Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 4 / 4

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 $H$.

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

$\boxed{BS(3,4)\text{ is not residually finite}.}$ We exhibit a nonidentity element killed by every finite quotient.

In any finite image, let $r$ be the order of the image of $a$. The relation conjugating $a^3$ to $a^4$ givesThus $\gcd (r,3)=\gcd (r,4)=1$. Since $3$ is invertible modulo $r$, the relation implies that the image of $ba{b}^{-1}$ is a power of the image of $a$. Therefore every finite image kills the group commutatorusing $[s,t]=s^{-1}{t}^{-1}st$.

View $BS(3,4)$ as an HNN extension of $⟨a⟩\cong \mathbb{Z}$, with associated subgroups $⟨a^3⟩$ and $⟨a^4⟩$. A pinch in an HNN extension would be $b{a}^{3j}{b}^{-1}$ or $b^{-1}{a}^{4j}b$. The word $w$ has none: its intervening exponents are $-1,1,1$, incompatible with the required divisibilities $3,4,3$. By Britton's lemma, $w\ne 1$. Hence finite quotients fail to separate this nonidentity element, proving the conclusion.

Both displayed matrices have determinant one and integer entries, so the generated group $M$ lies in ${\mathrm{SL}}_2(\mathbb{Z})$. For any nonidentity matrix $g\in M$, some entry of $g-I$ is a nonzero integer $k$. Choose a prime number $p$ not dividing $k$. The reduction modulo a prime in an integral matrix group gives a group homomorphismto a finite group in which $g$ is nonidentity. $\boxed{M\text{ is residually finite}.}$ No determination of the abstract subgroup generated by the two matrices is needed.