On coproducts in varieties, quasivarieties and prevarieties

Abstract

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 $S{\coprod }_{B}S$. But under various additional hypotheses on $B$, the question remains open.