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