Algebra Number Theory 3 (8), 847-879, (2009) DOI: 10.2140/ant.2009.3.847
KEYWORDS: coproduct of algebras in a variety or quasivariety or prevariety, free algebra on $n$ generators containing a subalgebra free on more than $n$ generators, amalgamation property, number of algebras needed to generate a quasivariety or prevariety, symmetric group on an infinite set, 08B25, 08B26, 08C15, 03C05, 08A60, 08B20, 20M30

If the free algebra $F$ on one generator in a variety $V$ of algebras (in the sense of universal algebra) has a subalgebra free on two generators, must it also have a subalgebra free on three generators? In general, no; but yes if $F$ generates the variety $V$.

Generalizing the argument, it is shown that if we are given an algebra and subalgebras, ${A}_{0}\supseteq \cdots \supseteq {A}_{n}$, in a prevariety ($\mathbb{S}\mathbb{P}$-closed class of algebras) $P$ such that ${A}_{n}$ generates $P$, and also subalgebras ${B}_{i}\subseteq {A}_{i-1}$ $\left(0<i\le n\right)$ such that for each $i>0$ the subalgebra of ${A}_{i-1}$ generated by ${A}_{i}$ and ${B}_{i}$ is their coproduct in $P$, then the subalgebra of $A$ generated by ${B}_{1},\dots ,{B}_{n}$ is the coproduct in $P$ of these algebras.

Some further results on coproducts are noted:

If $P$ satisfies the amalgamation property, then one has the stronger “transitivity” statement, that if $A$ has a finite family of subalgebras ${\left({B}_{i}\right)}_{i\in I}$ such that the subalgebra of $A$ generated by the ${B}_{i}$ is their coproduct, and each ${B}_{i}$ has a finite family of subalgebras ${\left({C}_{ij}\right)}_{j\in {J}_{i}}$ with the same property, then the subalgebra of $A$ generated by all the ${C}_{ij}$ is their coproduct.

For $P$ a residually small prevariety or an arbitrary quasivariety, relationships are proved between the least number of algebras needed to generate $P$ as a prevariety or quasivariety, and behavior of the coproduct operation in $P$.

It is shown by example that for $B$ a subgroup of the group $S=Sym\left(\Omega \right)$ of all permutations of an infinite set $\Omega $, the group $S$ need not have a subgroup isomorphic over $B$ to the coproduct with amalgamation $S\phantom{\rule{0.3em}{0ex}}{\coprod}_{B}S$. But under various additional hypotheses on $B$, the question remains open.