Any large group is SQ-universal. This result is part of the preliminary section of "Largeness and SQ-universality of Cyclically Presented Groups," by Gerald Williams.
The Details:

Definition 1: A group is large if it has a finite index subgroup that maps onto the free group $F_2$ of rank $2$.
Definition 2: A group $G$ is SQ-universal if every countable group can be embedded in a quotient group of $G$.

The Question:

Prove that any given large group is SQ-universal.

Thoughts:
Let $G$ be large. Then there exists a subgroup $H$ of $G$ such that $[G: H]$ is finite and there exists epimorphism $\theta: H\to F_2$. Thus, in a sense, $F_2$ can be seen as a subgroup of $G$. This would be sufficient (almost) provided that the following hold.

*

*The group $F_2$ is SQ-universal.

*If $A$ is a subgroup of a group $B$ and $A$ is SQ-universal, then $B$ is SQ-universal.

The former isn't very clear to me but the latter seems obvious.
Please help :)
 A: Theorem. Large $\Rightarrow$ SQ-universal.
Proof.


*

*The group $F_2$ is SQ-universal. This is an old result of Higman-Neumann-Neumann (in their paper where they first introduce HNN-extensions: G. Higman, B.H. Neumann and H. Neumann, Embedding theorems for groups, J. London Math. Soc. 24 (1949), 247-254).

*Being SQ-universal is closed under finite-index (suppose $H\leq_fG$, then $G$ is SQ-univesal if and only if $H$ is SQ-universal). This is a result of Peter Neumann*. The reference is P.M. Neumann: The SQ-universality of some finitely presented groups. J. Austral. Math. Soc. 16, 1-6 (1973). I have copied out the first direction of the proof, that if $G$ is SQ-univesal and $H\leq_f G$ then $H$ is SQ-universal, which unfortunately is not the direction you need. The opposite direction is longer; you can look at it in the paper if you wish. The  proof   of  the  lemma  depends  on  the  well-known  fact  that  every  countable group  can  be  embedded  in  a  countably  infinite  simple  group.  So, let $X$ be any  countable  group  and  let $S$ be  a  countable,  infinite  simple  group  which  has a subgroup  isomorphic  to $X$. Suppose first that $G$ is SQ-universal and that $H$ is a subgroup of finite index in $G$. There will exist a normal subgroup $N$ of $G$ such that $\overline{G} = G/N$ contains a subgroup $\overline{S}$ isomorphic to $S$. Now $\overline{H} = HN/N$ has finite index in $\overline{G}$. Therefore $H \cap \overline{S}$ has finite index in $\overline{S}$ and consequently contains a subgroup which is normal and of finite index in $\overline{S}$. Since $\overline{S}$ is simple and infinite it follows that $\overline{H} \cap \overline{S} = \overline{S}$. Thus $S$, and consequently also $X$, is isomorphic to a subgroup of $\overline{H}=HN/N\cong H/(H\cap N)$. Hence $H$ is SQ-universal.
Some remarks:


*

*If $H\twoheadrightarrow F_2$ then $H$ actually contains a copy of $F_2$. So when you say "Thus, in a sense, $F_2$ can be seen as a subgroup of $G$", this is actually quite concrete as $F_2\leq H\leq G$.

*There exist non-SQ-universal groups which contain an SQ-universal subgroup, so your second bullet point is wrong (that is, you need the finite index condition). For example, in the above paragraph from P. Neumann's paper it is pointed out that there is a simple group, $S$ say, which contains some SQ-universal group as a subgroup (e.g. $F_2\hookrightarrow S$). Then $S$ cannot be SQ-universal: to see this, suppose otherwise. By simplicity, every two-generated group embeds into $S$. There are uncountably many non-isomorphic two-generated groups. However, there are only countably many two-generated subgroups of $S$ (as there are only countably many pairs of elements $(s_1, s_2)\in S\times S$). This is a contradiction, and the result follows.

*Large groups were first studied by Steve Pride. The first result was with B. Baumslag**, where they proved that if a group $G$ has a presentation with two more generators than relators then $G$ is large; later, in the early '80s, his student Martin Edjvet studied them for his thesis ("The concept of largeness in group theory" - Edjvet is still around). Steve once told me that his motivation was to get some sort of partial ordering on finitely generated groups based on SQ-universality, and in particular two groups should be equal in this partial order if they are commensurable (that is, $A\sim B$ if there exists $C$ such that $C$ embeds with finite index into both $A$ and $B$). The ordering on finitely generated groups is as follows: $H\preceq G$ if there exists $K\lhd_fG$ such that $K\twoheadrightarrow H$. This partial order is called the largeness ordering. Note that large groups sit at the top of this order, as $F_2$ contains with finite index free subgroups of arbitrary finite rank.
*One of the sons of the above two Neumanns - the result is in fact dedicated to "the memory of (his) mother", H.Neumann.
**Brother of the great G. Baumslag.
