Residual finiteness

Let $G$ be a group and $g\in G$ with $g\neq 1$. Then we call $G$ residually finite (hereafter RF) if there exists a subgroup $K\subset G$ of finite index such that $g\notin K$. In other words, for every nontrivial element of $G$, there exists a finite index subgroup that does not contain that particular element.

Example. Finite groups are RF, since the trivial subgroup has finite index and does not contain any of the nontrivial elements of $G$.

Metaquestion. How general is the class of RF groups? In particular, which finitely generated/finitely presented groups are RF?

Remarks. Assume that $G$ is finitely generated.

(i) The definition can be strengthened so that we may assume that our finite index subgroup $K\subset G$ is normal in $G$. Indeed, if $G$ is finitely generated, then there are only finitely many subgroups of a given fixed index $k$. (See the exercise from the second lecture.) If $g_0\neq 1$, and $K\subset G$ is a subgroup of finite index that does not contain $g_0$, then

$\textrm{core}(K)=\bigcap_{g\in G}gKg^{-1}$

is a subgroup of finite index in $G$ which excludes $g_0$, since $gKg^{-1}$ and $K$ have the same index in $G$ for all $g\in G$, so this intersection is really the intersection of finitely many subgroups of finite index. Thus $\textrm{core}(K)$ also has finite index in $G$.

(ii) Equivalently, $G$ is RF if and only if for each $g\in G$ not the identity, there exists a homomorphism $\phi:G\to A$, where $A$ is a finite group, for which $\phi(g)$ is not the identity in $A$. Indeed, if $G$ is RF and $K_0$ is normal subgroup provided by (i), then $g$ does not die under the natural homomorphism from $G$ to $G/K_0$. Conversely, given such a homomorphism, the kernel of $\phi$ is a subgroup of finite index that does not contain $g$.

(iii) Also, $G$ is RF if and only if

$\bigcap_{K\subset G,\ [G:K]<\infty}K=\{1\}.$

That is, the intersection of all the subgroups of finite index in $G$ is the trivial subgroup. Were some nonidentity element $g$ to be contained in this intersection, then it would be contained in each subgroup of finite index in $G$, so this element prevents $G$ from being RF. Conversely, if this intersection is trivial, each nonidentity element of $G$ must be excluded from some finite index subgroup, so $G$ is RF.

(iv) If $G$ is RF and $g_1,\dots,g_n\in G$, then there exists a finite index subgroup $K\subset G$ with $g_j\notin K$ for all $1\leq j\leq n$. Here, just take the intersection of the finite index subgroup associated with each $g_j$. This is again a finite index subgroup of $G$.

Lemma 2: Let $G$ be a finitely generated group.

(i) If $G$ is RF and $H\subset G$ is a subgroup, then $H$ is RF.

(ii) If $H$ is RF and $H\subset G$ with $H$ finite index in $G$, then $G$ is RF.

That is, RF passes to all subgroups and also to supergroups of finite index.

Proof. For (i), choose $h\in H$. Considered as an element of $G$, there exists a homomorphism $\phi$ to a finite group $A$ for which $\phi(h)$ is not the identity. The restriction of $\phi$ to $H$ is also a homomorphism to a finite group for which the image of $h$ is nontrivial. The kernel of this restricted homomorphism is a subgroup of finite index in $H$ that does not contain $h$. (Note that we did not assume that $H$ is finitely generated.)

For (ii), choose $g\in G$. If $g\notin H$, then $H$ is a finite index subgroup not containing $g$, and we are done. If $g\in H$, then there exists a finite index subgroup $K\subset H$ that does not contain $g$. However, $K$ is also a finite index subgroup of $G$, so we are done. $\square$

Topological reformulation of RF

Now, we would like to connect RF with a topological property of a space with fundamental group $G$. Let $M$ be a compact manifold with universal covering $\widetilde{M}$ and $G=\pi_1(M)$. Accordingly, we assume throughout that $G$ is finitely generated. (We will see later that the manifold condition can be relaxed.)

Theorem 2: The group $G$ is RF if and only if the following condition holds: for every compact subset $C\subset\widetilde{M}$ there exists a finite sheeted covering $M_C\to M$ for which $C$ embeds homeomorphically in $M_C$.

Proof. Assume the topological condition holds and choose any $g\in G$ not the identity. This corresponds to a loop (based at a point dependent only upon our choice of universal covering) in $M$ which we also denote by $g$. To this loop, there also is a corresponding lift to a connected arc $a$ in $\widetilde{M}$ with distinct endpoints $x$ and $y$.

Let $C$ be the compact set $\{x,y\}$. Then, there exists a finite sheeted covering $M_C\to M$ for which $x$ and $y$ are distinct points of $M_C$. By the lifting property of covering spaces, this implies that if $K_C\subset G$ is the subgroup of finite index corresponding to the covering $M_C\to M$, then $g\notin K_C$. Therefore $G$ is RF.

Conversely, suppose that $G$ is RF and choose any subset $C\subset\widetilde{M}$. Then, since $G$ acts freely and properly discontinuous on $\widetilde{M}$, the set $T_C$ of those $g\in G$ (not the identity) for which $g C$ intersects $C$ nontrivially is finite. Now, choose $K_C\subset G$ of finite index containing none of the elements of $T_C$. If $M_C$ is the corresponding finite sheeted covering, we have that $hC\cap C=\emptyset$ for all $h\in\pi_1(M_C)$. That is, $C$ embeds homeomorphically in $M_C$. $\square$

Remark. We never really used here that $M$ was a manifold, only that its fundamental group acted properly discontinuous on the universal covering. (The action need not be free either, since this would only add another finite number of elements to our set $T_C$.) Thus, it suffices to assume that $M$ is Hausdorff and locally compact.

Examples.

(1) Finitely generated abelian groups are RF. (Exercise.)

(2) Selberg’s Lemma (Malcev-Selberg): If $G$ is a finitely generated linear group, that is, $G\subset\mathrm{GL}_N(\mathbb{C})$ for some $N$, then $G$ is RF.

Proof. We begin with the case $G\subset\mathrm{GL}_N(\mathbb{Z})$. Since RF passes to arbitrary subgroups, it suffices to prove that $\mathrm{GL}_N(\mathbb{Z})$ is RF.

Choose any $g\neq 1$. This means that the matrix $g-1$ has some nonzero entry, say $x$. Since $x$ is an integer, we can find some large prime $p$ that does not divide $x$. Now, consider the homomorphism

$\phi_p:\mathrm{GL}_N(\mathbb{Z})\to\mathrm{GL}_N(\mathbb{Z}/p\mathbb{Z})$

given by reducing the entries of a given matrix modulo $p$. This is a homomorphism because matrix multiplication is linear in the entries. (Exercise – make this precise.) Since $\mathbb{Z}/p\mathbb{Z}$ is a finite field, its general linear group is a finite group, i.e. the kernel $K_p$ of $\phi_p$ is a finite index subgroup of $\mathrm{GL}_N(\mathbb{Z})$, and it does not contain $g$ by construction. Thus $\mathrm{GL}_N(\mathbb{Z})$ is RF, and thus any subgroup thereof is also RF.

More generally, if $G\subset\mathrm{GL}_N(\mathbb{C})$ is finitely generated, we can arrange that $G\subset\mathrm{GL}_N(R)$, where $R\subset\mathbb{C}$ is the subring generated by the entries of a finite collection of generators for $G$. In particular, this is an integral domain, since it is a finitely generated subring of $\mathbb{C}$. Thus, given $x$ a nonzero element of $g-1$ as above, we can find a prime ideal $\mathcal{P}$ of $R$ that does not divide $x$, i.e. $x\notin\mathcal{P}$.

Now, $R/\mathcal{P}$ is a finite integral domain, that is, a finite field, and we can build a reduction homomorphism $\phi_{\mathcal{P}}$ analogous to the situation over the integers. The the image of the reduction homomorphism is the general linear group of a finite field, so it is a finite group. We now proceed exactly as above.  $\square$

This theorem is often referred to as Selberg’s Lemma, even though Malcev supposedly proved it first.

(3) Free groups $F_r$ of finite rank are linear, so they are RF. Notice that we have seen, in the exercise from the first lecture, that $F_r$ is a subgroup of $F_2$ for all $r\geq 0$, so it suffices to prove that $F_2$ is linear. One such representation the subgroup of $\mathrm{SL}_2(\mathbb{Z})$ generated by the two matrices

$a=\left(\begin{matrix} 1 & 2 \\ 0 & 1\end{matrix}\right)$

$b=\left(\begin{matrix} 1 & 0 \\ 2 & 1\end{matrix}\right).$

There are several ways to prove that this group is free of rank two. First, one use a bit of hyperbolic geometry and the action of $\mathrm{SL}_2(\mathbb{R})$ on the hyperbolic plane to prove that this group is the fundamental group of a hyperbolic manifold that is deformation equivalent to the rose with two petals. Also, one can use the so-called Ping-Pong Lemma, which says that a finite collection of homeomorphisms of a space that satisfy a certain set of conditions necessarily generate a free subgroup of the homeomorphism group.