If the free algebra
on one
generator in a variety
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
generates
the variety
.
Generalizing the argument, it is shown that if we are given an algebra and subalgebras,
, in a prevariety
(-closed class of
algebras)
such
that
generates
, and also
subalgebras
such that
for each
the
subalgebra of
generated by
and
is their coproduct
in
, then the
subalgebra of
generated by
is
the coproduct in
of these algebras.
Some further results on coproducts are noted:
If
satisfies
the amalgamation property, then one has the stronger “transitivity” statement, that if
has a finite family of
subalgebras
such that the
subalgebra of
generated
by the
is their coproduct,
and each
has a finite
family of subalgebras
with the same property, then the subalgebra of
generated
by all the
is their coproduct.
For
a
residually small prevariety or an arbitrary quasivariety, relationships
are proved between the least number of algebras needed to generate
as
a prevariety or quasivariety, and behavior of the coproduct operation in
.
It is shown by example that for
a subgroup of the group
of all
permutations of an infinite set
,
the group
need not have a subgroup isomorphic over
to the coproduct
with amalgamation
.
But under various additional hypotheses on
, the
question remains open.
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