This is the third note of our series dedicated to constructions using amalgamated products and HNN extensions.

In the early 1940’s, Hopf asked whether it is possible for a finitely-generated free group to be isomorphic to one of its proper quotients. Later, he proved that it is in fact impossible, but he left open the same question for finitely-generated groups. If a group G is said Hopfian precisely when it is not isomorphic to one of its proper quotients, or equivalently, that any epimorphism G \to G turns out to be an isomorphism, the question becomes : Does there exist a non-Hopfian finitely generated group?

The first example of non-Hopfian finitely-generated groups was only found in 1950 by B. H. Neumann, see his article A two-generator group isomorphic to a proper factor group. In 1951, Higman gave the first example of non-Hopfian finitely presented group using amalgamated products:

Theorem: (Higman) The group \langle a,s,t \mid sas^{-1}=a^2= tat^{-1} \rangle is not Hopfian.

Proof. Let D= \{ \ell / 2^m \mid \ell \in \mathbb{Z}, m \in \mathbb{N} \} denote the additive group of dyadic rationals. We first prove that

\Pi = \langle x_0,x_1, \dots \mid [x_i,x_{i+1}]=1, x_i^{2^i}=x_0, i \geq 0 \rangle

is a presentation of D. Clearly, the map sending x_i to 1/2^i for all i \geq 0 extends to an epimorphism \phi : \Pi \twoheadrightarrow D. It is sufficient to prove that \phi is one-to-one to conclude. Let

g:= x_{m_1}^{\ell_1} \cdots x_{m_r}^{\ell_r} \in \mathrm{ker}(\phi).

In \Pi, notice that for all i \leq j,

x_i^{2^j}= x_i^{2^i \cdot 2^{j-i}} = (x_i^{2^i})^{2^{j-i}} = x_0^{2^{j-i}}.

Therefore, if m:= \max(m_1,\dots,m_r),

x_0^{\ell_1 2^{m-m_1} + \cdots + \ell_r 2^{m-m_r}} = g^{2^m} \in \mathrm{ker}(\phi).

But 0= \phi \left( g^{2^m} \right) = \ell_1 2^{m-m_1} + \cdots + \ell_r 2^{m-m_r}, hence g^{2^m}=x_0^{0}=1. However, \Pi is an abelian group whose generators are sending to infinite-order element in D by \phi, so \Pi is torsion-free; in particular, we necessarily have g=1. We just found our presentation.

Afterwards, let G denote the semi-direct product D \rtimes \mathbb{Z} where \mathbb{Z} acts by multiplication by 2. Then

G= \langle x_0,x_1, \dots, t \mid [x_i,x_{i+1}]=1, x_i^{2^i}=x_0, tx_it^{-1}=x_{i-1}, i \geq 0 \rangle.

The third relation implies x_i = t^{-i}x_0t^i, so the presentation above may be simplified as

G= \langle x_0, t \mid [tx_0t^{-1},x_0]=1, tx_0t^{-1}=x_0^2 \rangle = \langle x_0,t \mid tx_0t^{-1}=x_0^2 \rangle.

Let G_1 = \langle a,s \mid sas^{-1} = a^2 \rangle and G_2 = \langle b,t \mid tbt^{-1}=b^2 \rangle be two copies of G. From the discution above, we know that a and b have an infinite order in G_1 and G_2 respectively (in fact, recognizing a HNN extension, it immediatly follows from Britton’s lemma). Therefore, we may define the amalgamated product

K:= G_1 \underset{\langle a \rangle = \langle b \rangle}{\ast} G_2 = \langle a,s,t \mid sas^{-1}=a^2= tat^{-1} \rangle.

Let \alpha : G_1 \to G_1 (resp. \beta : G_2 \to G_2) be a morphism sending a to a^2 and s to s (resp. b to b^2 and t to t). Noticing that \alpha and \beta agree on \langle a \rangle = \langle b \rangle, we may define \mu = \alpha \ast \beta : K \to K.

We first notice that \mu is onto. Indeed, \mu(s)=s, \mu(t)=t, \mu (s^{-1}as)= s^{-1}a^2s=a and \mu (t^{-1}bt)=t^{-1}b^2t=b.

Then, we notice that \mu is not one-to-one. Let g := s^{-1}ast^{-1}b^{-1}t. In K,

\mu(g)=s^{-1}a^2 st^{-1}b^{-2}t = ab^{-1}=1,

so g \in \mathrm{ker}(\mu). On the other hand,

g = \underset{ \in G_1 \backslash \langle a \rangle }{ \underbrace{ (s^{-1} as ) }} \cdot \underset{ \in G_2 \backslash \langle b \rangle }{ \underbrace{ (t^{-1} b t )^{-1} }} \neq 1.

We justify that s^{-1}as \notin \langle a \rangle (and in the same way that t^{-1}bt \notin \langle b \rangle) by saying that otherwise there would be a k \in \mathbb{Z} such that s^{-1}as =a^k, hence

a=sa^ks^{-1}=(sas^{-1})^k =a^{2k},

a contradiction with the fact that G_1 is torsion-free. \square

In 1962, Baumslag and Solitar found a simple family of non-Hopfian one-relator groups using HNN extensions.

Definition: Let n,m \in \mathbb{Z} \backslash \{0 \}. The Baumslag-Solitar group BS(n,m) is the HNN extension of \mathbb{Z} with respect to the subgroups n \mathbb{Z} and m \mathbb{Z}, and the obvious isomorphisms.

BS(n,m)= \langle a,t \mid ta^nt^{-1} = a^m \rangle.

In particular, BS(1,2) is the semi-direct product D \rtimes \mathbb{Z} we described above.

Theorem: If m,n \in \mathbb{Z} \backslash \{ 0\} are coprime with n \neq \pm 1, then BS(n,m) is not Hopfian.

Proof. Let \varphi : BS(n,m) \to BS(n,m) be the morphism sending a to a^n and t to t; \varphi is well-defined since

\varphi( ta^nt^{-1} a^{-m}) =ta^{n^2} t^{-1} a^{-mn} = (ta^nt^{-1})^n a^{-mn} = a^{mn} a^{-mn}= 1.

 First, \varphi is onto. Indeed, t = \varphi(t)= \mathrm{Im}(\varphi) and

\left\{ \begin{array}{l} a^n = \varphi(a) \in \mathrm{Im}(\varphi) \\ \\ a^m = ta^nt^{-1} = \varphi(t^{-1}at) \in \mathrm{Im}(\varphi) \end{array} \right. implies a=a^{pn+qm}= (a^n)^p \cdot (a^m)^q \in \mathrm{Im}(\varphi),

where p,q \in \mathbb{Z} are two integers such that pn+qm=1 (they exist because n and m are coprime).

Then, we notice that \varphi is not one-to-one. Because n \neq \pm 1, we deduce from Britton’s lemma that

[t^{-1}at, a]= tat^{-1} a ta^{-1} t^{-1} a^{-1} \neq 1,

On the other hand,

\varphi ([t^{-1}at, a]) = \underset{=a^m}{\underbrace{ ta^{n} t^{-1} }} a^{n} \underset{ = a^{-m} }{\underbrace{ ta^{-n}t^{-1} }} a^{-n} = 1,

so [t^{-1}at, a ] \in \mathrm{ker}(\varphi). \square