A group is said SQ-universal whenever every countable group can be embedded into a quotient of
. In some sense, SQ-universality is a “largeness property”. Motivating by this idea, we prove the two following properties:
Property 1: A SQ-universal group contain a non-abelian free group (of countable rank).
Proof. Let be a SQ-universal group. In particular, there exists a quotient
containing a free group of infinite rank
. Let
be a lift of
in
. If there existed a non-trivial relation between the
‘s then the same relation would hold between the
‘s in
. Therefore,
is a free subgroup of infinite rank in
.
Property 2: A countable SQ-universal group contains uncountably many normal subgroups.
Proof. Let be a countable SQ-universal group. Every quotient of
contains only countably many two-generated subgroups. However, we know that there exist uncountably many two-generated groups up to isomorphism: see for instance the notes Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann or Cantor-Bendixson rank in group theory: A theorem of B.H. Neumann. Therefore,
must have uncountably many quotients, and a fortiori uncountably many normal subgroups.
Although SQ-universality seems to be a very strong property, many groups turn out to be SQ-universal. We already saw in previous notes that the free group of rank two is SQ-universal; in fact, we gave two proofs, the one in Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann, and the other, completely elementary, in A free group contains a free group of any rank. Other examples come from the following trivial observation: a group with a SQ-universal quotient is itself SQ-universal. Therefore, any group with a non-abelian free quotient is SQ-universal; in particular, we deduce that any non-abelian free group, of any rank, is SQ-universal. Together with the following result, we will be able to exhib a lot of other examples of SQ-universal groups.
Theorem 3: A group containing a SQ-universal subgroup of finite-index is itself SQ-universal.
For an (almost) elementary proof, see P. Neumann’s article, The SQ-universality of some finitely presented groups, J. Aust. Math. Soc. 16, 1 (1973) .
Example 1: Let be a closed surface. Then either
is SQ-universal or
is a sphere
, a projective plane
, a torus
or a Klein bottle
.
Notice that we already dealt with surface groups in the previous note On subgroups of surface groups. In particular, we know that ,
,
are not SQ-universal. We also know that
contains
as a subgroup of finite-index two, so it cannot be SQ-universel.
From now on, let be a closed surface different from the surfaces mentioned above. If
is non-orientable, there exist a finite-cover
where
is orientable; in particular,
is a subgroup of finite-index in
, so, according to Theorem 3, we may suppose that
is orientable, that is
for some . Noticing that the following quotient is free,
,
we deduce that is SQ-universal.
Example 2: A right-angled Artin groups is either free abelian or SQ-universal.
The right-angled Artin group associated to a finite simplicial graph is defined by the presentation
,
where and
denote respectively the set of vertices and edges of
. In particular, if
is a complete graph with
vertices, then
is the free abelian group
; if
is a graph with
vertices and no edges, then
is the free group
.
For another example, if is the graph given above, then
is isomorphic to .
We claim that, if is not a complete graph, then
has a non-abelian free quotient, and in particular is a SQ-universal group.
Because is not complete, there exist two vertices
not linked by an edge. Let
denote the subgraph induced by
and its neighbors, and
the subgraph induced by
. Then
and
are non-empty, since
and
, they cover
, and
is not connected, because
and
are separated by
. Now, quotienting
by the subgroup
,
we get the free group . For an explicit example, let us consider the following graph:
So, the associated right-angled Artin group is
.
Quotienting by the subgroup , we get the free group
.
Example 3: If and
are two finite groups with
and
, then the free product
is a SQ-universal group.
We proved in the previous note A free group contains a free group of any rank that
is a free basis of the derived subgroup of
. Furthermore, the quotient
is finite, so that
is a non-abelian free subgroup of finite index in
. From Theorem 3, we deduce that
is SQ-universal.
For example, is SQ-universal, because it has the SQ-universal group
as a quotient. (See An elementary application of ping-pong lemma for an elementary proof of this fact.)