglione}
\address{Department of Mathematics\\
U.S. Naval Academy\\
Annapolis, MD 21402}
\author{S. Lipschutz}
\address{Department of Mathematics\\
Temple University\\
Philadelphia, PA 19122}
\author{D. Spellman}
\address{Department of Mathematics\\
Temple University\\
Philadelphia, PA 19122}
%\subjclass{Primary 20E26; Secondary 20E05, 20E06}
%\keywords{}

\begin{abstract}
This paper proposes the Tarski Problem for free groups in a Burnside
variety, \textbf{B}$_{n}$, where n is a sufficiently large odd integer so
that Adian's results hold. We note that just as in the case of absolutely
free groups it is easy to show that the nonabelian free groups in \textbf{B}$%
_{n}$ for n as above are all universlly equivalent.
\end{abstract}


\section{Introduction}

We bear in mind two problems which resisted solution for decades and
succumbed only to the fresh attacks of supremely talented mathematicians.
For our purposes we shall dub the questions (1) and (2) below as \textit{The
Tarski Problem} and \textit{The Burnside Problem} respectively.

\bigskip

(1) Are the nonabelian free groups elementarily equivalent?

(2) Must the finitely generated nonabelian free Burnside groups of fixed
finite exponent be finite?

\bigskip

Of course the answers to (1) and (2) are now known to be (1) yes and (2) no
respectively. (1) was solved independently by Kharlampovich and Myasnikov on
the one hand and by Sela on the other. Kharlampovich and Myasnikov applied 
\textit{algebraic geometry over groups} invented by G. Baumslag, Myasnikov
and Remeslennikov while Sela invented \textit{Diophantine geometry in groups}
to solve the Tarski problem.{\large \ }

{\large \ }Let's focus on algebraic geometry over groups. By way of analogy
let us consider a Noetherian integral domain R. The closed subsets in the 
\textit{Zariski topology} on affine n-space over R are precisely the affine
algebraic subsets. In order for the analog of this scenario to go through in
groups \ the proper notions of domain and Noetherian had to be defined. One
consequence of the correct definition of domain is that every \ nonabelian
CSA group (See Section 3) is a domain. Since free groups are CSA, nonabelian
free groups are domains. The correct notion of Noetherian is \textit{%
equationally Noetherian} (See Section 6). Happily free groups are also
equationally Noetherian. One felicitous consequence of this confluence of
facts is the existence and uniqueness (in the usual sense) of the
decomposition of a closed set into a finite union of irreducible affine
algebraic subsets. (We need to reserve the word \textit{variety} for an
equational class in this paper.) While the groups free in the variety of all
groups are elementarily equivalent (provided their rank exceeds 1) it is
easy to see that the corresponding result is false for groups free in many
other varieties. For example, one can distinguish the free nilpotent groups
(of fixed class) of different finite ranks by first order sentences. Perhaps
the variety of all groups is unique in this regard? Or is it!

We now turn our attention to Question (2). A negative answer was first
provided by Adian who showed that, for all sufficiently large odd n, the
nonabelian free groups in the variety $\mathbf{B}_{n}$ determined by the law 
$x^{n}=1$ were all infinite. Soon afterward Sirvanjan proved that, just as
in the variety of all groups, the free group of countably infinite rank in
the variety $\mathbf{B}_{n}${\large \ }embeds in its group free of rank 2
for all sufficiently large odd n. From this it easily follows that the
nonabelian free groups in the varieties $\mathbf{B}_{n}$, for all
sufficiently large fixed odd n, have the same universal theory in the sense
of first order logic. For our purposes we can say the most when n = p is a
sufficiently large prime. In any event, if n is a sufficiently large odd
integer, the free groups in $\mathbf{B}_{n}$ are CSA; hence, the nonabelian
free groups in such varieties are domains. We do not know whether or not
they are equationally Noetherian. The Tarski problem relativized to such $%
\mathbf{B}_{n}$ will require new techniques for its solution. This paper
provides some halting first steps and should be viewed as an invitation to
our colleagues to ponder possible approaches.

\section{Preliminaries}

We let $\omega $ be the first limit ordinal, which we identify with the
first infinite cardinal $\aleph _{0}$. If $H$ and $G$ are groups we say that 
$G$ is $\mathit{H}$\textit{-inclusive} provided it contains a subgroup
isomorphic to $H$; $G$ is $\mathit{H}$\textit{-exclusive} provided it is not 
$H$-inclusive. For each positive integer $n$, $C_{n}$ shall be a group
cyclic of order $n$. If $G$ is a group and $\mathbf{H}$ is a nonempty class
of groups, then we say that $\mathbf{H}$ \textit{separates} $G$ provided for
every $g\in G\backslash \{1\}$ there is a group $H_{g}\in \mathbf{H}$ and a
homomorphism $\varphi _{g}:G\rightarrow H_{g}$ such that $\varphi
_{g}(g)\neq 1$; we say that $\mathbf{H}$\textit{\ discriminates }$G$
provided for every finite nonempty subset $S\subseteq G\backslash \{1\}$
there is a group $H_{S}\in \mathbf{H}$ and a homomorphism $\varphi
_{S}:G\rightarrow H_{S}$ such that $\varphi _{S}(g)\neq 1$ for all $g\in S$.
In the case $\mathbf{H}=\{H\}$ is a singleton we say that $H$ separates
(discriminates) $G$ for $\mathbf{H}$ separates (discriminates) $G$. In the
case that $\mathbf{H}$ separates (discriminates) $G$ by epimorphisms the
language $G$ is \textit{residually (fully residually)} $\mathbf{H}$ is
sometimes used.

Let $L_{0}$ be the first order language with equality containing a binary
operation symbol $\bullet $, a unary operation symbol $^{-1}$ and a constant
symbol $1$. We remark that being first order means that the variables are
interpreted as varying over individual elements of the domain of discourse -
never over subsets nor functions. Thus an $\mathit{L}_{0}$\textit{-structure}
is a set $G$ provided with a distinguished constant $1\in G$, a unary
operation $G\rightarrow G$, $g\mapsto g^{-1}$ and a binary operation $%
G^{2}\rightarrow G$, $(g,h)\mapsto gh$. A \textit{universal sentence} of $%
L_{0}$ is one of the form $\forall \mathbf{x}\varphi (\mathbf{x})$ where $%
\mathbf{x}$ is a tuple of distinct variables and $\varphi (\mathbf{x})$ is a
formula of $L_{0}$ containing no quantifiers and containing free at most the
variables in $\mathbf{x}$. A \textit{law} in $L_{0}$ is a universal sentence
of the form $\forall \mathbf{x}(s(\mathbf{x})=t(\mathbf{x}))$ where $\mathbf{%
x}$ is a tuple of distinct variables and $s(\mathbf{x})$ and $t(\mathbf{x})$
are terms of $L_{0}$ containing at most the variables in $\mathbf{x}$. The
model class of a set of laws of $L_{0}$ is a \textit{variety} of $L_{0}$%
-structures. The following three sentences are laws.

$\gamma _{1}$: $\forall x_{1},x_{2},x_{3}((x_{1}\bullet x_{2})\bullet
x_{3}=x_{1}\bullet (x_{2}\bullet x_{3}))$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
Associative Law

$\gamma _{2}$: $\forall x(x\bullet 1=x)$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Identity
Law

$\gamma _{3}$: $\forall x(x\bullet x^{-1}=1)$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Inverse
Law

We shall refer to the set $\{\gamma _{1},\gamma _{2},\gamma _{3}\}$ as 
\textit{the group axioms}. \ The model class $\mathbf{O}$ of the group
axioms is \textit{the variety of \ all groups}. Every variety under
consideration in this paper shall be a subvariety of $\mathbf{O}$. \ If $%
\gamma $ is the conjunction $\gamma _{1}\wedge \gamma _{2}\wedge \gamma _{3}$
of the group axioms and $\sigma $ and $\tau $ are sentences of $L_{0}$ then
we shall say that $\sigma $ and $\tau $ are \textit{equivalent modulo the
group axioms }provided the sentence $\gamma \rightarrow (\sigma
\leftrightarrow \tau )$ is true. Every law in $L_{0}$ is equivalent modulo
the group axioms to one of the form $\forall \mathbf{x}(w(\mathbf{x})=1)$
where $\mathbf{x}$ is a tuple of distinct variables and $w(\mathbf{x})$ is a
word in at most the variables in $\mathbf{x}$. Henceforth we shall omit the
universal quantifiers and abbreviate the law $\forall \mathbf{x}(w(\mathbf{x}%
)=1)$ as simply $w(\mathbf{x})=1$. From this point on we tacitly assume the
group axioms. The \textit{trivial variety} $\mathbf{E}$, determined by the
law $x=1$, is the isomorphism class of the one element group. All other
varieties of groups are \textit{nontrivial}. Every nontrivial variety $%
\mathbf{V}$ of groups admits, for each cardinal $r$, groups free of rank $r$
relative to $\mathbf{V}$. We adopt the notation $F_{r}(\mathbf{V})$ for a
fixed, but arbitrary, group free of rank $r$ relative to $\mathbf{V}$. Any
two such groups (for fixed $\mathbf{V}$ and $r$) are isomorphic.

\smallskip

\noindent \textbf{Convention: }If $\mathbf{E}$ is the trivial variety and $r$
is any cardinal, then $F_{r}(\mathbf{E})$ is the trivial group $1$.

\smallskip

If $G$ and $H$ are groups then we say that $G$ \textit{universally covers }$%
H $ provided every universal sentence of $L_{0}$ true in $G$ is also true in 
$H $. Note that $H\leq G$ is a sufficient condition for $G$ to universally
cover $H$. We say that $G$ and $H$ are \textit{universally equivalent} or 
\textit{have the same universal theory} and write $G\equiv _{\forall }H$
provided $G$ universally covers $H$ and $H$ universally covers $G$. \ An 
\textit{existential sentence }of $L_{0}$ is one of the form $\exists \mathbf{%
x}\varphi (\mathbf{x})$ where $\mathbf{x}$ is a tuple of distinct variables
and $\varphi (\mathbf{x})$ is a formula of $L_{0}$ containing no quantifiers
and containing free at most the variables in $\mathbf{x}$. A \textit{%
primitive sentence }of $L_{0}$ is an existential sentence of $L_{0}$
equivalent modulo the group axioms to one of the form $\exists \mathbf{x}%
(\wedge _{j}(u_{j}(\mathbf{x})=1)\wedge \wedge _{k}(v_{k}(\mathbf{x})\neq
1)) $ where $\mathbf{x}$ is a tuple of distinct variables and the $u_{j}(%
\mathbf{x})$ and $v_{k}(\mathbf{x)}$ are words in at most the variables in $%
\mathbf{x}$. Since the negation of a universal sentence of $L_{0}$ is
equivalent to an existential sentence of $L_{0}$ and vice-versa $G\equiv
_{\forall }H$ may be paraphrased as asserting that every universal sentence
and every existential sentence of $L_{0}$ true in $G$ is also true in $H$
(and vice-versa). It is easy to see that if $G_{1}\equiv _{\forall }G\equiv
_{\forall }G_{2}$, then $G_{1}\leq H\leq G_{2}$ is a sufficient condition
for $G\equiv _{\forall }H$. A somewhat more sophisticated sufficient
condition is the following. $H\leq G$ and every finite system%
\begin{eqnarray*}
u_{j}(\mathbf{x}) &=&1,1\leq j\leq J \\
v_{k}(\mathbf{x}) &\neq &1,1\leq k\leq K
\end{eqnarray*}%
of equations and inequations ( in finitely many variables $\mathbf{x}%
=(x_{1},...,x_{n})$) which has a solution in $G$ must also have a solution
in $H$. To see that this is so observe that $G$ universally covers $H$; so,
it will suffice to show that every existential sentence of $L_{0}$ true in $%
G $ must also be true in $H$. Now we may assume that the \textit{matrix} $%
\varphi (\mathbf{x})$ of the existential sentence $\exists \mathbf{x}\varphi
(\mathbf{x})$ is written in \textit{disjunctive normal form} modulo the
group axioms $\vee _{i}(\wedge _{j}(u_{i,j}(\mathbf{x})=1)\wedge \wedge
_{k}(v_{i,k}(\mathbf{x})\neq 1))$. The sentence is then equivalent to the
disjunction $\vee _{i}\exists \mathbf{x}(\wedge _{j}(u_{i,j}(\mathbf{x}%
)=1)\wedge \wedge _{k}(v_{i,k}(\mathbf{x})\neq 1))$ of primitive sentences.
Since a disjunction is true provided at least one of the disjuncts is, it
suffices to prove that every primitive sentence of $L_{0}$ true in $G$ is
also true in $H$; hence, the sufficiency of the above criterion is
established.

We say that two groups $G$ and $H$ are \textit{elementarily equivalent} and
write $G\equiv H$ provided $G$ and $H$ satisfy the same sentences of $L_{0}$%
. A theorem of Vaught (Theorem 4 of Chapter 6, Section 38 in [G]) asserts
that , if $\mathbf{V}$ is any variety and $r$ and $s$ are infinite
cardinals, then $F_{r}(\mathbf{V})\equiv F_{s}(\mathbf{V})$. In particular, $%
F_{r}(\mathbf{V})\equiv _{\forall }F_{s}(\mathbf{V})$ when $r$ and $s$ are
infinite.

Let $n$ be a positive integer. The \textit{Burnside variety} $\mathbf{B}_{n}$
of exponent $n$ is the variety of groups determined by the law $x^{n}=1$.
Adian proved that, for all sufficiently large odd $n$,

(B1) $F_{r}(\mathbf{B}_{n})$ is infinite for all $r\geq 2$; moreover, every
finite subgroup of $F_{r}(\mathbf{B}_{n})$ is cyclic.

and

(B2) The centralizer of every nontrivial element in $F_{r}(\mathbf{B}_{n})$
is cyclic for all $r\geq 2.$

Sirvanjan proved that, for all sufficiently large odd $n$,

(B3) $F_{\omega }(\mathbf{B}_{n})$ embeds in $F_{2}(\mathbf{B}_{n})$.

We shall call an integer $n>0$ an \textit{Adian-Sirvanjan integer} provided
(B1), (B2) and (B3) hold for $\mathbf{B}_{n}$. A prime Adian-Sirvanjan
integer $p$ shall be an \textit{Adian-Sirvanjan prime}.

Every member $G$ of a variety $\mathbf{V}$ of groups is a homomorphic image
of a group $F_{r}(\mathbf{V})$ free in $\mathbf{V}$. If there is an
epimorphism $\psi :F_{r}(\mathbf{V})\rightarrow G$ such that $r$ is finite
and $Ker(\psi )$ is the normal closure in $F_{r}(\mathbf{V})$ of finitely
many elements of $F_{r}(\mathbf{V})$, then $G$ is \textit{finitely presented
relative to }$\mathbf{V}$.

\section{\textbf{Varieties and Discrimination}}

We begin by remarking that, although we chose to live in the world of
groups, the results of this section go through in the context of universal
algebra.

Let $\mathbf{V}$ be a nontrivial variety of groups. Let $F_{\omega }(\mathbf{%
V})$ be a group free of countably infinite rank relative to $\mathbf{V}$.
Suppose $\{a_{1},a_{2},...\}=\{a_{n+1}:n<\omega \}$ freely generates $%
F_{\omega }(\mathbf{V})$ relative to $\mathbf{V}$.

\begin{definition}
\ [N] \ A group $G\in \mathbf{V}$ \textbf{discriminates}\textit{\ }$\mathbf{V%
}$ provided $G$ discriminates $F_{\omega }(\mathbf{V})$.
\end{definition}

Now $G\in \mathbf{V}$ discriminates $\mathbf{V}$ just in case, given
finitely many elements%
\begin{equation*}
w_{k}(a_{1},...,a_{n})\neq 1
\end{equation*}
in $F_{\omega }(\mathbf{V})$, there is a homomorphism $\psi :F_{\omega }(%
\mathbf{V})\rightarrow G$ such that $\psi (w_{k}(a_{1},...,a_{n}))\neq 1$
for all $k$. This is equivalent to the following. Given finitely many words $%
w_{k}(x_{1},...,x_{n})$ such that none of the equations%
\begin{equation*}
w_{k}(x_{1},...,x_{n})=1
\end{equation*}
is a law in $\mathbf{V}$, there is a tuple $(g_{1},...,g_{n})\in G^{n}$ such
that $w_{k}(g_{1},...,g_{n})\neq 1$ for all $k$.

\bigskip

\noindent \textbf{Convention: }\textit{The trivial group }$1$\textit{\
discriminates the trivial variety }$E$\textit{.}

\bigskip

\begin{definition}
Let $\mathbf{V}$ be a variety of groups. $\mathbf{V}$ is \textbf{finitely
discriminable} provided there is a finitely generated group $G\in \mathbf{V}$
such that $G$ discriminates $\mathbf{V}$.
\end{definition}

\begin{lemma}
$\mathbf{V}$ is finitely discriminable if and only if there is a positive
integer $r$ such that $F_{r}(\mathbf{V})$ discriminates $\mathbf{V}$.
\end{lemma}

\noindent \textit{Proof: \ }If $G=F_{r}(\mathbf{V})$ discriminates $\mathbf{V%
}$ for some integer $r>0$, then $\mathbf{V}$ is discriminated by an $r$%
-generator member; hence, it is finitely discriminable. Suppose $\mathbf{V}$
is finitely discriminable. \ Suppose $r$ is a positive integer and $%
G=\langle b_{1},...,b_{r}\rangle \in \mathbf{V}$ discriminates $\mathbf{V}$.
Let $F_{r}(\mathbf{V})$ be freely generated relative to $\mathbf{V}$ by $%
a_{1},...,a_{r}$. Then we get an epimorphism $\psi :F_{r}(\mathbf{V}%
)\rightarrow G$, $a_{i}\mapsto b_{i}$, $1\leq i\leq r$.

Suppose that $w_{k}(x_{1},...,x_{n})$ are finitely many words such that none
of the equations $w_{k}(x_{1},...,x_{n})=1$ is a law in $\mathbf{V}$. Then
there are elements $g_{i}=u_{i}(b_{1},...,b_{r})$, $1\leq i\leq n$ in $G$
such that $w_{k}(u_{1}(b_{1},...,b_{r}),...,u_{n}(b_{1},...,b_{r}))\neq 1$
for all $k$. It follows that $%
w_{k}(u_{1}(a_{1},...,a_{r}),...,u_{n}(a_{1},...,a_{r}))\neq 1$ in $F_{r}(%
\mathbf{V})$ for all $k$. Hence, $F_{r}(\mathbf{V})$ discriminates $\mathbf{V%
}$. $\blacksquare $

\begin{definition}
Let $\mathbf{V}$ be a finitely discriminable variety of groups. Then $\min
\{1\leq r<\omega :F_{r}(\mathbf{V})$ discriminates $\mathbf{V}\}$ is the 
\textbf{index of discrimination} of $\mathbf{V}$. If $m$ is the index of
discrimination of $\mathbf{V}$, then $D(\mathbf{V})=\{F_{r}(\mathbf{V}%
):m\leq r<\omega \}.$
\end{definition}

\begin{theorem}
\ [GS] Let $\mathbf{V}$ be a finitely discriminable variety of groups. Let $%
r\geq 1$ be a cardinal. Then $F_{r}(\mathbf{V})\equiv _{\forall }F_{s}(%
\mathbf{V})$ for all cardinals $s\geq r$ if and only if $F_{r}(\mathbf{V})$
discriminates $\mathbf{V}$. In particular, if $m$ is the index of
discrimination of $\mathbf{V}$, then $F_{m}(\mathbf{V})\equiv _{\forall
}F_{s}(\mathbf{V})$ for all $m\leq s\leq \omega $.
\end{theorem}

\begin{theorem}
Let $\mathbf{V}$ be a finitely discriminable variety of groups with index of
discrimination $m$. Let $G\in \mathbf{V}$ be $F_{m}(\mathbf{V})$ inclusive.
If $D(\mathbf{V})$ discriminates $G$, then $G\equiv _{\forall }F_{m}(\mathbf{%
V})$.
\end{theorem}

\noindent \textit{Proof: \ }Assume $D(\mathbf{V})$ discriminates $G$. It
will suffice to show that, if%
\begin{eqnarray*}
u_{j}(x_{1},...,x_{n}) &=&1\ \ \ \ 1\leq j\leq J \\
v_{k}(x_{1},...,x_{n}) &\neq &1\ \ \ \ 1\leq k\leq K
\end{eqnarray*}%
has a solution $(g_{1},...,g_{n})\in G^{n}$, then it has a solution over $%
F_{m}(\mathbf{V})$. Since $D(\mathbf{V})$ discriminates $G$ there is an
integer $r\geq m$ and a homomorphism $\psi :G\rightarrow F_{r}(\mathbf{V})$
such that $\psi (v_{k}(g_{1},...,g_{n}))\neq 1$ for all $1\leq k\leq K$. It
follows that the primitive sentence $\exists x_{1},...,x_{n}(\wedge
_{j}(u_{j}(x_{1},...,x_{n})=1)\wedge \wedge _{k}(v_{k}(x_{1},...,x_{n})\neq
1))$ is true in $F_{r}(\mathbf{V})$. But since $F_{r}(\mathbf{V})\equiv
_{\forall }F_{m}(\mathbf{V})$ the above primitive sentence must also be true
in $F_{m}(\mathbf{V})$. Hence, the system has a solution over $F_{m}(\mathbf{%
V})$. $\blacksquare $

\begin{corollary}
Let $\mathbf{V}$ be a finitely discriminable variety of groups with index of
discrimination $m$. Let $G\in \mathbf{V}$ be finitely presented relative to $%
\mathbf{V}$ and suppose that $G$ is $F_{m}(\mathbf{V})$ inclusive. Then $%
G\equiv _{\forall }F_{m}(\mathbf{V})$ if and only if $D(\mathbf{V})$
discriminates $G$.
\end{corollary}

\noindent \textit{Proof: \ }One direction follows immediately from the
theorem. Suppose $G\in \mathbf{V}$ is finitely presented relative to $%
\mathbf{V}$, is $F_{m}(\mathbf{V})$ inclusive and $G\equiv _{\forall }F_{m}(%
\mathbf{V})$. \ Suppose $<a_{1},...,a_{n};R_{1},...,R_{J}>_{\mathbf{V}}$is a
finite presentation of $G$ relative to $\mathbf{V}$. Let $%
w_{k}(a_{1},...,a_{n})$, $1\leq k\leq K$ be finitely many nontrivial
elements of $G$. Then the primitive sentence $\exists x_{1},...,x_{n}(\wedge
_{j}(R_{j}(x_{1},...,x_{n})=1)\wedge \wedge _{k}(w_{k}(x_{1},...,x_{n})\neq
1))$ holds in $G$; hence, it holds in $F_{m}(\mathbf{V})$ and there is $%
(b_{1},...,b_{n})\in F_{m}(\mathbf{V})^{n}$ such that%
\begin{eqnarray*}
R_{j}(b_{1},...,b_{n}) &=&1\ \ \ \ 1\leq j\leq J \\
w_{k}(b_{1},...,b_{n}) &\neq &1\ \ \ \ 1\leq k\leq K.
\end{eqnarray*}

It follows that the assignment $a_{i}\mapsto b_{i}$, $1\leq i\leq n$ extends
to a homomorphism $\psi :G\rightarrow F_{m}(\mathbf{V})$ such that $\psi
(w_{k}(a_{1},...,a_{n}))\neq 1$, $1\leq k\leq K$. $\blacksquare $

\section{\textbf{The Variety }$\mathbf{O}$ \textbf{of All Groups}}

In this section we merely review known results about universally free groups
(see Definition \ref{uf}). These results will be contrasted later with
results for the groups $G\equiv _{\forall }F_{2}(\mathbf{B}_{p})$ where $p$
is an Adian-Sirvanjan prime. If $\mathbf{O}$ is the variety of all groups
and $r\geq 1$ is a cardinal, then we write $F_{r}$ for $F_{r}(\mathbf{O})$. $%
F_{\omega }$ embeds in $F_{2}$. For example, the commutator subgroup $%
[F_{2},F_{2}]$ of $F_{2}$ is free of countably infinite rank. Now let $2\leq
r\leq \omega $. Then $F_{\omega }\cong \lbrack F_{2},F_{2}]\leq F_{r}\leq
F_{\omega }$ from which it follows that $F_{r}\equiv _{\forall }F_{s}$ for
all cardinals $2\leq r<s$. Thus $2$ is the index of discrimination of $%
\mathbf{O}$. The universal equivalence of the nonabelian free groups
suggests the possibility of their elementary equivalence. Of course their
universal equivalence is a far cry from their elementary equivalence.

Suppose $R$ is a commutative ring with $1$. \ Within the category of unital $%
R$-modules an object $P$ is \textit{projective} just in case every short
exact sequence%
\begin{equation*}
0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0
\end{equation*}%
splits. This is easily seen to be equivalent to $P$ being a direct summand
in a free $R$-module. \ Now suppose $\mathbf{V}$ is a variety of groups. We
define a group $P\in \mathbf{V}$ to be \textit{projective} relative to $%
\mathbf{V}$ provided every short exact sequence of groups in $\mathbf{V}$,%
\begin{equation*}
1\rightarrow K\rightarrow G\rightarrow P\rightarrow 1,
\end{equation*}%
splits. Essentially the same proof shows that this is equivalent to $P$
being a retract of a group free in $\mathbf{V}$. By the Neilsen-Schreier
subgroup theorem, a group $P$ is projective relative to the variety of all
groups if and only if it is free. The Neilsen-Schreier subgroup theorem also
implies that a group is freely separated (discriminated) if and only if it
is residually (fully residually) free. Suppose $G$ is a nonabelian
residually free group. Suppose $gh\neq hg$ in $G$. \ Then their commutator $%
[g,h]=g^{-1}h^{-1}gh$ is nontrivial. Thus, there is a free group $F_{r}$ and
an epimorphism $\psi :G\rightarrow F_{r}$ such that $[\psi (g),\psi
(h)]=\psi ([g,h])\neq 1$. Hence, $F_{r}$ is nonabelian and $r\geq 2$. \
Since $F_{r}$ is projective it is a retract in $G$ and $F_{2}\leq F_{r}\leq
G $; so, $G$ is $F_{2}$-inclusive. Nonabelian free groups, of course,
satisfy the existential sentence $\exists x,y(xy\neq yx)$. Other properties
of nonabelian free groups are that they are CT and even CSA. Here a group is 
\textit{CT} or \textit{commutative transitive} provided the centralizers of
nontrivial elements coincide with the maximal abelian subgroups. That is
rendered by the universal sentence%
\begin{equation*}
\forall x,y,z(((y\neq 1)\wedge (xy=yx)\wedge (yz=zy))\rightarrow (xz=zx)).
\end{equation*}

Moreover, a group is \textit{CSA} or \textit{conjugately separated abelian}
provided maximal abelian subgroups are malnormal. That is equivalent to
being CT and satisfying the universal sentence%
\begin{equation*}
\forall x,y,z(((x\neq 1)\wedge (xy=yx)\wedge
(z^{-1}yzx=xz^{-1}yz))\rightarrow (xz=zx)).
\end{equation*}

Free groups are CSA.

\begin{lemma}
\label{B}[B] A CT residually free group is CSA.
\end{lemma}

\noindent \textit{Proof: \ }Suppose $G$ is CT and residually free. Let $a\in
G\backslash \{1\}$ and suppose $b,g^{-1}bg\in C_{G}(a)=\{x\in G:ax=xa\}$.
Suppose to deduce a contradiction that $g\notin C_{G}(a)$. Then $[g,a]\neq 1$%
. Thus there is a free group $F$ and an epimorphism $\psi :G\rightarrow F$
such that $[\psi (g),\psi (a)]=\psi ([g,a])\neq 1$. But this is impossible.
From $[\psi (g),\psi (a)]\neq 1$ we have $\psi (a)\neq 1$. Moreover $\psi
(b),\psi (g)^{-1}\psi (b)\psi (g)\in C_{F}(\psi (a))$ and $F$ is CSA. Hence, 
$\psi (g)$ commutes with $\psi (a)$ - a contradiction. Hence $g\in C_{G}(a)$
and $G$ is CSA. $\blacksquare $

\bigskip

\begin{theorem}
\ [B] Let $G$ be a nonabelian residually free group. The following three
conditions are equivalent in pairs.
\end{theorem}

\begin{enumerate}
\item $G$\textit{\ is fully residually free.}

\item $G$\textit{\ is CT.}

\item $G$\textit{\ is CSA.}
\end{enumerate}

\smallskip

\noindent \textit{Proof: }Lemma \ref{B} has already established the
equivalence of (2) and (3). It will suffice to show that (1) and (2) are
equivalent.

\noindent Suppose $G$ is fully residually free. Let $b\in G\backslash \{1\}$
and suppose $a,c\in C_{G}(b)$. Assume to deduce a contradiction that $ac\neq
ca$ Then $[a,c]\neq 1$. Thus there is a free group $F$ and an epimorphism $%
\psi :G\rightarrow F$ such that $\psi (b)\neq 1$ and $[\psi (a),\psi
(c)]=\psi ([a,c])\neq 1$. But this is impossible. $\psi (a)\psi (b)=\psi
(b)\psi (a)$, $\psi (b)\psi (c)=\psi (c)\psi (b)$ and $F$ is CT; hence, $%
\psi (a)\psi (c)=\psi (c)\psi (a)$ - a contradiction. Therefore $ac=ca$ and $%
G$ is CT.

Now suppose $G$ is CT. The proof will proceed by induction on the
cardinality $n$ of $S=\{g_{1},...,g_{n}\}\subseteq G\backslash \{1\}$ no
element of which is annihilated in a free homomorphic image. \ The result is
true when $n=1$ since $G$ is residually free. Now suppose $n>1$ and the
result is true for all $1\leq k<n$. Suppose first that $S$ is not contained
in an abelian subgroup of $G$. Then some pair of elements of $S$, which we
may take to be $g_{n-1}$ and $g_{n}$, does not commute. Thus $%
T=\{g_{1},...,g_{n-2},[g_{n-1},g_{n}]\}$ is contained in $G\backslash \{1\}$
and, by inductive hypothesis, there is a free group $F$ and an epimorphism $%
\psi :G\rightarrow F$ such that $\psi $ does not annihilate any element of $%
T $. But then $\psi $ cannot annihilate any element of $S$ either. It
remains to treat the case where the $g_{i}$ commute in pairs ,which
hypothesis we now assume. Since $G$ is a residually free CT group it is CSA
by Lemma \ref{B}. We claim that there is some $g\in G$ such that $%
g^{-1}g_{n}g$ does not commute with $g_{n-1}$. Otherwise, since $G$ is CSA, $%
\ g_{n-1}$ would be central in $G$. But a nonabelian CT group must be
centerless; so, we have arrived at a contradiction. The claim is
established. Pick one such $g$. Hence $U=%
\{g_{1},...,g_{n-2},[g_{n-1},g^{-1}g_{n}g]\}$ is contained in $G\backslash
\{1\}$. By inductive hypothesis there is a free group $F$ and an epimorphism 
$\psi :G\rightarrow F$ such that $\psi $ does not annihilate any element of $%
U$. But then $\psi $ cannot annihilate any element of $S$ either. That
completes the induction. $\blacksquare $

\bigskip

\begin{definition}
\label{uf}A group $G\equiv _{\forall }F_{2}$ is \textbf{universally free}.
\end{definition}

\bigskip

\begin{theorem}
\ [R] A finitely generated group is universally free if and only if it is
nonabelian and fully residually free.
\end{theorem}

\bigskip

\begin{theorem}
\ [KM1] A finitely generated fully residually free group is finitely
presented.
\end{theorem}

\bigskip \noindent

\begin{definition}
\label{CE}Let $G$ be a CT group and let $a\in G\backslash \{1\}$. Let $%
M=C_{G}(a)$. Then the HNN-extension%
\begin{equation*}
\langle G,t;rel(G),t^{-1}mt=m~~\forall m\in M\rangle
\end{equation*}
\end{definition}

\textit{is a} \textbf{free rank 1 centralizer extension of}\textit{\ }$G$%
\textit{.}

\smallskip

The above construction preserves fully residually freeness. In particular,
we have

\smallskip

\begin{theorem}
If $G_{0}$ is fully residually free then so is every free rank 1 centralizer
extension of $G_{0}$.
\end{theorem}

\smallskip

\noindent \textit{Proof: }By Definition \ref{CE}, we take $b\in
G_{0}\backslash \{1\}$ and let $M=C_{G_{0}}(b)$. So that our free rank 1
centralizer extension of $G_{0}$ is the group $G$,%
\begin{equation*}
G=\langle G_{0},t;~rel(G_{o}),t^{-1}zt=z~\forall z\in M\rangle .
\end{equation*}%
Start off by viewing $G$ as the amalgamated free product%
\begin{equation*}
G=G_{0}\ast _{M}(M\times \langle t;~\rangle .
\end{equation*}%
We need to show that $G$ is fully residually free. For that purpose let $%
g_{1},...,g_{k}$ be finitely many nontrivial elements of $G$. Using the
normal form for free products with amalgamation (see [MKS]), we may write
for each $j=1,...,k$%
\begin{equation*}
g_{j}=b_{0,j}t^{m_{1,j}}b_{1,j}...b_{N(j)-1,j}t^{m_{N(j),j}}z_{j}
\end{equation*}%
where $N(j)\geq 0,b_{i,j}\in G_{0}\backslash M,~m_{i,j}\in \mathbb{Z}%
\backslash \{0\},$ and $z_{j}\in M$. Note that $b_{i,j}\in G_{0}\backslash M$
is equivalent to $[b_{i,j},b]\neq 1$. Now since $G_{0}$ is fully residually
free, there is a free group $F$ and an epimorphism $\varphi
:G_{0}\rightarrow F$ such that $[\varphi (b_{i,j}),\varphi (b)]=\varphi
([b_{i,j},b])\neq 1$ for all $i,j$. This forces $\varphi (b_{i,j})\neq 1$
and $\varphi (b)\neq 1$.

Let $C_{F}(\varphi (b))=\langle u\rangle $. Suppose that $\varphi
(z_{j})=u^{e_{j}}$ for all $j,~1\leq j\leq k$. Now for each positive integer 
$n\in \mathbb{N}$, we may define an extension $\psi _{n}:G\rightarrow F$ of $%
\varphi $ by $\psi _{n}\mid _{G_{0}}=\varphi ,~\psi _{n}(t)=u^{n}$.

Now fix a $j,~1\leq j\leq k$. Could we have $\psi _{n}(g_{j})=1$ for
infinitely many $n\in \mathbb{N}$? Suppose to deduce a contradiction that
there were infinitely many $n\in \mathbb{N}$ such that $\psi _{n}(g_{j})=1$.
Then%
\begin{equation*}
\varphi (b_{0,j})u^{m_{1,j}n}\varphi (b_{1,j})...\varphi
(b_{N(j)-1,j})u^{m_{N(j),j}n+e_{j}}=1
\end{equation*}%
for infinitely many values of $n$. But then by G. Baumslag's
\textquotedblleft Big Powers Lemma\textquotedblright\ (see Proposition 1
[GB]), we then conclude that 
\begin{equation*}
\varphi (b_{i,j})u=u\varphi (b_{i,j})
\end{equation*}%
for some $i$, with $0\leq i\leq N(j)-1$. Thus for that $\varphi (b_{i,j})$
we must have that $\varphi (b_{i,j})\in C_{F}(u)=C_{F}(\varphi (b))$ and so $%
[\varphi (b_{i,j}),\varphi (b)]=1$. This contradicts our choice of $\varphi $%
.

The above contradiction shows that the set%
\begin{equation*}
S_{j}=\left\{ n\in \mathbb{N}:\psi _{n}(g_{j})\neq 1\right\}
\end{equation*}%
is a cofinite subset of $\mathbb{N}$ (i.e., its complement $S_{j}^{^{\prime
}}=\mathbb{N}\backslash S_{j}$ is finite). Since this is so for all $j,1\leq
j\leq k$, we must have the finite intersection%
\begin{equation*}
S_{1}\cap \cdots \cap S_{k}\neq \phi .
\end{equation*}%
(Note if $S_{1}\cap \cdots \cap S_{k}$ were empty, then $(S_{1}\cap \cdots
\cap S_{k})^{\prime }=S_{1}^{^{\prime }}\cup \cdots \cup S_{k}^{^{\prime }}=%
\mathbb{N}$ - which is impossible since $S_{1}^{^{\prime }}\cup \cdots \cup
S_{k}^{^{\prime }}$ is a finite union of finite sets.)

Choose $n\in S_{1}\cap \cdots \cap S_{k}$. Then $\psi _{n}(g_{j})\neq 1$ for
all $j$ with $1\leq j\leq k$. Hence $G$ is fully residually free. $%
\blacksquare $

\medskip

\begin{theorem}
\ [KM1] A finitely generated group $G$ is fully residually free if and only
if \ there is a finite rank free group $G_{0}$ and a finite sequence $%
G_{0}\leq G_{1}\leq ...\leq G_{n}$ of free rank 1 centralizer extensions
such that $G$ is isomorphic to a finitely generated subgroup of $G_{n}$.
\end{theorem}

\bigskip

\section{\textbf{The Burnside Varieties}}

Let $n$\ be an Adian-Sirvanjan integer. Then $F_{\omega }(\mathbf{B}_{n})$
embeds in $F_{2}(\mathbf{B}_{n})$. So, if $2\leq r\leq \omega $, then $F_{2}(%
\mathbf{B}_{n})$ embeds in $F_{r}(\mathbf{B}_{n})$ which, in turn, embeds in 
$F_{2}(\mathbf{B}_{n})$. It follows that $F_{r}(\mathbf{B}_{n})\equiv
_{\forall }F_{2}(\mathbf{B}_{n})$ for all $2\leq r\leq \omega $. Thus $2$ is
the index of discrimination of $\mathbf{B}_{n}$. Moreover, since the
centralizer of every nontrivial element in $F_{r}(\mathbf{B}_{n})$ is
cyclic, the relatively free groups $F_{r}(\mathbf{B}_{n})$ are CT. Now
suppose that $p$ is an Adian-Sirvanjan prime.

\bigskip

\begin{lemma}
Suppose $G\in \mathbf{B}_{p}$ and $G$ is $C_{p}\times C_{p}$ exclusive. If $%
G $ is CT, then $G$ is CSA.
\end{lemma}

\noindent \textit{Proof: \ }Suppose $G\in \mathbf{B}_{p}$, $G$ is $%
C_{p}\times C_{p}$ exclusive and $G$ is CT. Let $a\in G\backslash \{1\}$.
Then $C_{G}(a)=\langle a\rangle \cong C_{p}$. Suppose $g^{-1}bg\in \langle
a\rangle $ for some $1\neq b\in \langle a\rangle $. Since $\langle a\rangle
=\langle b\rangle $ we may assume $b=a$. Then $g^{-1}ag=a^{m}$ and $%
a=g^{-p}ag^{p}=a^{m^{p}}$. But we may compute exponents modulo $p$ and, by
Fermat's Little Theorem, $a^{m^{p}}=a^{m}$. Therefore, $a^{m}=a$ and $%
g^{-1}ag=a$. So $g\in C_{G}(a)=\langle a\rangle $. Hence, $G$ is CSA. $%
\blacksquare $

\begin{corollary}
The relatively free groups $F_{r}(\mathbf{B}_{p})$ are CSA.
\end{corollary}

\noindent \textit{Proof: }Since the centralizer of every nontrivial element
is isomorphic to $C_{p}$, one has that $F_{r}(\mathbf{B}_{p})$ is CT and $%
C_{p}\times C_{p}$ exclusive. $\blacksquare $\textit{\ }

More generally, if $n$ is any Adian-Sirvanjan integer, then the free groups $%
F_{r}(\mathbf{B}_{n})$ are CSA. To see that let $G=F_{r}(\mathbf{B}_{n})$
where $r\geq 2$. Let $a\in G\backslash \{1\}$. Let $C_{G}(a)=\langle
b\rangle $. Suppose $g^{-1}\langle b\rangle g$ intersects $\langle b\rangle $
nontrivially. Since $G$ is CT we must have, in that event, $g^{-1}\langle
b\rangle g=\langle b\rangle $. Consider the subgroup $H=\langle g,b\rangle
\leq G$. Observe $\langle b\rangle $ is normal in $H$ and the quotient $%
H/\langle b\rangle $ is generated by the image of $g$ - an element of finite
order. \ Then $|H|=|\langle b\rangle |[H:\langle b\rangle ]<\infty $. Then $%
H $ must be cyclic. But $C_{G}(a)=\langle b\rangle \leq H$ is maximal
cyclic. Hence, $H=\langle b\rangle $ and $g\in \langle b\rangle =C_{G}(a)$.
It follows that $C_{G}(a)$ is malnormal in $G$. Hence, $G$ is CSA.

\bigskip

\begin{theorem}
Suppose $G\in \mathbf{B}_{p}$ is CT and $C_{p}\times C_{p}$ exclusive. If $G$
is separated by $D(\mathbf{B}_{p})$, then $G$ is discriminated by $D(\mathbf{%
B}_{p})$.
\end{theorem}

\noindent \textit{Proof: \ }The proof will be by induction on the
cardinality $n$ of $S=\{g_{1},...,g_{n}\}\subseteq G\backslash \{1\}$ no
element of which is annihilated by a homomorphism into a group free in $%
\mathbf{B}_{p}$. We have the result for $n=1$ since $G$ is separated by $D(%
\mathbf{B}_{p})$. Now suppose $n>1$ and the result is true for all $k\,\,$%
with $1\leq k<n$. Assume first that $S$ is contained in an abelian subgroup
of $G$. Then $\langle g_{1}\rangle =...=\langle g_{n}\rangle \cong C_{p}$
and so $g_{i}=g_{1}^{m_{i}}$ where $p$ does not divide $m_{i}$ for $2\leq
i\leq n$. Now, since $D(\mathbf{B}_{p})$ separates $G$, there is $r\geq 2$
and a homomorphism $\psi :G\rightarrow F_{r}(\mathbf{B}_{p})$ such that $%
\psi (g_{1})\neq 1$. But then $\psi (g_{i})=\psi (g_{1})^{m_{i}}\neq 1$ for
all $2\leq i\leq n$. Hence, $\psi $ does not annihilate any element of $S$.
Now suppose at least one pair of elements of $S$ does not commute. We may
assume that $g_{n-1}$ and $g_{n}$ do not commute. Then $T=%
\{g_{1},...,g_{n-2},[g_{n-1},g_{n}]\}$ is contained in $G\backslash \{1\}$
By inductive hypothesis there is $r\geq 2$ and a homomorphism $\psi
:G\rightarrow F_{r}(\mathbf{B}_{p})$ such that $\psi $ does annihilate any
element of $T$. But then $\psi $ does not annihilate any element of $S$
either. That completes the induction. $\blacksquare $

\bigskip

Observe that, since $F_{2}(\mathbf{B}_{p})$ satisfies the universal sentence 
$\forall x,y(((x\neq 1)\wedge (xy=yx))\rightarrow \vee
_{k=0}^{p-1}(y=x^{k})) $, every group $G\equiv _{\forall }F_{2}(\mathbf{B}%
_{p})$ is $C_{p}\times C_{p}$ exclusive. More generally, if $n$ is any
Adian-Sirvanjan integer and $G\equiv _{\forall }F_{2}(\mathbf{B}_{n})$,
then, for every $g\in G\backslash \{1\}$, one has that $C_{G}(g)$ is cyclic.
To see that consider the universal sentences

(u1) $\forall x_{1},x_{2}((x_{1}x_{2}=x_{2}x_{1})\rightarrow \vee _{0\leq
k_{1},k_{2},m_{1},m_{2}<n}((x_{1}=(x_{1}^{m_{1}}x_{2}^{m_{2}})^{k_{1}})%
\wedge (x_{2}=(x_{1}^{m_{1}}x_{2}^{m_{2}})^{k_{2}}))$ and

(u2) $\forall x_{1},...,x_{n+1}(\wedge
_{i<j}(x_{i}x_{j}=x_{j}x_{i})\rightarrow \vee _{i<j}(x_{i}=x_{j}))$.

\bigskip

Both hold in $F_{2}(\mathbf{B}_{n})$ since every abelian subgroup of $F_{2}(%
\mathbf{B}_{n})$ is cyclic of order at most $n$. Hence they both hold in $G$%
. (u1) asserts that every abelian subgroup is locally cyclic and (u2)
asserts that every abelian subgroup has at most $n$ elements.

Suppose that $n$ is a composite Adian-Sirvanjan integer. Let $1<d<n$ be a
divisor of $n$. Let $B$ be freely generated in $\mathbf{B}_{n}$ by $%
\{a_{1},a_{2},a_{3}\}$ and let $A$ be the subgroup generated (necessarily
freely) by $\{a_{1},a_{2}\}$. Let $G$ be the subgroup of $B$ generated by $%
\{a_{1},a_{2},a_{3}^{\frac{n}{d}}\}$. Since$F_{2}(\mathbf{B}_{n})=$ $A\leq
G\leq B=F_{3}(\mathbf{B}_{n})\equiv _{\forall }F_{2}(\mathbf{B}_{n})$ we
have $G\equiv _{\forall }F_{2}(\mathbf{B}_{n})$. But the finitely generated
group $G$ cannot be free in $\mathbf{B}_{n}$ since its abelianization,
isomorphic to $C_{n}\times C_{n}\times C_{d}$, has order $n^{2}d$ which is
not a power of $n$. For an Adian-Sirvanjan prime $p$ are there any finitely
generated $G\equiv _{\forall }F_{2}(\mathbf{B}_{p})$ which are not free in $%
\mathbf{B}_{p}$? We ponder that question in the next section.

We conclude this section with a proof that cyclicity of finite subgroups is
inherited by models of the universal and existential theory of the free
Burnside groups.

\bigskip

\begin{theorem}
Let $n$ be an Adian-Sirvanjan integer and assume $G\equiv _{\forall }F_{2}(%
\mathbf{B}_{n})$. Then every finite subgroup of $G$ is cyclic.
\end{theorem}

\noindent \textit{Proof: \ }For each positive ineger $N$ the universal
sentence 
\begin{equation*}
\forall x_{1},...,x_{N}(\wedge _{i\leq j}\vee
_{k}(x_{i}x_{j}=x_{k})\rightarrow \wedge _{i<j}(x_{i}x_{j}=x_{j}x_{i}))
\end{equation*}%
holds in $F_{2}(\mathbf{B}_{n})$ since every finite subgroup is abelian.
Thus every finite subgroup of $G$ is abelian. But we have already seen that
every abelian subgroup of such $G$ is cyclic. $\blacksquare $

\section{\textbf{\ A Possible Non-Free Model and a Question of Philip Hall}}

\bigskip

If $G$ is a group and $H\leq G$ let $H^{G}$ be the normal closure of $H$ in $%
G$. If $H,K\leq G$ let $[H,K]$ be the subgroup generated by $%
\{[h,k]:(h,k)\in H\times K\}$. If $G_{1}$ and $G_{2}$ are groups let $%
G_{1}\ast G_{2}$ be their free product. Let $\mathbf{V}$ be a variety of
groups and $G$ be a group. Let $V(G)$ be the intersection of the family of
subgroups $K$ normal in $G$ such that $G/K\in \mathbf{V}$. Then $V(G)$, the 
\textit{verbal subgroup} of $G$ corresponding to $\mathbf{V}$, is fully
invariant in $G$ and is the least normal subgroup $K$ in $G$ such that $%
G/K\in \mathbf{V}$. We define (following Hanna Neumann [N] and Magnus,
Karrass, Solitar [MKS]) the \textit{verbal product }$G_{1}\ast _{\mathbf{V}%
}G_{2}$ as $\Gamma /([G_{1},G_{2}]^{\Gamma }\cap V(\Gamma ))$ where $\Gamma
=G_{1}\ast G_{2}$.

If $G_{1},G_{2}\in \mathbf{V}$ then $G_{1}\ast _{\mathbf{V}}G_{2}\in \mathbf{%
V}$ and each of $G_{1}$ and $G_{2}$ embeds in $G_{1}\ast _{\mathbf{V}}G_{2}$%
. Moreover $\ast _{\mathbf{V}}$ restricted to groups in $\mathbf{V}$ is the
coproduct in $\mathbf{V}$. That means essentially that if $G_{1},G_{2},G\in 
\mathbf{V}$ then every pair of homomorphisms $\psi _{i}:G_{i}\rightarrow G$, 
$i=1,2$, uniquely determines a homomorphism $\psi :G_{1}\ast _{\mathbf{V}%
}G_{2}\rightarrow G$.

Among other results one has that $F_{r}(\mathbf{V})$ is the verbal product
relative to $\mathbf{V}$ of $r$ copies of $F_{1}(\mathbf{V})$. Moreover, if $%
G_{1},G_{2}\in \mathbf{V}$, then each of $G_{1}$ and $G_{2}$ is a retract of 
$G_{1}\ast _{\mathbf{V}}G_{2}$. We observe $G_{1}$ is a retract since if $%
\psi :G_{1}\ast _{\mathbf{V}}G_{2}\rightarrow G_{1}$ is the homomorphism
determined by $\psi _{1}:G_{1}\rightarrow G_{1}$ and $\psi
_{2}:G_{2}\rightarrow G_{1}$ where $\psi _{1}$ is the identity automorphism
and $\psi _{2}$ is the trivial map $\psi _{2}(x)=1$ for all $x$, then $\psi $
is a retraction from $G_{1}\ast _{\mathbf{V}}G_{2}$ onto $G_{1}$. Similarly $%
G_{2}$ is a retract of $G_{1}\ast _{\mathbf{V}}G_{2}$. Furthermore, $%
G_{1}\ast _{\mathbf{V}}G_{2}$ is generated by the embedded images of $G_{1}$
and $G_{2}$ and, if $H_{i}\leq G_{i}$, $i=1,2$, then the subgroup $\langle
H_{1},H_{2}\rangle $ of $G_{1}\ast _{\mathbf{V}}G_{2}$ has the verbal
product decomposition $H_{1}\ast _{\mathbf{V}}H_{2}$.

Now let $p$ be an Adian-Sirvanjan prime. Let $\ast _{p}$denote the verbal
product with respect to the variety $\mathbf{B}_{p}$ and let $\mathbf{K}_{p}$
be the Kostrikin variety of locally finite groups satisfying the law $%
x^{p}=1 $. Let $\kappa _{p}$ be the verbal subgroup operator corresponding
to $\mathbf{K}_{p}$. Suppose $r\geq 2$ is finite. Then $F_{r}(\mathbf{K}%
_{p})=F_{r}(\mathbf{B}_{p})/\kappa _{p}(F_{r}(\mathbf{B}_{p}))$ is a finite
group. \ Since $\kappa _{p}(F_{r}(\mathbf{B}_{p}))$ has finite index in the
finitely generated group $F_{r}(\mathbf{B}_{p})$ it must itself be finitely
generated. \ However, $\kappa _{p}(F_{r}(\mathbf{B}_{p}))$ is perfect (i.e.
it coincides with its commutator subgroup). Therefore it cannot be free in $%
\mathbf{B}_{p}$. This is so since the abelianization of $F_{d}(\mathbf{B}%
_{p})$ for any cardinal $d\geq 1$ is a vector space of dimension $d$ over
the $p$ element field. Now let $F=F_{4}(\mathbf{B}_{p})$ be freely generated
relative to $\mathbf{B}_{p}$ by $a_{1},a_{2},a_{3}$ and $a_{4}$. Let $A$ be
the subgroup generated (necessarily freely) by $a_{3}$ and $a_{4}$ and let $%
B $ be the subgroup generated (necessarily freely) by $a_{1}$ and $a_{2}$.
Let $C=\kappa _{p}(A)$ so that $C$ is finitely generated and perfect. \ $%
F=B\ast _{p}A$. Consider the subgroup $G=\langle B,C\rangle =B\ast _{p}C$.
Now $F_{2}(\mathbf{B}_{p})=B\leq G\leq F=F_{4}(\mathbf{B}_{p})\equiv
_{\forall }F_{2}(\mathbf{B}_{p})$. It follows that the finitely generated
group $G$ is universally equivalent to $F_{2}(\mathbf{B}_{p})$. Observe
that, if $G$ were free in $\mathbf{B}_{p}$, then the retract $C$ of $G$
would be projective relative to $\mathbf{B}_{p}$.

Here we observe a connection with Problem 21 of [N], which problem is
attributed to Philip Hall. The question posed is the following. Suppose $%
\mathbf{V}$ is a variety of groups of exponent zero or prime power. If $P\in 
\mathbf{V}$ is projective relative to $\mathbf{V}$ must $P$ be free in $%
\mathbf{V}$? Note that a positive answer to Philip Hall's question in the
case \ of exponent an Adian-Sirvanjan prime would imply that $G$ is not free
in $\mathbf{B}_{p}$.

Kov\'{a}cs and Newman [KN] report that Philip Hall's question has a negative
answer in the case of exponent zero; however, to the best of our knowledge
the question remains open for prime power exponent. Other conditions would
also imply that $G$ is not free in $\mathbf{B}_{p}$. Suppose we define the
Rank of a group to be the minimum cardinality of a set of generators. A
consequence of the Grushko-Neumann Theorem asserts that $Rank(G_{1}\ast
G_{2})=Rank(G_{1})+Rank(G_{2})$. Now $C=C^{\prime }\leq G^{\prime }$ so $%
G/G^{\prime }\cong C_{p}\times C_{p}$. If $G$ were free it would have rank $%
2 $ and, since it is nonabelian, $Rank(G)=2$ also under the assumption of
freeness. But, if it were the case that $Rank(G_{1}\ast
_{p}G_{2})=Rank(G_{1})+Rank(G_{2})$ for all $G_{1},G_{2}\in \mathbf{B}_{p}$,
we would have $Rank(G)=Rank(B)+Rank(C)=2+Rank(C)$ and $Rank(C)>0$ since $%
C\neq 1$. Thus, if the analog of the Grushko-Neumann corollary holds for $%
\mathbf{B}_{p}$, then $G$ cannot be free in $\mathbf{B}_{p}$. Suppose the
finite rank free groups $F_{r}(\mathbf{B}_{p})$, $2\leq r<\omega $ \ are
Hopfian. As just argued, if $G$ were free in $\mathbf{B}_{p}$, then $%
rank(G)=2$. Say $G$ were freely generated by $b_{1}$and $b_{2}$. Now let $%
\psi :G\rightarrow G$ be the endomorphism determined by $\psi
_{1}:B\rightarrow G$, $\psi _{2}:C\rightarrow G$ where $\psi _{1}$ is the
homomorphism determined by $a_{i}\mapsto b_{i}$, $i=1,2$, and $\psi _{2}$ is
the trivial map $\psi _{2}(x)=1$ for all $x$. Then $\psi $ is an
epi-endomorphism with $1\neq C\leq Ker(\psi )$. That would contradict the
Hopf property. Hence, if the free groups $F_{r}(\mathbf{B}_{p})$, $2\leq
r<\omega $ are Hopfian, then $G$ cannot be free in $\mathbf{B}_{p}$. As far
as we know it also is an open question as to whether or not these free
groups $F_{r}(\mathbf{B}_{p})$ are Hopfian.

\pagebreak

\section{\protect\bigskip \textbf{Questions}}

Let $G$ be a group and let $n$ be a positive integer. Let $\langle
x_{1},...,x_{n};~\rangle $ be free on the $n$ distinct elements $%
x_{1},...,x_{n}$. Let $w\in G\ast \langle x_{1},...,x_{n};~\rangle $. View
the formal expression $w=1$ as an equation over $G$ in the variables $%
x_{1},...,x_{n}$. Now every assignment $x_{i}\mapsto g_{i}\in G$, $i=1,...,n$%
, extends to a unique retraction $\psi :G\ast \langle
x_{1},...,x_{n};~\rangle \rightarrow G$. Call the tuple $(g_{1},...,g_{n})%
\in G^{n}$ a \textit{solution to }$w=1$ provided $w\in Ker(\psi )$. For each
subset $S\subseteq G\ast \langle x_{1},...,x_{n};~\rangle $, let $%
V_{G}(S)\subseteq G^{n}$ be the solution set to the system $w=1$, $w\in S$
of equations. $G$ is \textit{equationally Noetherian} provided for every
positive integer $n$ and every subset $S\subseteq G\ast \langle
x_{1},...,x_{n};~\rangle $ there is a finite subset $S_{0}\subseteq S$ such
that $V_{G}(S)=V_{G}(S_{0})$.

\bigskip

\textbf{Question 1:}\textit{\ If }$n$\textit{\ is an Adian-Sirvanjan integer
and }$r\geq 2$\textit{\ is an integer must }$F_{r}(B_{n})$\textit{\ be
equationally Noetherian?}

\bigskip

\textbf{Question 2}(Philip Hall)\textbf{:}\textit{\ If }$V$\textit{\ is a
variety of groups of prime power exponent and }$P\in V$\textit{\ is
projective relative to }$V$\textit{\ must }$P$\textit{\ be free in }$V$%
\textit{?}

\bigskip

\textbf{Question 3:}\textit{\ Suppose we define the Rank of a group to be
the minimum cardinality of a set of generators. Let }$p$\textit{\ be an
Adian-Sirvanjan prime and let }$\ast _{p}$\textit{\ be the coproduct in }$%
B_{p}$\textit{. If }$G_{1},G_{2}\in B_{p}$\textit{\ must }$Rank(G_{1}\ast
_{p}G_{2})=Rank(G_{1})+Rank(G_{2})$\textit{?}

\bigskip

\textbf{Question 4: }\textit{If }$p$\textit{\ is an Adian-Sirvanjan prime
and }$r\geq 2$\textit{\ is an integer must }$F_{r}(B_{p})$\textit{\ be
Hopfian?}

\textit{\bigskip }

\textbf{Question 5: }\textit{If }$n$ \textit{is an Adian-Sirvanjan integer
and }$H\in \mathbf{B}_{n}$ \textit{is finitley generated and universally
equivalent to }$F_{2}(\mathbf{B}_{n})$ \textit{must }$H$ \textit{be
embeddable in some }$F_{r}(\mathbf{B}_{n})$?

\bigskip

\textbf{Question 6: }\textit{If }$n$\textit{\ is an Adian-Sirvanjan integer
and }$2\leq r<s\leq \omega $\textit{\ must }$F_{r}(B_{n})\equiv F_{s}(B_{n})$%
\textit{?}

\bigskip

\section{References}

\noindent \lbrack A] S.I. Adian,"Classification of periodic words and their
application in group theory," \textit{Springer Lecture Notes in Mathematics
806, Burnside Groups}, J.L. Mennicke, Editor, Springer-Verlag, Berlin
(1980), 1 -40.

\bigskip

[B] B. Baumslag,"Residually free groups," \textit{Proc. London Math. Soc.
(3) 17} (1967), 402 - 418.

\bigskip

[GB] G. Baumslag, \textquotedblleft On generalised free
products,\textquotedblright\ \textit{Math. Z. }78 (1962), 423-438.

\bigskip

[BMR] G. Baumslag, A.G. Myasnikov and V.N. Remeslennikov,"Algebraic geometry
over groups I. Algebraic sets and ideal theory," \textit{J. Alg. 219}
(1999), 16 - 79.

\bigskip

[G] G. Gr\"{a}tzer, \textit{Universal Algebra}, Van Nostrand, Princeton
(1968).

\bigskip

[GS] A.M. Gaglione and D. Spellman,"The persistence of universal formulae in
free algebras," \textit{Bull. Austral. Math. Soc. 36 }(1987), 11 - 17.

\bigskip

[K] A.I. Kostrikin,"The Burnside problem," \textit{Izv. akad. Nauk. Ser.
Mat. 23 }(1959), 3 - 34.

\bigskip

[KM1] O. Kharlampovich and A.G. Myasnikov,"Irreducible affine varieties over
a free group: II Systems in quasi-quadratic triangular form and description
of residually free groups," \textit{J. Alg. 200 }(1998), 517 - 570.

\bigskip

[KM2] O. Kharlampovich and A.G. Myasnikov,"Tarski's problem about the
elementary theory of free groups has a positive solution," \textit{Electron.
Res. Announc. Amer. Math. Soc. 4 }(December 1998), 101 - 108.

\bigskip

[KN] L.G. Kov\'{a}cs and M.F. Newman,"Hanna Neumann's problems on varieties
of groups," \textit{Springer Lecture Notes in Mathematics 372, Proc.
Internat. Conf. Theory of Groups, Canberra} (1973), 417 - 431.

\bigskip

[MKS] W.Magnus, A. Karrass and D. Solitar, \textit{Combinatorial Group Theory%
}, Dover, Mineola (2004).

\bigskip

[N] H. Neumann, \textit{Varieties of Groups}, Springer - Verlag, New York
(1967).

\bigskip

[Ne] P.M. Neumann,"Splitting groups and projectives in varieties of groups," 
\textit{QJ Maths (Oxford) 18} (1967), 325 - 352.

\bigskip

[R] V.N. Remeslennikov,"$\exists $-Free groups," \textit{Siberian J. Math.
30 (6) }(1989), 153 - 157.

\bigskip

[Se] Z. Sela,"Diophantine geometry over groups VI: The elementary theory of
a free group," \textit{GAFA}, in press.

\bigskip

[Si] V.L. Sirvanjan,"Imbedding of the group $B(\infty ,n)$ in the group $%
B(2,n)$," \textit{Math. USSR-Izv. 4} (1977), 181 - 199.