Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 4

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.

By the Fundamental theorem of finitely generated abelian groups,with $T$ finite. If a nonzero element has a nonzero component in $T$, projection to $T$ separates it. Otherwise some integer coordinate is a nonzero $n$; choose a prime $p$ not dividing $n$ and reduce that coordinate modulo $p$. This gives a finite quotient in which the element survives, so every finitely generated abelian group is residually finite.

If $G$ is generated by $d$ elements, a homomorphism $G\to Q$ is determined by the images of those generators. There are at most $|Q{|}^d$ such choices, so only finitely many homomorphisms exist.

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.

Enumerate $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$. The universal property of a free group gives an endomorphismSince $Y$ generates, $\alpha$ is surjective. The finitely generated free group is residually finite and hence Hopfian by the preceding part, so $\alpha$ is an automorphism. An automorphism sends a free basis to a free basis; therefore $Y$ is a basis of $F(X)$.

For the free basis $X=\{x_0,x_1,x_2,\dots \}$, defineThe universal property of a free group extends this assignment to an endomorphism of $F(X)$. It is surjective because every $x_n$ is the image of $x_{n+1}$, but it is not injective because $x_0\ne 1$ lies in its kernel. Thus $F(\mathbb{N})$ is not Hopfian.