  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 ... \supseteq A_n, in a prevariety (SP-closed class
of algebras) P such that A_n generates P, and also subalgebras B_i\subseteq
A_{i-1} (0<i\leq n) 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, ..., B_n is the coproduct in P of these algebras.
  Some further results on coproducts are noted:
  If P satisfies the amalgamation property, one has the stronger "transitivity"
statement: if A has a finite family of subalgebras (B_i)_{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 (C_{ij})_{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(\Omega) 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 \cP_B S. But
under weak additional hypotheses on B, the question remains open.
