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.)