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 can be embedded into a two-generator group . Moreover, it can be supposed that any torsion element of is conjugated to a element of .
Proof. Let be the elements of (not necessarily pairwise distinct), and let , be two copies of the free group of rank two.
We first define the free product . Clearly,
freely generates a subgroup , because its projection into is the free subgroup generated by
In the same way,
freely generates a subgroup . Therefore, we may define the amalgamated free product by identifying with . By construction, is generated by
Because and , in fact turns out to be generated by only the three elements .
Then, notice that and are two subgroups of isomorphic to the free group of rank two. Indeed,
Therefore, if denotes the isomorphism between and sending to and to , we may define the associated HNN extension ; let denotes its stable letter. Now is generated by , but and , so is in fact generated by .
Consequently, we just embedded into the two-generator group . Moreover, we notice that a torsion element of is conjugated to an element of because a torsion element of an amalgamated product or of a HNN extension is necessarily conjugated to an element of a factor (and because and are torsion-free).
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 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 nonisomorphic two-generator groups. To conclude, it is sufficient to exhibit a family of nonisomorphic two-generator groups.
Let denote the sequence of consecutive primes. For any subset , we define
and the two-generator group associated to given by the previous theorem.
If , say there exists , (and a fortiori ) has an element of order ; however, (and a fortiori ) don’t. Therefore, and are not isomorphic.