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If $g$ has infinite order in a hyperbolic group, the map $n\mapsto g^n$ is a quasi-isometric embedding of the integers. Equivalently $|g^n|\ge c|n|-C$ for some $c>0$. This is the homogeneous quasigeodesic theorem: local-to-global control of metric geodesics in a hyperbolic Cayley graph gives a quasigeodesic orbit for a suitable conjugate of a positive power; undoing conjugation and passing between consecutive powers changes only uniform constants. It implies $\tau (g)>0$. The theorem and its hypotheses are given in Löh, Theorem 6.5.9. It is not true for arbitrary finitely generated groups, as the Baumslag-Solitar group relation $ab{a}^{-1}=b^3$ shows.

A non-Hopfian group admits a surjective endomorphism which is not injective. The Baumslag-Solitar group $BS(2,3)$ has such an endomorphism $a\mapsto a^2$, $t\mapsto t$: its image contains $a^2$ and $a^3$, while the nonidentity commutator $[a,ta{t}^{-1}]$ lies in its kernel by Britton's lemma.

Suppose $G$ is generated by $d$ elements. An index of a subgroup $n$ gives a transitive action of $G$ on $n$ cosets. After labelling them by $1,\dots ,n$, this is a homomorphism $G\to S_n$, and the subgroup is the stabilizer of one label. There are at most $(n!{)}^d$ homomorphisms, since images of the generators determine the map, and at most $n$ point stabilizers for each map. Thus there are finitely many subgroups of each specified finite index, with the coarse bound $n(n!{)}^d$.

The finite residual isIt is equivalently the intersection of the kernels of all homomorphisms to finite groups. Intersecting all finite-index subgroups gives the same subgroup, since each contains its normal core, a normal subgroup of finite index. A residually finite group is one with $R_G=1$, equivalently every nonidentity element survives in a finite image.

A Hopfian group is one whose surjective endomorphisms are all injective, hence automorphisms. Let $\theta :G\to G$ be surjective, retaining the finite-generation hypothesis. For each $n$, pulling back by $\theta$ sends a normal subgroup of index $n$ to another normal subgroup of the same index. It is injective on this finite set: equality of two preimages implies equality of the subgroups because $\theta$ is surjective. Therefore it is a permutation. For every finite-index normal subgroup $N$, some $m\ge 1$ satisfiesIt follows that $\ker \theta \subseteq \ker {\theta }^m\subseteq N$. Intersecting over $N$ proves finite residual contains surjective endomorphism kernels:If $G$ is also a residually finite group, this forces every surjective endomorphism to have trivial kernel, proving that a finitely generated residually finite group is Hopfian.

For the specified Baumslag-Solitar group, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ and writeIn the abelianization, the defining relation gives $2a=3a$, so the image of $a$ is zero, while $t$ generates an unrestricted infinite cyclic group. Thus $G/G^{\prime }\cong \mathbb{Z}$ through the exponent sum of $t$. The kernel is generated by all $a_i$: any word can be rewritten as a product of conjugates of $a$ followed by the final power of $t$. Alternatively $a=[t,a^2]\in G^{\prime }$, so all $a_i\in G^{\prime }$, while killing them leaves a cyclic quotient. Therefore

Define $\theta (a)=a^2$ and $\theta (t)=t$. This respects the defining relation becauseThe image contains $a^2$, $t$, and $t{a}^2{t}^{-1}=a^3$, hence contains $a=a^3(a^2{)}^{-1}$ and is all of $G$. It is not injective. Indeedmaps to $[a^3,a^2]=1$. But it is a reduced word in the HNN extension with associated subgroups $⟨a^2⟩$ and $⟨a^3⟩$. Its potential subwords $t{a}^{\pm 1}{t}^{-1}$ do not have even exponent, and its potential subword $t^{-1}at$ does not have exponent divisible by three. It has no pinch in an HNN extension, so Britton's lemma gives $c\ne 1$. We have explicitly exhibited a surjective, noninjective endomorphism.

It remains to treat the whole second commutator subgroup, not just this one kernel element. SetThese kernels form an increasing sequence of normal subgroups, so $K$ is a normal subgroup. Every iterate is surjective, and the preceding finite residual argument gives $K\subseteq R_G$.

Conjugating the defining relation gives $a_{i+1}^2=a_i^3$ for every integer $i$. Repeated substitution gives, for $j>i$,Also ${\theta }^k(a_i)=a_i^{2^k}$. Thus when $k\ge j-i$,Every pair of the generating elements of $G^{\prime }$ therefore commutes modulo $K$. The image of $G^{\prime }$ in $G/K$ is abelian, which means $G^{\prime \prime }\subseteq K$. This proves the full statement using iterated endomorphism kernels in BS(2,3):

The first Baumslag-Solitar group iswhere the second infinite cyclic group acts on the first by inversion. The presentation of a semidirect product proves this identification; both copies of $\mathbb{Z}$ are residually finite groups, since reduction modulo a suitable positive integer separates any nonzero integer. The normal factor is finitely generated, so the residual finiteness of semidirect products theorem applies. $\boxed{BS(1,-1)\text{ is residually finite}.}$

Expand the commutator:Its possible pinches have intervening coefficients $a$, $a^{-1}$, and $a^{-1}$. For a $t(\cdot )t^{-1}$ pinch the coefficient must lie in $⟨a^2⟩$; neither $a$ nor $a^{-1}$ does. For the $t^{-1}(\cdot )t$ pinch it must lie in $⟨a^3⟩$, and again $a^{-1}$ does not. Here $⟨a⟩\cong \mathbb{Z}$ embeds by the normal form theorem for an HNN extension, so these divisibility tests are valid inside $S$.

The word is a reduced sequence in an HNN extension with four stable letters. Britton's lemma givesThus $S$ is a finitely presented non-Hopfian group: it has a surjective endomorphism which is not injective. This group is the Baumslag-Solitar group $BS(2,3)$.

We construct the output presentation explicitly, using HNN extensions and an amalgamated free product. Throughout, $P$ also denotes the group given by the input finite group presentation, and $X=\{x_1,\dots ,x_n\}$. Every word introduced below is a literal computable word in the displayed generators; testing whether $w=1$ is never part of the construction.

First fix the auxiliary Higman groupWe need two facts: $⟨a_1,a_3⟩$ is a free group of rank two, and the normal closure of $a_1$ is $Q$. The latter follows immediately from the relations: setting $a_1=1$ forces successively $a_2=a_2^2$, $a_3=a_3^2$, and $a_4=a_4^2$, hence all three are $1$.

Here is a normal-form verification of the first fact. The three-generator groupis an amalgamated free product of two Baumslag-Solitar groups $BS(1,2)$ along their infinite cyclic subgroups generated by $a_2$. Their base cyclic groups embed by Britton's lemma, so this amalgamation is legitimate. Nonzero powers of $a_1$ in the first factor lie outside $⟨a_2⟩$, as shown by the homomorphism recording the exponent of its stable letter $a_1$. Nonzero powers of $a_3$ in the second factor also lie outside $⟨a_2⟩$: the stable-letter exponent map first forces a putative equality $a_3^r=a_2^s$ to have $s=0$, and then injectivity of the base cyclic group forces $r=0$. The normal form theorem for an amalgamated free product therefore makes $⟨a_1,a_3⟩$ a rank-two free group in $Q_{123}$.

The same argument applies to $Q_{341}$, formed from the other two cyclic conjugation relations. The full $Q$ is $Q_{123}{\ast }_{⟨a_1,a_3⟩}{Q}_{341}$, so that rank-two free group embeds in $Q$ as well. In particular $a_1$ has infinite order.

Now take the free productand setIntroduce $n+2$ letters $u_0,\dots ,u_{n+1}$, and define the finite presentationFinally impose the Higman relations and identifyThus the algorithm outputsThe symbols $c,C,g_j$ in this display are abbreviations for the explicit words above, not extra generators. Renaming the finitely many new generators to avoid $X$ makes this a uniform effective construction.

Use the fact that infinite cyclic subgroups of hyperbolic groups are undistorted: if $z$ has infinite order in a hyperbolic group, then its stable word lengthis strictly positive. The limit exists by subadditivity and the Fekete lemma. It is invariant under conjugation, because conjugating $z^m$ changes its length by at most twice the conjugator's length. It also satisfies $\tau (z^k)=|k|\tau (z)$.

In the given Baumslag-Solitar group, $b$ has infinite order. To verify this without presuming the presentation's normal form, let $a$ act on the real line by $t\mapsto 3t$ and $b$ by $t\mapsto t+1$. These affine bijections satisfy $ab{a}^{-1}=b^3$, and every nonzero power of $b$ is a nonidentity translation. Thus the element $b$ in the presented group cannot have finite order.

If an injective group homomorphism to $G$ existed, its images $z$ of $b$ and $u$ of $a$ would satisfy $uz{u}^{-1}=z^3$, with $z$ still of infinite order. Consequentlyforcing $\tau (z)=0$, a contradiction. ThereforeEquivalently, the relation gives $a^{k}b{a}^{-k}=b^{3^k}$, so powers of $z$ with exponent $3^k$ would have length at most $2k|u|+|z|$. This contradicts the linear lower bound for an undistorted infinite cyclic subgroup. This alternative also makes the exponential compression obstruction explicit.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.

Introduce $b=t^{-1}at$. The presentation may be written as the HNN extensionof the Baumslag-Solitar group $\mathrm{BS}(1,2)=⟨a,b\mid ba{b}^{-1}=a^2⟩$, with stable letter $t$ identifying the infinite cyclic subgroups $⟨a⟩$ and $⟨b⟩$. Britton's lemma embeds the base group in the HNN extension. In particular, $a$ has infinite order, so $G$ is infinite. It is also nonabelian because $ba{b}^{-1}=a^2\ne a$.

Suppose that $a$ acted as a hyperbolic isometry of a tree, with translation length $\tau >0$. Nonzero powers have the same axis andTranslation length is invariant under conjugacy, whereas the defining relation in the Baumslag-Solitar group says that $a^m$ and $a^n$ are conjugate. Hence $m\tau =n\tau$, contradicting $m\ne n$. Therefore $a$ acts elliptically in every combinatorial tree action.

Suppose $G=A\ast B$ were a nontrivial free product. Its Bass-Serre tree action has trivial edge stabilizers and no global fixed vertex. Part c makes $a$ elliptic. Since $a$ has infinite order, the fixed set of every nonzero power $a^r$ is a single vertex: it is nonempty, while fixing two vertices would fix the intervening edge and put the infinite-order element $a^r$ in a trivial edge stabilizer.

Let this vertex be $v$. The relation $b{a}^m{b}^{-1}=a^n$ givesand both sides are the singleton $\{v\}$. Thus $b$ also fixes $v$. Since $a$ and $b$ generate the Baumslag-Solitar group, the entire group fixes $v$, contradicting the Bass-Serre action of a nontrivial free product. Hence no such decomposition exists.

The HNN extension is the Baumslag-Solitar groupIts Bass-Serre tree has vertices $G/⟨u⟩$ and oriented edges $G/⟨u⟩$, with the two endpoint maps induced by the identity embedding and the index-two embedding $n\mapsto 2n$. At each vertex there is one incident edge on the identity side and two on the index-two side. The underlying unoriented tree is therefore infinite and $3$-regular, with an orientation in which every vertex has one incoming and two outgoing edges, up to reversing the convention.