A group G is said SQ-universal whenever every countable group can be embedded into a quotient of G. 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 G be a SQ-universal group. In particular, there exists a quotient \overline{G} containing a free group of infinite rank \langle \overline{x_1}, \overline{x_2}, \ldots \rangle. Let x_i be a lift of \overline{ x_i } in G. If there existed a non-trivial relation between the x_i‘s then the same relation would hold between the \overline{x_i}‘s in \overline{G}. Therefore, \langle x_1,x_2 , \ldots \rangle is a free subgroup of infinite rank in G. \square

Property 2: A countable SQ-universal group contains uncountably many normal subgroups.

Proof. Let G be a countable SQ-universal group. Every quotient of G 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, G must have uncountably many quotients, and a fortiori uncountably many normal subgroups. \square

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 \mathbb{F}_2 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 \Sigma be a closed surface. Then either \pi_1(\Sigma) is SQ-universal or \Sigma is a sphere \mathbb{S}^2, a projective plane \mathbb{P}^2, a torus \mathbb{T}^2 or a Klein bottle \mathbb{K}.

Notice that we already dealt with surface groups in the previous note On subgroups of surface groups. In particular, we know that \pi_1(\mathbb{S}^2) \simeq \{ 1 \}, \pi_1(\mathbb{T}) \simeq \mathbb{Z}^2, \pi_1( \mathbb{P}^2) \simeq \mathbb{Z}_2 are not SQ-universal. We also know that \pi_1(\mathbb{K}) contains \pi_1(\mathbb{T}) \simeq \mathbb{Z}^2 as a subgroup of finite-index two, so it cannot be SQ-universel.

From now on, let \Sigma be a closed surface different from the surfaces mentioned above. If \Sigma is non-orientable, there exist a finite-cover \Sigma' \to \Sigma where \Sigma' is orientable; in particular, \pi_1(\Sigma') is a subgroup of finite-index in \pi_1(\Sigma), so, according to Theorem 3, we may suppose that \Sigma is orientable, that is

\pi_1(\Sigma)= \langle a_1, b_1, \ldots, a_g, b_g \mid [a_1,b_1] \cdots [a_g, b_g] = 1 \rangle

for some g \geq 2. Noticing that the following quotient is free,

\pi_1( \Sigma) / \langle \langle a_1 b_1^{-1}, \ldots, a_g b_g^{-1} \rangle \rangle = \langle a_1, \ldots, a_g \mid \ \rangle \simeq \mathbb{F}_g,

we deduce that \pi_1(\Sigma) 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 \Gamma is defined by the presentation

A(\Gamma)= \langle v \in V ( \Gamma ) \mid [v_1,v_2] = 1, \ (v_1,v_2) \in E(\Gamma) \rangle,

where V ( \Gamma) and E(\Gamma) denote respectively the set of vertices and edges of \Gamma. In particular, if \Gamma is a complete graph with n vertices, then A(\Gamma) is the free abelian group \mathbb{Z}^n; if \Gamma is a graph with n vertices and no edges, then A(\Gamma) is the free group \mathbb{F}_n.


For another example, if \Gamma is the graph given above, then

A(\Gamma) = \langle a,b,c,d,e \mid [a,b]= [a,c] = [a,d] = [e,b]=[e,c]=[e,d]=1 \rangle

is isomorphic to \mathbb{F}_2 \times \mathbb{F}_3.

We claim that, if \Gamma is not a complete graph, then A(\Gamma) has a non-abelian free quotient, and in particular is a SQ-universal group.

Because \Gamma is not complete, there exist two vertices v_1,v_2 not linked by an edge. Let \Gamma_1 denote the subgraph induced by v_1 and its neighbors, and \Gamma_2 the subgraph induced by V(\Gamma) \backslash \{ v_1 \}. Then \Gamma_1 and \Gamma_2 are non-empty, since v_1 \in \Gamma_1 and v_2 \in \Gamma_2, they cover \Gamma, and \Gamma \backslash (\Gamma_1 \cap \Gamma_2) is not connected, because v_1 and v_2 are separated by \Gamma_1 \cap \Gamma_2. Now, quotienting A(\Gamma) by the subgroup

\langle \langle v \in V(\Gamma_1) \cap V(\Gamma_2), \ uv^{-1} \ (u,v \in V(\Gamma_1)), \ uv^{-1} \ (u,v \in V( \Gamma_2 ) ) \rangle \rangle,

we get the free group \langle a , b \mid \ \rangle. For an explicit example, let us consider the following graph:


So, the associated right-angled Artin group is

\langle a, b , c, d \mid [a,b] = [a,c] = [b,c] = [b,d] = [c,e] = [d,e] = 1 \rangle.

Quotienting by the subgroup \langle \langle b,c, de^{-1} \rangle \rangle, we get the free group \langle a,d \mid \ \rangle.

Example 3: If A and B are two finite groups with |A| \geq 2 and |B| \geq 3, then the free product A \ast B is a SQ-universal group.

We proved in the previous note A free group contains a free group of any rank that

\{ [a , b ] \mid a \in A \backslash \{ 1 \}, b \in B \backslash \{ 1 \} \}

is a free basis of the derived subgroup D of A \ast B. Furthermore, the quotient (A \ast B) / D \simeq A \times B is finite, so that D is a non-abelian free subgroup of finite index in A \ast B. From Theorem 3, we deduce that A \ast B is SQ-universal.

For example, SL(2,\mathbb{Z}) is SQ-universal, because it has the SQ-universal group

PSL(2,\mathbb{Z}) \simeq \mathbb{Z}_2 \ast \mathbb{Z}_3

as a quotient. (See An elementary application of ping-pong lemma for an elementary proof of this fact.)