What is a residually finite group?

We say that a group $G$ is residually finite if for each $g\in G$ that is not equal to the identity of $G$, there exists a finite group $F$ and a group homomorphism
$$\varphi:G\to F$$ such that $\varphi(g)$ is not the identity of $F$.

The definition does not change if we require that $\varphi$ be surjective. Therefore, a group $G$ is residually finite if and only if for each $g\in G$ that is not the identity, there exists a finite index normal subgroup $N$ of $G$ such that $g\not\in N$.

Hence, if $G$ is residually finite, then the intersection of all finite-index normal subgroups is trivial. The converse holds, too (why?).

Examples and Non-examples of Residually Finite Groups

Before we examine why residually finite groups are interesting, let's first take a look at some examples. The following groups are residually finite:

  1. Finite groups are residually finite
  2. $\Z$ is residually finite: if $z\in Z$ is a nonzero integer, then the reduction map $\Z\to \Z/(|z| + 1)$ does not send $z$ to zero
  3. More generally, any finitely-generated abelian group is residually finite
  4. The automorphism group of a residually finite group is residually finite. For example, $\Aut(\Z\times \Z)\cong \GL_2(\Z)$ is residually finite (though it is also easy to see this directly)
  5. Free groups are residually finite
  6. John Hempel proved that fundamental groups of 2-manifolds are residually finite (Proc. Amer. Math. Soc. 32 (1972), 323)

It seems like quite a lot of groups are residually finite. So we need some examples of groups that aren't residually finite, right? Here they are:

  1. The additive group of rational numbers $\Q$ is not residually finite. That's because it's divisible. If $D$ is a divisible group and $F$ is finite, then every homomorphism $\varphi:D\to F$ is in fact trivial! That's because there exists a positive integer $n$ such that $g^n$ is the identity for all $g\in F$. So, if $x\in D$, there exists a $y\in D$ such that $y^n = x$ and so $\varphi(x) = \varphi(y^n) = \varphi(y)^n = 1_F$.
  2. The subgroup of ${\rm Sym}(Z)$ generated by the translation map $n\mapsto n + 1$ and the permutation $(0~1)$ is finitely generated and not residually finite.

Residually Finite Groups and the Word Problem

If a group $G$ is given by a presentation $\langle X~|~ R\rangle$ where $X$ is a set of generators and $R$ is a set of relations (meaning that $G$ is the quotient of the free group on $X$ by the smallest normal subgroup containing the words in the set $R$), then a natural question to ask is: given a word in the symbols in $X$ and their formal inverses, does the word represent the identity element? If there is a guaranteed to terminate algorithm that answers this question for any word, then the group $G$ is said to have solvable word problem.

Let's take an example: let $G$ be the group presented by $\langle x,y~|~ xyx^2 \rangle$. Does the word $xy^2xy$ represent the identity element of $G$? If you take a few moments to try a prove this, you'll see that it is actually very difficult. That's the word problem for you: it is a hard problem. That's because working combinatorially with presented groups is tough.

But presented groups arise naturally in mathematics as the fundamental groups of manifolds, so it makes sense to try and figure them out.

If a group $G$ has a presentation $\langle X~|~R\rangle$ such that $X$ and $R$ are both finite sets, then $G$ is called finitely presented. Because you have a finite number of relations and generators, you can enumerate the words that represent the identity in $G$ such that if $w$ is any word that represents the identity, $w$ will appear on this list in a finite amount of time.

Great! But that still doesn't solve the word problem. That's because if $w$ does not represent the identity, merely enumerating all the words that represent the identity of successively longer lengths still won't tell you that $w$ does not represent the identity.

That's where residually finite groups come into play. All finite groups of a given size are enumerable in a finite amount of time, and all homomorphisms from a finitely-presented group to a given finite group can be enumerated in a finite amount of time. Therefore, if $G$ is residually finite and is finitely presented, by enumerating all these homomorphisms, you will eventually find one where the given word $w$ is sent to a nontrivial element.

So, enumerating all words that represent the identity and all homomorphisms to a finite group at the same time is an algorithm that is guaranteed to determine whether a given word represents the identity in a finitely-presented residually finite group.

As in the case with modules, finitely-presented groups are better behaved that finitely-generated groups. Meskin in (Proc. Amer. Math. Soc. 43 (1974), 8–10.) gave an example of a finitely-generated residually finite group that has unsolvable word problem!

Hopfian Groups

There is another popular concept in group theory: we call a group $G$ Hopfian if every surjective homomorphism $G\to G$ is also injective. The idea of Hopfian also generalises a property of finite groups in a different way: of course, every finite group is Hopfian.

Residually finite groups are not always Hopfian. For example, a free group on infinitely many generators is residually finite but not Hopfian. That's because you can take any surjective map $X\to X$ on an infinite set that is not injective and extend it to a map the corresponding free groups.

However, if $G$ is residually finite and finitely generated, then it is Hopfian. Indeed, suppose $G$ is residually finite and finitely generated, and that $\varphi:G\to G$ is surjective. Let $K$ be any normal subgroup of $G$ of finite index. Define a map
\begin{align}
\Phi: {\rm Hom}(G,G/K)&\longrightarrow {\rm Hom}(G,G/K)\\
f&\longmapsto f\circ\varphi.\end{align} Since $G$ is finitely generated and $G/K$ is finite, the set ${\rm Hom}(G,G/K)$ is finite. Since $\varphi$ is surjective, $\Phi$ is injective and thus also surjective. Therefore, there exists a $f:G\to G/K$ such that $\Phi(f)$ is the quotient map $G\to G/K$.

This means that the kernel of $\varphi$ is contained in $K$. Since $K$ was arbitrary, the kernel of $\varphi$ is contained in the intersection of all finite-index normal subgroups. This intersection is trivial because $G$ is residually finite. Thus, the kernel of $\varphi$ is trivial and hence $\varphi$ is injective. QED

This result is quite interesting. For example, you can't find a surjective homomorphism $\Z\times \Z/2\to\Z\times\Z/2$ that is surjective but not injective. You can prove this directly but it is not altogether obvious.

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