= Solution
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
$$
\boxed{\#\{H\leq G:[G:H]=n\}\leq n(n!)^d<\infty.}
$$
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 $\ker\psi\leq N$. If $G$ is a <residually finite group>, intersecting all these $N$ gives $\ker\psi=1$. Hence $\psi$ is an <automorphism>. \b[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>, then
$$
\boxed{K\rtimes H\text{ is residually finite}\iff K\text{ and }H\text{ are residually finite}.}
$$
More 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 $h\ne1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne1$, choose a finite-index <normal subgroup> $U\trianglelefteq K$ 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 $\rho:H\to\operatorname{Aut}(K/C)$ be the induced action. The map
$$
K\rtimes H\longrightarrow (K/C)\rtimes\rho(H),\qquad (x,y)\longmapsto(xC,\rho(y))
$$
is 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 $H$.
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