3} from free algebras to general
coproducts, we found that the context that made the
argument work was that of coproducts in a {\em prevariety}.
Theorem~\ref{T.P-indp} does not give us the corresponding statement
for general $A_0,\dots,A_n,\,B_1,\dots,B_n$
with the prevariety $\Pv$ replaced by a variety $\V,$ and
the condition that $\Pv$ be generated by $A_n$ as a prevariety replaced
by the condition that $\V$ be generated by $A_n$ as a variety,
since for a variety $\V$ and an algebra $A\in\V,$
the condition that $\V$ be generated
by $A$ as a variety is weaker than the condition that
it be generated by $A$ as a prevariety.
(E.g., the variety of abelian groups is generated by the
infinite cyclic group as a variety, but not as a prevariety, since all
groups in the prevariety it generates must be torsion-free.)
But I don't have a counterexample to the modified statement.
\begin{question}\label{Q.variety}
\textup{(i)} \ Does Theorem~\ref{T.P-indp} remain true if
``prevariety'' is everywhere replaced by ``variety''?
\vspace{.2em}
If not, or if the question proves difficult,
one might examine some special cases, such as\\[.2em]
\textup{(ii)} \ If in Corollary~\ref{C.A_i=A}
we replace ``prevariety'' by ``variety'',
and add the assumption that $A$ is free of rank~$1$ in that variety
\textup{(}but not that the $B_i$ are free\textup{),}
does the statement still hold?\\[.2em]
\textup{(iii)} \ If $\V$ is a variety, and $A$ an algebra that
generates~$\V$ as a variety, and that contains as a
subalgebra a coproduct of two copies of itself in~$\V,$ must
it contain a coproduct of three copies of itself in~$\V$?
\end{question}
More likely to have positive answers, since quasivarieties
are more like prevarieties than varieties are, is
\begin{question}\label{Q.qv}
Same questions \textup{(i), (ii), (iii)} as above, but with
``variety'' everywhere replaced by ``quasivariety''
\textup{(}necessarily, of finitary algebras\textup{)}.
\end{question}
Looking back further, to \S\ref{S.free},
the mixture of positive and negative results there suggests
\begin{question}\label{Q.algorithm}
Is there a nice criterion for whether
a $\!3\!$-tuple $(a,\**3}
to hold for the $\!3\!$-tuple $(\**1$ and $N>1,$ one may ask
for a criterion for an $\!N\!$-tuple $(a_1,\dots,a_N)$
of words in $n$ letters $p_1,\dots,p_n$ to witness the existence of a
free subalgebra on $N$ generators in any relatively free algebra on one
generator that contains a free subalgebra
$\langle\m>0$ and $N>1,$ one may ask
how to decide whether a given $\!N\!$-tuple of terms in $m$
variables and $n$ operation symbols each of arity $m$ witnesses
the result of Corollary~\ref{C.m1$ are more complicated
than words in unary operation symbols, there seems to be less
likelihood of a simple answer.\textup{)}
\end{question}
The next four sections are related to, but
do not depend on, the material above, except for the definitions.
Section~\ref{S.amalg} recalls what it means for a category
of algebras to have the {\em amalgamation property}, obtains some
equivalent statements, and then shows that for prevarieties
with that property, one has stronger results on when a family
of subalgebras of an algebra generates a subalgebra isomorphic to
their coproduct than those that we have seen to hold in general.
In a different direction, motivated by the fact
that the prevarieties considered in \S\ref{S.P-indp}
were by hypothesis each generated by a single algebra,
\S\S\ref{S.compat1}-\ref{S.comfort} show that the
number of algebras needed to generate a prevariety has important
consequences for the behavior of coproducts therein.
The brief section \S\ref{S.Sym}, which is included in this note
only for convenience, answers a different question about coproducts,
also raised in~\cite{embed}, concerning subgroups of
the full symmetric group on an infinite set.
\section{The amalgamation property, and its consequences for $\!\Pv\!$-independence.}\label{S.amalg}
In any class of algebras that admits coproducts with amalgamation
(pushouts), it is well known and easy to verify
that the {\em amalgamation property}
(definition recalled in~(\ref{d.amalg}) below)
is equivalent to the condition that for all pairs of one-to-one
maps with common domain, $A\to B$ and $A\to C,$ the coprojection
maps of $B$ and $C$ into the coproduct with amalgamation
$B\,\cP_A\,C$ are also one-to-one.
The next lemma gives some further consequences of that
property, in the same vein.
We formulate it in a context more general than that of
categories of algebras, though less sophisticated than
that \mbox{of~\cite[\S6]{KMPT}}.
In that lemma, the functor $U:\mathbf{C}\to\mathbf{Set}$
plays the role of the underlying set functor of a category of
algebras, but we shall not need to assume it faithful,
as one does when defining the concept of a concrete category.
One other notational remark: So far, I have generally written
$\coprod_\Pv^{}$ for ``coproduct in the category $\Pv$''; but when
discussing coproducts with amalgamation of an object, we
will use the subscript position for that object, leaving the
category to be understood from the context.
I will follow this mixed practice for the rest of the paper,
showing the category when no amalgamation is involved.
(The superscript position, which might otherwise be assigned to one
of these, is used here for index sets over which coproducts are taken.
If there were danger of ambiguity, we could write
$\coprod_{\Pv,\1),$ the conditions of Theorem~\ref{T.absdird}
hold, but $C_d$ satisfies $(\forall\,x,\,y,\,z)
\linebreak[0]\ ax\,{=}\,x{\implies}\linebreak[0]y\,{=}\,z,$ so the
trivial algebra
is not $\!\Pv\!$-compatible with any nontrivial algebra.
There are also examples where~(\ref{d.trivinnontriv})
holds, but where the equivalent conditions of
Theorem~\ref{T.absdird} and Corollary~\ref{C.dirdwtriv}
do not; for example, the variety of groups
or monoids with one distinguished element, or of lattices with two
distinguished elements: a counterexample to~(\ref{d.allcompatwtriv})
is given by any group or monoid with distinguished element that is
not the identity, or any lattice with a pair of distinguished
elements that are not equal.
These same examples also show that the analog of the
implication (\ref{d.qv1gen})$\!\implies\!$(\ref{d.setcompat}) does not
hold for varieties, with ``generated as a variety'' in place of
``generated as a quasivariety'', since every variety
satisfies the analog of~(\ref{d.qv1gen}).\vspace{.2em}
The next corollary is a result promised in
the comment following Lemma~\ref{L.dirsys}.
\begin{corollary}\label{C.subindep}
Suppose $\Pv$ is a prevariety generated by a single algebra,
or, more generally, satisfying\textup{~(\ref{d.absdirgen})},
and let $(B_i)_{i\in I}$ be a family of nontrivial algebras in $\Pv.$
\textup{(}Again, if $\Pv$
satisfies\textup{~(\ref{d.trivinnontriv}),}
the restriction ``nontrivial'' can be dropped.\textup{)}
Then
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.JtoI11}
For every $J\subseteq I,$ the natural map
$\coprod_\Pv^{i\in J}B_i\to\coprod_\Pv^{i\in I}B_i$ is one-to-one.
\end{minipage}\end{equation}
%
Hence
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.subfamilyindp}
Any subfamily of a $\!\Pv\!$-independent family of
subalgebras of a nontrivial algebra $A$ is\\
$\!\Pv\!$-independent.
\end{minipage}\end{equation}
%
\end{corollary}
\begin{proof}
To see~(\ref{d.JtoI11}), note that $\coprod_\Pv^{i\in I}B_i\cong
(\coprod_\Pv^{i\in J}B_i)\,\cP_\Pv\,(\coprod_\Pv^{i\in I-J}B_i),$ with
the natural map $\coprod_\Pv^{i\in J}B_i\to\coprod_\Pv^{i\in I}B_i$
corresponding to the first coprojection under this decomposition.
By the implication (\ref{d.absdirgen})$\Rightarrow$(\ref{d.setcompat})
(or its modified version
if $\Pv$ satisfies~(\ref{d.trivinnontriv})),
the indicated coproducts over $J$ and $I-J$ are $\!\Pv\!$-compatible,
hence that coprojection map is one-to-one, as claimed.
(A slight hiccup in this argument: If $J$ or $I-J$ is empty,
can we be sure the coproduct over that subset, namely the
initial algebra, is nontrivial?
No, but if it is trivial, and if $\Pv$ is not the trivial prevariety,
then since the initial algebra can be mapped into every
algebra,~(\ref{d.trivinnontriv}) holds, and so we
are in the case where we don't need nontriviality.)
To get~(\ref{d.subfamilyindp}), recall that the statement that
$(B_i)_{i\in I}$ is an independent family of subalgebras
of $A$ means that the subalgebra of $A$
generated by these subalgebras can be identified with their coproduct.
If none of the $B_i$ is trivial, then this observation together
with~(\ref{d.JtoI11}) immediately gives the desired conclusion.
If at least one of the $B_i$ is trivial, then since by assumption
$A$ is not,~(\ref{d.trivinnontriv}) holds, and by the parenthetical
addendum to the first part of this corollary,
we again have~(\ref{d.JtoI11}) and can proceed as before.
\end{proof}
% In the final statement of the above result, we can delete
% the condition that $A$ be nontrivial if we replace ``subfamily''
% by ``nonempty subfamily''.
% If we do not make that adjustment, then the case where $\Pv$ is the
% variety of unital associative algebras over a fixed
% field $k,$ $A$ is the zero algebra, $(B_i)_I$ consists
% of that same algebra listed one or more times, and $J$ is
% the empty subset of the index-set $I,$ is a counterexample.
% (The coproduct of the empty subfamily of $(B_i)_{i\in I}$
% is the initial algebra $k$ of $\Pv,$ which maps
% noninjectively into~$A.)$\vspace{.2em}
%
% We have proved essentially the same
% conclusion,~(\ref{d.subfamilyindp}),
% under two different hypotheses in Corollaries~\ref{C.indincind}
% and~\ref{C.subindep}, so it is natural to ask whether one of these
% hypotheses implies the other.
% The answer is not quite as simple as that.
% Let $\Pv$ be a prevariety, and $\mathbf{S}$ the set of
% quotient algebras of the initial algebra of $\Pv.$
% If $\Pv$ satisfies the amalgamation property (or more
% generally, if it satisfies~(\ref{d.SAB,ST}) with $S$ restricted
% to run over $\mathbf{S}),$ then for each $S\in\mathbf{S},$
% the subprevariety $\Pv_S$ consisting of the trivial algebra and all
% algebras in which the image of the initial object of $\Pv$ is
% isomorphic to $S$ will satisfy~(\ref{d.2compat}) (though
% $\Pv$ itself need not).
% As in the proof of Corollary~\ref{C.indincind}, this implies
% that $\Pv_S$ satisfies~(\ref{d.JtoI11}) for all nonempty $J;$ and we
% get~(\ref{d.subfamilyindp}) on taking for $S$ in this statement the
% image of the initial algebra in $A.$
% (In contrast to the situation of Corollary~\ref{C.indincind},
% we do not have to exclude empty $J$ here, since the empty coproduct
% in $\Pv_S$ is $S,$ which by assumption embeds in $A.)$
%
% One does not have an implication in the opposite direction,
% from the conditions of this section to,
% say~(\ref{d.SAB,ST}) with $S$ restricted to run over $\mathbf{S}.$
% Indeed, the varieties of Propositions~\ref{P.2_not_3}
% and~\ref{P.not_px,pqx,qqx} both satisfy~(\ref{d.trivinnontriv})
% and~(\ref{d.allcompatwtriv}), since every algebra has
% the trivial subalgebra $\{0\};$ but the force of those propositions
% is that those varieties do not satisfy the conclusion
% of Corollary~\ref{C.cPincP}, though we saw that that conclusion holds
% whenever~(\ref{d.SAB,ST}) holds for $S$ running
% over~$\mathbf{S}.$\vspace{.2em}
%
% On a different topic, we introduced Theorem~\ref{T.absdird} as
% a generalization of Theorem~\ref{T.compat1} to
% non-residually-small prevarieties in the case $\kappa=1.$
% Are there corresponding generalizations for higher $\kappa$?
% A natural common generalization
% of~(\ref{d.*kgen}) and~(\ref{d.absdirgen}) would be
% %
% \begin{equation}\begin{minipage}[c]{35pc}\label{d.*kabsdirgen}
% $\Pv$ is generated as a prevariety by a union of $\leq\kappa$
% absolutely directed classes of algebras.
% \end{minipage}\end{equation}
%
% An analogous common generalization of~(\ref{d.*kdecomp})
% and~(\ref{d.setcompat}) (a sort of ``(\ref{d.*kdecomp}) without the
% assumption that there are enough subdirectly irreducible algebras'')
% might say that given any set $\X$ of algebras in $\Pv,$
% one can choose a subdirect decomposition in $\Pv$ of each member of
% $\X,$ and collect the factor algebras obtained from all these
% into $\leq\kappa$ sets, each of which is $\!\Pv\!$-compatible.
% This easily follows from~(\ref{d.*kabsdirgen}),
% but I don't know whether the converse is true.
% A statement that {\em is} equivalent to~(\ref{d.*kabsdirgen})
% is that $\Pv$ is the join of $\leq\kappa$ subprevarieties
% each satisfying~(\ref{d.setcompat}); but this seems
% an uninsightful corollary of the existing theorem.
%
% For quasivarieties (which we know always have ``enough subdirectly
% irreducible algebras''), we can do somewhat better:
On a different topic, let us note
the extent to which Theorem~\ref{T.compat1} does and does
not go over from prevarieties to quasivarieties.
\begin{corollary}\label{C.qv*k}
If $\Pv$ is a {\em quasivariety}, then \textup{(}even without the
residual smallness assumption of Theorem~\ref{T.compat1}\textup{)},
condition\textup{~(\ref{d.*kdecomp})} implies
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.qv*k}
$\Pv$ can be generated as a quasivariety by $\leq\kappa$ algebras.
\end{minipage}\end{equation}
%
The reverse implication holds
if $\kappa$ is finite, but not for any infinite $\kappa.$
\end{corollary}
\begin{proof}
(\ref{d.*kdecomp})$\Rightarrow$(\ref{d.qv*k}): A quasivariety
$\Pv$ is generated as a prevariety, and hence as a quasivariety,
by its subdirectly irreducible algebras
\cite[Theorem~3.1.1]{VAG}, hence, as in
the proof of Theorem~\ref{T.absdird}, we can find a {\em set}
$\X$ of these that generates it as a quasivariety.
By~(\ref{d.*kdecomp}) we can write $\X$ as
$\bigcup_{\alpha\in\kappa}\X_\alpha$ where each $\X_\alpha$
is $\!\Pv\!$-compatible.
If for each $\alpha\in\kappa$ we let $A_\alpha$ be an algebra in $\Pv$
containing embedded images of all members of $\X_\alpha,$ then $\Pv$
is generated as a quasivariety by this set of $\kappa$ algebras.
For the converse assertion when $\kappa$ is a natural number $n,$
let $\Pv$ be generated as a quasivariety by $A_1,\dots,A_n.$
Then $\Pv=\Su\Pr\Pr_\mathrm{\!ult}\{A_1,\dots,A_n\},$ where
$\Pr_\mathrm{\!ult}$
denotes closure under ultraproducts; thus, each $\!\Pv\!$-subdirectly
irreducible object of $\Pv$ is embeddable in a member
of $\Pr_\mathrm{\!ult}\{A_1,\dots,A_n\}.$
Moreover, the operator $\Pr_\mathrm{\!ult}$ respects finite
decompositions; that is, any ultraproduct of a family of structures
indexed by a finite union of sets, $(A_i)_{i\in I_1\cup\dots\cup I_n},$
can be written as an ultraproduct of one of the subfamilies
$(A_i)_{i\in I_m}.$
Hence $\Pr_\mathrm{\!ult}\{A_1,\dots,A_n\}=
\Pr_\mathrm{\!ult}\{A_1\}\cup\dots\cup\Pr_\mathrm{\!ult}\{A_n\}.$
The class of subdirectly irreducible algebras in $\Pv$ that are
embeddable in members of a given $\Pr_\mathrm{\!ult}\{A_i\}$
is contained in the $\!1\!$-generator quasivariety
$\Su\Pr\Pr_\mathrm{\!ult}\{A_i\},$ hence by the implication
(\ref{d.qv1gen})$\Rightarrow$(\ref{d.setcompat}), each of these $n$
classes has the property that all its subsets are $\!\Pv\!$-compatible.
This gives~(\ref{d.*kdecomp}).
On the other hand, given any infinite $\kappa,$ let $\V$ be the
variety of sets given with a $\!\kappa\!$-tuple of zeroary operations
(constants) $c_\alpha$ $(\alpha\in\kappa).$
Since as a quasivariety, $\V$ is generated by its finitely
presented objects \cite[Proposition~2.1.18]{VAG},
and there are only $\kappa$ of these, it satisfies~(\ref{d.qv*k}).
On the other hand, since all operations are zeroary, every
equivalence relation on a $\!\V\!$-algebra is a congruence,
so the subdirectly irreducible algebras are precisely
the $\!2\!$-element algebras.
We have one of these for each $\!2\!$-class equivalence
relation on $\kappa,$ and one more corresponding to the
partition of that set into $\kappa$ and $\emptyset.$
This gives $2^{\kappa}$ subdirectly irreducible algebras,
no two of which are $\!\V\!$-compatible.
\end{proof}
\section{Afterthoughts on $\!\Pv\!$-compatible algebras.}\label{S.comfort}
Perhaps the concept of ``$\Pv\!$-compatible algebras''
is not the best handle on the phenomena we have been examining;
or at least should be complemented by another way of looking at them.
Suppose that for algebras $A$ and $B$ in $\Pv,$ we say that $A$
is ``comfortable'' with $B$ in $\Pv$ if the coprojection
$A\to A\cP_\Pv\,B$ is one-to-one; equivalently, if $A$ is
$\!\Pv\!$-compatible with some homomorphic image of $B$ in $\Pv.$
This relation is not in general symmetric; e.g., in the
variety of associative unital rings, each ring $\mathbb{Z}/n\mathbb{Z}$
is comfortable with $\mathbb{Z},$ but not vice versa.
Algebras $A$ and $B$ are $\!\Pv\!$-compatible if and only if each is
comfortable with the other.
(So the relations of $\!\Pv\!$-compatibility and of being
comfortable in $\Pv$ may be characterized in terms of one another.)
More generally, an arbitrary family of
algebras is $\!\Pv\!$-compatible if and only if each
is comfortable with the coproduct of the others.
If $\Pv$ is generated as a prevariety by a class of algebras $\X,$ we
see that an algebra $A$ is comfortable in $\Pv$ with an
algebra $B$ if and only if homomorphisms into members
of $\X$ that contain homomorphic images of $B$ separate points of $A.$
Hence, if we classify algebras $B\in\Pv$ according to which
algebras $A$ are comfortable with them,
then algebras $B_1$ and $B_2$ will belong to the same equivalence
class under this relation if the subclass of $\X$
consisting of algebras containing homomorphic images of $B_1$
coincides with the subclass of those containing images of $B_2.$
(We do not assert the converse.)
In particular, if $\Pv$ is residually small, so that
$\X$ can be taken to be a set, the number of these equivalence classes
has the cardinality of a set.
More generally, if $\Pv$ is generated by the union of $\kappa$ classes
of algebras, each absolutely directed under the relation $\preceq$ of
Theorem~\ref{T.absdird}, the same reasoning shows that it will have
at most $2^\kappa$ equivalence classes under this equivalence relation.
On the other hand, if we classify algebras according to which
other algebras they {\em are comfortable with}, we may, so far as I can
see, get up to $2^{2^\kappa}$ classes.
For any algebra $A$ in $\Pv,$ the class of algebras
which are comfortable with $A$ forms a subprevariety of $\Pv.$
The class of algebras that $A$ is comfortable with likewise yields
a subprevariety on throwing in the trivial algebra.
(A stronger statement, also easy to
see, is that this class is closed under
taking subalgebras and under taking products with arbitrary algebras
in $\Pv;$ equivalently, that if this class contains an algebra $B,$ then
it contains every algebra in $\Pv$ admitting a homomorphism to $B.)$
\section{On infinite symmetric groups: an answer and a question.}\label{S.Sym}
This last section does not depend on any of the preceding material.
It was shown in \cite{debruijn} (cf.~\cite{embed})
that for $\Omega$ an infinite set,
the group $S=\mathrm{Sym}(\Omega)$ of all permutations of $\Omega$
contains a coproduct of two copies of itself (from which it was
deduced by other properties of that group that it
contains a coproduct of $2^{\mathrm{card}(\Omega)}$ copies of itself).
In \cite[Question~4.4]{embed}, I asked, inter alia, whether, for every
subgroup $B$ of $S,$ if we regard $S$ as
a member of the variety of groups given with homomorphisms of $B$ into
them, $S$ contains a coproduct of two copies of itself
{\em in that variety}.
The answer is negative.
To see this, pick any $x\in\Omega$ and let $B$ be the
stabilizer in $S$ of $x.$
Writing elements of $S$ to the left of
their arguments and composing them accordingly, we see that
the partition of $S$ into left cosets of $B$
classifies elements according to where they send $x,$ and that
for each $y\in\Omega,$ the coset sending $x$ to $y$
has elements of finite order; e.g.,
if $y\neq x,$ the $\!2\!$-cycle interchanging $x$ and $y.$
On the other hand, I claim that if
$S_1$ and $S_2$ are any two groups with a common subgroup $B$
proper in each, then in the coproduct with amalgamation $S_1\cP_B\,S_2$
there are left cosets of $B$ containing no elements of finite order.
Indeed, the standard normal form in that coproduct shows that
each left coset is generated by a possibly empty alternating
string of left coset representatives of $B$ in $S_1$ and $S_2.$
When that string is nonempty and has even length,
one sees that elements of finite order cannot occur.
Hence for $S=\mathrm{Sym}(\Omega)$ and $B$ as above, $S,$
as a group containing $B,$ cannot contain a copy of $S_1\cP_B\,S_2.$
So let us modify our earlier question.
\begin{question}\label{Q.B~~ E(mbedding) P(roperty) (Mal'cev)
% 68 "local embeddability"
% (84-85: "color family": {mappable into member of finite list})
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residual smallness, and injectivity},
Studia Sci. Math. Hungar. {\bf 18} (1982) 79--140.
~MR~{\bf 85k}:18003.
\bibitem{HNq} L.\,G.\,Kov\'{a}cs and M.\,F.\,Newman,
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pp. 417--431 in
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SLNM, v.\,372, Springer, 1974.
MR~{\bf 50}\#4750.
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Springer GTM, v.5,~1971.
MR~{\bf 50}\#7275.
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Algebra i Logika Sem. {\bf 5} (1966) 3--9.
~MR~{\bf 34}\#5728.
\bibitem{M+M+T} Ralph N. McKenzie, George F. McNulty and
Walter F. Taylor, {\em Algebras, Lattices, Varieties. Vol. I},
Wadsworth \& Brooks/Cole, 1987.
~MR~{\bf 88e}:08001.
\bibitem{HN} Hanna Neumann,
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Ergebnisse der Mathematik und ihrer Grenzgebiete,
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MR~{\bf 35}\#6734.
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{\em On isomorphic free algebras},
Fund. Math. {\bf 50} (1961/1962) 35--44.
MR~{\bf 25}\#2017.
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MR~{\bf 2008m}:06011.
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\end{thebibliography}
\end{document}
~~