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.)