This is the first note of a series dedicated to constructions using amalgamated products and HNN extensions. We begin with an embedding theorem (an alternative proof can be found in our previous note A free group contains a free group of any rank):

Theorem: (Higman-Neumann-Neumann) Any countable group $G$ can be embedded into a two-generator group $E$. Moreover, it can be supposed that any torsion element of $E$ is conjugated to a element of $G$.

Proof. Let $g_0=1, g_1,g_2, \dots$ be the elements of $G$ (not necessarily pairwise distinct), and let $U= \langle u,v \mid \ \rangle$, $B= \langle a,b \mid \ \rangle$ be two copies of the free group of rank two.

We first define the free product $A= G \ast U$. Clearly,

$\{g_0u, g_1vuv^{-1}, g_2 v^2uv^{-2}, \dots \}$

freely generates a subgroup $H \subset A$, because its projection into $U$ is the free subgroup generated by

$\{u, vuv^{-1}, v^2uv^{-2}, \dots \}$.

In the same way,

$\{ a, bab^{-1}, b^2ab^{-2}, \dots \}$

freely generates a subgroup $K \subset B$. Therefore, we may define the amalgamated free product $P= A \underset{H=K}{\ast} B$ by identifying $g_iv^iuv^{-i}$ with $b^iab^{-i}$. By construction, $P$ is generated by

$\{g_0,g_1,g_2, \dots, u,v,a,b \}$.

Because $g_iv^ivu^{-i}=b^ia b^{-i}$ and $u=g_0u=a$, in fact $P$ turns out to be generated by only the three elements $\{v,a,b\}$.

Then, notice that $\langle a,v \rangle$ and $\langle a,b \rangle$ are two subgroups of $P$ isomorphic to the free group of rank two. Indeed,

$\langle a,b \rangle \simeq B$ and $\langle v,a \rangle \simeq \langle v,g_0u \rangle \simeq \langle u,v \rangle \simeq U$.

Therefore, if $\varphi$ denotes the isomorphism between $\langle a,v \rangle$ and $\langle a,b \rangle$ sending $v$ to $a$ and $a$ to $b$, we may define the associated HNN extension $E$; let $t$ denotes its stable letter. Now $E$ is generated by $\{v,a,b,t \}$, but $tvt^{-1}=a$ and $tat^{-1}=b$, so $E$ is in fact generated by $\{v,t\}$.

Consequently, we just embedded $G$ into the two-generator group $E$. Moreover, we notice that a torsion element of $E$ is conjugated to an element of $G$ because a torsion element of an amalgamated product or of a HNN extension is necessarily conjugated to an element of a factor (and because $U$ and $B$ are torsion-free). $\square$

As a corollary, we deduce the following theorem that we already proved in the note Cantor-Bendixson rank in group theory: A theorem of B.H. Neumann, using the space of marked groups and small cancellation groups.

Theorem: (B.H. Neumann) There exist exactly $2^{\aleph_0}$ nonisomorphic two-generator groups.

Proof. First, because any two-generator group is a quotient of a free group of rank two, there are at most $2^{\aleph_0}$ nonisomorphic two-generator groups. To conclude, it is sufficient to exhibit a family of $2^{\aleph_0}$ nonisomorphic two-generator groups.

Let $p_1 < p_2 < \cdots$ denote the sequence of consecutive primes. For any subset $I \subset \mathbb{N}$, we define

$A_I = \bigoplus\limits_{i \in I} \mathbb{Z}_{p_i}$,

and $G_I$ the two-generator group associated to $A_I$ given by the previous theorem.

If $I \neq J$, say there exists $i \in I \backslash J$, $A_I$ (and a fortiori $G_I$) has an element of order $p_i$; however, $A_J$ (and a fortiori $G_J$) don’t. Therefore, $G_I$ and $G_J$ are not isomorphic. $\square$