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Let $G$ be a finitely generated group and $\theta :G\to G$ a surjective endomorphism. Pullback by $\theta$ permutes the finite set of normal subgroups of any prescribed finite index. For each such $N$, some iterate satisfies ${\theta }^{-m}(N)=N$, so $\ker \theta \subseteq \ker {\theta }^m\subseteq N$. Intersecting proves the inclusion. Consequently every finitely generated residually finite group is a Hopfian group.

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):

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

Let $\alpha :G\to G$ be surjective and suppose that $1\ne g\in \ker \alpha$. By residual finiteness, choose $q:G\to Q$ with $Q$ finite and $q(g)\ne 1$. The preceding part makes the sequencerepeat, so $q{\alpha }^i=q{\alpha }^j$ for some $i<j$. Surjectivity of ${\alpha }^i$ permits cancellation on the right and gives $q=q{\alpha }^{j-i}$. But $g\in \ker {\alpha }^{j-i}$, which would imply $q(g)=1$, a contradiction. Thus $\alpha$ is injective. Every finitely generated residually finite group is therefore a Hopfian group.