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The finite residual consists of all elements killed by every homomorphism to a finite group. A group is a residually finite group precisely when its finite residual is trivial. Intersecting all finite-index subgroups gives the same result: the normal core of each such subgroup is normal of finite index and lies inside it.

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.

Enumerate all words $w_i$ on two generators in a free pro-p group. Choose powers $p^{e_i}$ so large that ${\sum }_i{\tau }^{p^{e_i}}<2\tau -1$ for some $1/2<\tau <1$, and impose the closed normal relations $w_i^{p^{e_i}}=1$. The Golod-Shafarevich infinitude criterion for pro-p groups makes the resulting pro-$p$ group infinite. Its abstract subgroup generated by the two generator images is dense and infinite; each element is represented by an enumerated word and has finite order. The ambient finite quotients separate its elements, so it is also a residually finite group. The abstract subgroup, rather than the whole topological group, is the desired finitely generated torsion group.

Let $K=\ker \theta$. The subgroup correspondence for a surjective group homomorphism sends a subgroup $B\le H$ to ${\theta }^{-1}(B)$ and a subgroup $A\le G$ containing $K$ to $\theta (A)$. Inverse images are subgroups containing $K$, and images are subgroups. Surjectivity gives $\theta ({\theta }^{-1}(B))=B$. Conversely, if $K\le A$ and $\theta (g)\in \theta (A)$, choose $a\in A$ with $\theta (g)=\theta (a)$. Then $g{a}^{-1}\in K\le A$, so $g\in A$. Thus ${\theta }^{-1}(\theta (A))=A$. These two assignments are inverse bijections.

For $C={\theta }^{-1}(B)$, the map of left coset setsis well defined and onto. If $\theta (g_1)B=\theta (g_2)B$, then $\theta (g_2^{-1}{g}_1)\in B$, so $g_2^{-1}{g}_1\in C$ and $g_{1}C=g_{2}C$. Hence it is injective, provingNo normality of $B$ is required; these are coset sets, not necessarily quotient groups.

The abelianization is $\overline{G}=G/[G,G]$, where the commutator subgroup is generated by the commutators $[x,y]=x^{-1}{y}^{-1}xy$. Conjugating a commutator gives another commutator, so this subgroup is normal. In the quotient all commutators are trivial, and therefore the quotient is abelian. If $N⊲G$ and $G/N$ is abelian, every commutator lies in $N$, so $[G,G]\le N$. Henceis well defined and satisfies $\pi =\tilde{\pi }\circ q$, where $q:G\to \overline{G}$ is the natural map. It is uniquely determined because $q$ is onto. This is the universal factorization property of abelianization.

Now write $S_i$ for the unique index-$i$ subgroup of $S$. Since conjugation preserves index, as shown in part (ii), $g{S}_i{g}^{-1}=S_i$ for every $g\in S$. Thus every $S_i$ is normal. For each prime $p$ the quotient $S/S_p$ has order $p$ and is cyclic, so there is a surjection $S\to C_p$. Being an abelian quotient, it factors through $\overline{S}$.

The abelianization is finitely generated because $S$ is. By the Fundamental theorem of finitely generated abelian groups,where $T$ is finite. If $r=0$, choose a prime $p>|T|$. The finite group $T$ could not surject onto the order-$p$ group $C_p$, contradicting the preceding surjections. Hence $r\ge 1$, and projection from a free coordinate supplies

Every subgroup of finite index in $\overline{S}$ has an inverse image of the same index in $S$, so there cannot be two different subgroups of any given finite index in $\overline{S}$. If $r\ge 2$, reduction modulo a prime $p$ in the first and second free coordinates gives two distinct surjective maps to $C_p$ with different kernels: the first basis vector lies in one kernel but not the other. Both kernels have index $p$, a contradiction. Therefore $r=1$.

If $T\ne 1$, choose a prime $p$ dividing its order. The decomposition of a finite abelian group into cyclic factors gives a surjection $\chi :T\to C_p$. In $\mathbb{Z}\oplus T$, the mapshave distinct index-$p$ kernels, again distinguished by $(1,0)$. This is another contradiction, so $T=1$. We have proved uniqueness of finite-index subgroups forces infinite cyclic abelianization:

A residually finite group is one in which every nonidentity element has nonidentity image under some homomorphism to a finite group. Equivalently, the intersection of all finite-index normal subgroups is trivial: a finite separating map has such a kernel, and conversely quotienting by such a normal subgroup gives a finite separating map.

Choose the surjection $\sigma :S\to \overline{S}\cong \mathbb{Z}$. For each positive integer $i$, the subgroup ${\sigma }^{-1}(i\mathbb{Z})$ has index $i$ and hence equals $S_i$. Its kernel $[S,S]$ is therefore contained in every finite-index subgroup. Indeed,If $S$ is residually finite, this intersection is trivial. Thus $[S,S]=1$ and $\sigma$ is an isomorphism. The residually finite group with one subgroup of each finite index is infinite cyclic conclusion isResidual finiteness is used exactly to eliminate the commutator subgroup that all finite quotients would otherwise fail to detect.

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):

We construct an infinite residually finite torsion group using the Golod-Shafarevich infinitude criterion for pro-p groups. Its useful form says that a pro-$p$ presentation on $d$ generators with relators $r_i$ gives an infinite pro-p group wheneverHere $D$ is the Magnus degree of a free pro-p group element: map each generator to $1+X_j$ in a noncommutative power-series algebra over ${\mathbb{F}}_p$, and take the lowest degree of $w-1$. The criterion permits infinitely many relators with a convergent weight sum. It is a consequence of the Golod-Shafarevich inequality for the completed group algebra; see Ershov's survey, Theorem 3.1.

Take $p=2$, $d=2$, and $\tau =3/4$. Let $F$ be the free pro-p group on $a,b$, and enumerate all ordinary finite words $w_1,w_2,\dots$ in these generators and their inverses. Choose integers $e_i$ sufficiently large thatForm the quotient pro-p groupwhere the bar means the closed normal closure. If the Magnus image of $w_i$ is $1+v_i$, characteristic two givesfor nonidentity $w_i$; identity relators contribute zero weight. It follows thatThus the Golod-Shafarevich infinitude criterion for pro-p groups proves that $P$ is infinite.

Let $G=⟨\bar{a},\bar{b}⟩$ be the abstract subgroup generated by the two images inside $P$. This subgroup is dense because the two images topologically generate the quotient, and it is infinite: a finite subgroup of a Hausdorff profinite group is closed, so could not be dense in infinite $P$. Every element of $G$ is the image of some ordinary word $w_i$, and the imposed relation makes its order divide $2^{e_i}$. Therefore $G$ is a two-generated torsion group. Finally $P$ is an inverse limit of finite 2-groups. Any nonidentity element of $G$ is nonidentity in $P$ and hence has a nonidentity image in one of those finite groups. Restricting that map to $G$ proves that it is a residually finite group. We have obtainedThe abstract subgroup $G$ is essential here: the argument establishes finite order for ordinary words, not for every topological limit in the ambient pro-p group.

There is no surjective homomorphism from this group to an infinite soluble group. In fact finitely generated soluble torsion groups are finite. To prove this, let $S$ be finitely generated, soluble and torsion, and induct on its derived length. In the abelian case, finitely many generators of finite orders generate a finite group. In general the abelianization $S/S^{\prime }$ is a finitely generated abelian torsion group, hence finite. Its kernel $S^{\prime }$ therefore has finite index and is finitely generated by Schreier's lemma. Explicitly, for a finite coset transversal $T$ containing $1$ and a finite symmetric generating set $X$, the finitely many elements $tx{\overline{tx}}^{-1}$ generate the finite-index subgroup, where the bar denotes the chosen representative of the coset. Now $S^{\prime }$ is torsion and has smaller derived length, so induction makes it finite. Hence $S$ is finite as well. Every quotient of $G$ is finitely generated and torsion, so any soluble quotient must be finite.

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. $\boxed{BS(1,-1)\text{ is residually finite}.}$

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 $H$.

A group $G$ is residually finite when, for every $1\ne g\in G$, there are a finite group $Q$ and a group homomorphism $q:G\to Q$ such that $q(g)\ne 1$. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.

Use the injective group homomorphism $F_2\hookrightarrow {\mathrm{SL}}_2(\mathbb{Z})$ from part (a). If $1\ne g\in F_2$, its image is an integral matrix $M\ne I$. Choose a prime number $p$ that does not divide one nonzero entry of $M-I$. The reduction modulo a prime in an integral matrix group homomorphismthen sends $M$ to a nonidentity element. Its target is a finite group, so the composite map separates $g$ from the identity. Hence $F_2$ is a residually finite group.

Every finitely generated free group is a residually finite group. For $F_2$, the ping-pong lemma embeds it into ${\mathrm{SL}}_2(\mathbb{Z})$, and reduction modulo a prime in an integral matrix group separates each nonidentity element in a finite quotient.

Every right-angled Artin group is a residually finite group. One route is its faithful integral linear representation, followed by reduction modulo suitable primes.

If $K$ is a finitely generated group, then the semidirect product $K⋊H$ is a residually finite group exactly when both $K$ and $H$ are residually finite groups. To separate an element whose $H$ coordinate is nontrivial, use a finite quotient of $H$. For $1\ne k\in K$, choose a finite-index normal subgroup $U$ omitting $k$, and intersect all subgroups of index at most $[K:U]$ to obtain a characteristic finite-index subgroup $C\le U$. The induced map into $(K/C)⋊\mathrm{im}(H\to \mathrm{Aut}(K/C))$ has finite target and separates $k$. More generally, finite generation can be replaced by a separating family of finite-index normal subgroups invariant under the given $H$ action.