## Residual Finiteness and Hopfian Groups

1. Residual Properties

Definition 1 Let ${\mathcal{P}}$ be a property of a group (e.g., finite, nilpotent, free). A group is residually ${\mathcal{P}}$ if for all ${g \in G \backslash \{1\}}$ there exists a normal subgroup ${N \unlhd G}$ such that

1. ${g \notin N}$
2. ${G/N \in \mathcal{P}}$

For example if we take ${\mathcal{P}}$ to be the class of finite groups then we get

Definition 2 (RF1) A group ${G}$ is residually finite if for all ${g\neq1\in G}$ there exists a finite index normal subgroup ${N \unlhd G}$ such that ${g \notin N}$

Equivalently,

Definition 3 (RF2) ${G}$ is residually finite if for all ${g \neq 1 \in G}$ there exists a homomorphism ${\phi:G\rightarrow F}$ such that ${\phi(g) \neq 1}$ and ${F}$ is finite.

Motivation: Our aim in making the above definition is that such groups are well approximated by their ${\mathcal{P}}$-quotients (that is, quotients in ${\mathcal{P}}$). The hope is that we may extract properties of the class ${\mathcal{P}}$ to groups that are residually ${\mathcal{P}}$. For example, the word problem is easy in the class of finite groups. Since residually finite groups are well approximated by their finite quotients we might hope that the word problem is solvable in the class of residually finite groups. This turns out to be the case.

Theorem 4 A finitely presented residually finite group ${G}$ has solvable word problem.

Proof (McKinsey’s Algorithm): We have already discussed a procedure that lists all words ${w=1 \in G}$. It remains to construct a procedure which detects non-trivial words.

1. List all finite groups (of order $n$) by writing down all possible multiplication tables.
2. For each finite group ${F}$, list the finite set of maps from generators of ${G}$ to to elements of ${F}$. This extends to a group homomorphism iff all relators of ${G}$ map to ${1_F}$. Thus we produce a list of homomorphisms from ${G \rightarrow F}$. Do this for all finite groups.
3. Since ${G}$ is residually finite, if a word ${w}$ is non-trivial in ${G}$ it will get sent to a non-trivial element in ${F}$ for some finite group ${F}$, and hence this procedure will terminate.

$\Box$

2. Examples

So this class of residually finite groups has solvable word problem. We better have some examples of residually finite groups then! Below we give some equivalent definitions of residual finiteness and use them to find some examples of residually finite groups.

Firstly we show that free groups are residually finite.

Proposition 5 Let ${F=F(A)}$ be the group freely generated by the set ${A}$. Then ${F}$ is residually finite.

Proof: Let ${x =x_1\cdots x_n \in F}$ be a freely reduced non-trivial word on ${A\cup A^{-1}}$. We will produce a homomorphism ${F \rightarrow S_{n+1}}$ such that the image of ${x}$ is non-trivial, whence ${F}$ satisfies (RF2).

Define ${f:A \rightarrow S_{n+1}}$ by

1. ${f(a)=id}$ if ${a \notin \{x_1^{\pm1}, \cdots x_n^{\pm1}\}}$
2. ${f(a)}$ is any permutation sending ${i \mapsto i+1}$ for all ${i}$ such that ${x_i=a}$ and sending ${j+1 \mapsto j}$ for all ${j}$ such that ${x_j^{-1}=a}$.

This extends uniquely to a homomorphism since ${F}$ is free. Moreover, by construction ${f(x):1 \mapsto n+1}$ thus is non-trivial.

$\Box$

Example of proof: ${F_2=\langle a,b \rangle, x=x_1x_2x_3=aba^{-1}}$. We have

$\displaystyle f(a) = \left(\begin{array}{cccc}1 & 2 & 3 & 4 \\2 & 1 & 4 & 3\end{array}\right) \quad f(b)=\left(\begin{array}{cccc}1 & 2 & 3 & 4 \\1 & 3 & 2 & 4\end{array}\right)\notag \ \ \ \ \ (1)$

First observe that ${f(a^{-1})=f(a)^{-1}=f(a)}$. Thus

$\displaystyle f(aba^{-1}):1 \mapsto 2 \mapsto 3 \mapsto 4 \notag \ \ \ \ \ (2)$

and is thus non-trivial.

The following definition shows that we can remove the stipulation that our finite index subgroups be normal.

Definition 6 (RF3) ~

1. ${G}$ is residually finite if the intersection of all finite index subgroups is trivial
2. ${\forall \ g \in G \backslash \{1\} \ \exists}$ a finite index subgroup ${H \leq G}$ not containing ${g}$

Proof: This follows from Poincaré’s theorem which says that any subgroup of index ${n}$ contains a normal subgroup, called its normal core, of index dividing ${n!}$. In particular it is of finite index. [The normal core is defined as the kernel of map ${G \rightarrow Sym(n)}$ induced by the action of ${G}$ on the left coset space ${G/H}$.] $\Box$

This allows us to give a topological interpretation of residual finiteness.

Definition 7 (RF4) ${G}$ is residually finite if there exists a 2-complex ${K}$ with ${\pi_1(K)=G}$ such that for any non-trivial loop ${\alpha}$ in ${K}$ there exists a finite sheeted cover ${\widetilde{K}}$ of ${K}$ such that the lift ${\widetilde{\alpha}}$ of ${\alpha}$ to ${\widetilde{K}}$ is not a closed loop.

Proposition 8 ${\mathbb{Z}}$ is residually finite.

Proof: ${\pi_1(S^1)=\mathbb{Z}}$. Consider the loop ${\alpha}$ that wraps around ${S^1}$ ${m}$-times. Take the standard ${m+1}$ sheeted cover of ${S^1}$. Then ${\alpha}$ lifts to a path that is not closed. $\Box$

If ${A}$ and ${B}$ are residually finite then so is ${A \times B}$. It is also immediate that all finite groups are residually finite. Therefore, by the structure theorem for finitely generated abelian groups, we get the following result.

Proposition 9 All finitely generated abelian groups are residually finite.

Definition 10 (RF5) ${G}$ is residually finite if the intersection of all finite index (normal) subgroups is trivial.

Proof: This is clearly equivalent to (RF3). $\Box$

Proposition 11 ${SL(n, \mathbb{Z})}$ is residually finite

Proof: We consider the congruence subgroups ${SL(n,\mathbb{Z})[m]}$ defined to be the kernel of the map

$\displaystyle SL(n,\mathbb{Z}) \rightarrow SL(n, \mathbb{Z}/m\mathbb{Z}) \notag \ \ \ \ \ (3)$

It is immediate that ${SL(n,\mathbb{Z})[m]}$ is a finite index normal subgroup. Now observe that

$\displaystyle \mathcal{I}:=\bigcap_{m \geq3} SL(n, \mathbb{Z})[m] = \{1\} \notag \ \ \ \ \ (4)$

for if ${A \in \mathcal{I}}$ then all off diagonal entries must be congruent to ${0\mod m}$ for all ${m \geq 3}$, and hence must be 0. Since ${\det A =1}$, the diagonal entries are ${\pm1}$. Since they are each congruent to ${1\mod m}$ for all ${m\geq 3}$ however, they must in fact be 1. Thus ${A}$ is the identity matrix.

$\Box$

In fact, we have the following powerful result.

Theorem 12 (Malcev) Finitely generated linear groups are residually finite.

In fact, all the examples we have seen so far are linear groups, and thus are corollaries to the above theorem. There do exist non-linear residually finite groups. A particularly nice example is the following one-relator group.

Theorem 13 (Drutu-Sapir) The group ${\langle a, t | a^{t^2}=a^2 \rangle}$ is residually finite and non-linear.

We now introduce the profinite topology and profinite completions. This will give us yet another way to define residual finiteness. This will open the door to the study of profinite groups (we will not, however, walk through it in this blog post).

Definition 14 Any group ${G}$ can be made into a topological space by taking as a basis of open neighborhoods of the identity the collection of all finite index normal subgroups of ${G}$. This gives the profinite topology on ${G}$

Definition 15 (RF6) A group ${G}$ is residually finite if its profinite topology is Hausdorff.

Proof: Suppose ${G}$ satisfies (RF1). Let ${x \neq y \in G}$. Then since ${xy^{-1} \neq 1 \in G}$ there exists a finite index ${N}$ such that ${xy^{-1} \notin N}$. Thus ${xN \cap yN = \varnothing}$. One sees that this argument can readily be reversed.

$\Box$

Yet another way to phrase the definition is in terms of profinite completions. First we recall the notion of an inverse limit. An inverse system of groups is a family ${\{G_i\}_{i \in I}}$ of groups ${G_i}$, where ${I}$ is a directed set (i.e., a poset such that ${\forall \ i, j \in I, \ \exists \ k \in I}$ such that ${i \leq k, j \leq k}$), together with a collection of maps ${\{\phi_{ij}:G_i \rightarrow G_j\}_{i\geq j}}$ for all ${i \geq j}$ satisfying

1. ${\phi_{ii} = id}$
2. ${\phi_{ij}\phi_{jk}=\phi_{ik}}$ for all ${i \geq j \geq k}$

Given such data we can define the inverse limit (sometimes called the projective limit) as

$\displaystyle \varprojlim G_i = \left\{ (g_i) \in \prod_{i \in I} G_i : \phi_{ij}(g_i)=g_j \ \forall \ i \geq j \right\} \notag \ \ \ \ \ (5)$

Example. Fix a group ${G}$. The collection of finite index normal subgroups ${\mathcal{N}}$ is a directed set with partial order given by containment. We thus define the inverse system ${\{G/N : N \in \mathcal{N}\}}$ with maps ${\phi_{NM}:G/N \rightarrow G/M}$ given by ${gN \mapsto gM}$ whenever ${M \leq N}$. The inverse limit of this system is the profinite completion of ${G}$

$\displaystyle \widehat{G} = \varprojlim_{\mathcal{N}} G/N = \left\{ (gN) \in \prod_{N \in \mathcal{N}} G/N : \phi_{NM}(gN)=gM \right\} \notag \ \ \ \ \ (6)$

There is an obvious map ${G \rightarrow \widehat{G}}$ defined by ${g \mapsto (gN)_{N \in \mathcal{N}}}$. If this map is injective then ${G}$ is residually finite.

Definition 16 (RF7) ${G}$ is residually finite if ${G}$ injects into it’s profinite completion.

Proof: Suppose ${G}$ is residually finite. Then for all ${g\neq1 \in G}$ there exists ${N \lhd G}$ of finite index such that ${gN \neq N}$. Therefore the natural map ${G \rightarrow \widehat{G}}$ by ${g \mapsto (gN)_{N \in \mathcal{N}}}$ is an injection.

We prove the converse by contradiction. Suppose ${G}$ is not residually finite. Then there exists ${g\neq1 \in G}$ such that ${g \in N}$ for all ${N \in \mathcal{N}}$. But then ${g \mapsto (gN)_{N \in \mathcal{N}} = id}$ and ${g}$ is in the kernel.

$\Box$

3. Stability Properties

We have seen some examples of residually finite groups. In this section we give some techniques that allow us to construct new residually finite groups from old ones.

Proposition 17 Residual finiteness is closed under taking subgroups.

Proof: Suppose ${G}$ is residually finite. Then by (RF5)

$\displaystyle \bigcap_{N \unlhd G} N = \{1\} \notag \ \ \ \ \ (7)$

Let ${H \leq G}$, then, by the second isomorphism theorem ${N \cap H \unlhd H}$ and thus

$\displaystyle \bigcap_{M \unlhd H} M \leq \bigcap_{N\unlhd G} (N \cap H)= \{1\} \notag \ \ \ \ \ (8)$

$\Box$

Proposition 18 Any inverse limit of residually finite groups is residually finite.

Proof: Let ${x = (x_i)_{i \in I} \in \varprojlim X_i}$ be a nontrivial element in an inverse limit of residually finite groups ${X_i}$. It is non-trivial, therefore some ${x_i\in X_i}$ is non-zero. But ${X_i}$ is residually finite, so ${\exists \phi_i:X_i \rightarrow F}$ a finite group such that ${\phi_i(x_i) \neq 1 \in F}$. So we just project to the ${i}$th factor and use residual finiteness there.

$\Box$

Remark

1. Profinite groups are inverse limits of finite groups, and thus are residually finite.
2. This gives us an alternative way to see that (RF7) is an equivalent definition. Indeed, ${\widehat{G}}$ is an inverse limit of finite groups, and is therefore residually finite by 18. If ${G}$ injects into its profinite completion then ${G}$ is residually finite by Proposition 17.

Proposition 19 (Baumslag – Residual finiteness is closed under passing from a group to its groups of automorphisms) Let ${G}$ be a finitely generated residually finite group. Then ${Aut(G)}$ is residually finite.

Proof: Let ${A=Aut(G)}$, and let ${a\neq1 \in A}$. Then there exists a ${g \in G}$ such that ${a(g)\neq g}$. Define

$\displaystyle h = a(g)g^{-1} \notag \ \ \ \ \ (9)$

Then, since ${h\neq1 \in G}$ residually finite, by (RF1) there exists a finite index normal subgroup ${N \leq G}$ such that ${h \notin N}$. We recall the following result of Marshall Hall, and a the definition of a characteristic subgroup.

Fact.(M.Hall) In a finitely generated group there are only finitely many finite index subgroups of a fixed index ${n}$.

Definition. A characteristic subgroup ${K \leq G}$ is one that is invariant under all automorphisms of ${G}$. (Reminder: normal subgroups are defined as being invariant under conjugation, and thus characteristic subgroups are normal. The converse, however, is not true in general.)

Define ${K}$ to be the intersection of all subgroups of index ${[G:N]}$. This is

1. a characteristic subgroup – by construction.
2. ${h \notin K}$ – indeed since ${N}$ is of index ${[G:N]}$ in ${G}$, and ${N}$ doesn’t contain ${h,}$ nor does ${K}$.
3. ${K}$ is finite index in ${G}$ – indeed, the intersection of finite index subgroups is finite index.

Since $K$ is characteristic, we get that ${A}$ induces a finite group ${\overline{A} (=A/L}$) of automorphisms of the finite group ${G/K}$. Since ${h \notin K}$ we see that ${a}$ induces a non-trivial automorphism of ${\overline{A}}$ i.e. ${a \notin L}$. Thus ${A}$ is residually finite.

$\Box$

4. Hopfian Groups

We have seen many ways to show that a group is residually finite. We now give some ways to show that one is not.

Definition 20 A group ${G}$ is Hopfian if either of the following (equivalent statements) hold:

1. ${G}$ is not isomorphic to any of its proper quotients
2. Any epimorphism ${\pi: G \twoheadrightarrow G}$ is an isomorphism

(related via first isomorphism theorem!!! ${1 \rightarrow K \rightarrow G \xrightarrow{\pi} G \rightarrow 1}$

Theorem 21 (Mal’cev) Let ${G}$ be a finitely generated residually finite group. Then ${G}$ is Hopfian.

Proof: Suppose ${G \cong G/N}$. As $G$ is finitely generated, it is a fact of Marshal Hall that the number of subgroups of index $m$ in $G$ is finite, and,

#subgroups of finite index m in G/M = #subgroups of index m in G

the correspondence being

$\displaystyle H_1/N, \ldots, H_r/N \quad \leftrightarrow \quad H_1, \ldots, H_r \notag \ \ \ \ \ (11)$

Note: it is immediate that ${N \leq H_i \ \forall \ i}$. Therefore ${N}$ is contained in all finite index subgroups of ${G}$ and thus

$\displaystyle N \leq \bigcap_{K \leq G, [G:K]<\infty} K = \{1\}\notag \ \ \ \ \ (12)$

where the last equality follows because ${G}$ is residually finite (RF5). Therefore ${N = \{1\}}$ and ${G}$ is Hopfian.

$\Box$

Remark

1. First we point out that the hypothesis that ${G}$ is finitely generated is necessary. Here are two examples of groups that are residually finite but not Hopfian.
1. Let ${G = \oplus_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}}$. It is clear that ${G}$ is residually finite (any element lives in a finitely generated abelian group, and these are residually finite, as we have seen). To see ${G}$ is non Hopfian, consider the surjection ${G \rightarrow G}$ given by ${(a_1, a_2, a_3, \ldots) \mapsto (a_2, a_3, a_4, \ldots)}$. It has non-trivial kernel.
2. Infinitely generated free groups are residually finite but are not Hopfian.
2. The converse of this theorem is false. There are examples of finitely generated groups which are Hopfian but not residually finite. We will see some below.

Here is a simple application.

Theorem 22 (Nielsen) Suppose ${F}$ is a free group of finite rank ${n}$ generated by a set ${X}$. Then ${X}$ freely generates ${F}$

Proof: Suppose ${F}$ is freely generated by a set ${Y}$. Any map ${\phi: Y \twoheadrightarrow X}$ induces an epimorphism on $F$. But since ${F}$ is free and thus residually finite, it is Hopfian, and thus ${\phi}$ is an automorphism. Hence ${X}$ freely generates ${F}$.

$\Box$

One way then to show that a finitely generated group is not residually finite is to show that it is non-Hopfian. The following is an example of a more general theorem of Baumslag and Solitar. It is based on the following definition of the Baumslag-Solitar group

$\displaystyle BS(m,n) = \langle a, b | a^{-1}b^ma = b^n \rangle \notag \ \ \ \ \ (13)$

Proposition 23 (Baumslag-Solitar) The (finitely generated) Baumslag-Solitar group ${BS(2,3)}$ is non-Hopfian and thus not residually finite.

Proof: Let ${G = BS(2,3)}$. Define ${\mu: G \rightarrow G}$ by ${\mu(a)=a}$ and ${\mu(b)=b^2}$. Then

1. ${\mu}$ is an epimorphism ${G \twoheadrightarrow G}$ – indeed can get ${a}$ and ${b}$ in the image

$a = \mu(a), b = \mu([a, b^{-1}]))$)

2. ${\mu}$ has non-trivial kernel – for example,

$(a^{-1}bab^{-1})^2b^{-1}$

Thus ${G/\ker\mu \cong G}$ and we are done.

$\Box$

Here is the full theorem.

Theorem 24 (Baumslag-Solitar, Meskin) The groups ${BS(m,n)}$ are

1. Hopfian if $m$ and $n$ have the same prime factors
2. Residually finite if $|m|$ or $|n|=1$, or $|m|=|n|$

Therefore this construction contains examples which are

1. Non-Hopfian: for example $BS(2,3)$
2. Hopfian, non-residually finite: for example $BS(2,4)$
3. Residually finite: for example $BS(1,n)$.

The groups ${BS(m,n)}$ are examples of a more general construction, called HNN-extensions.

5. HNN extensions

We begin with a motivating question. Given two isomorphic subgroups ${A\cong B \subseteq G}$, does there exist a group ${\widetilde{G} \geq G}$ in which ${A}$ and ${B}$ are conjugate? The question was answered affirmatively by Graham Higman, B. H. Neumann and Hanna Neumann, after whom the groups are named.

Definition 25 (HNN-extension) Let ${A, B \leq G}$ and ${\phi:A \rightarrow B}$ an isomorphism. Define

$\displaystyle G_\phi = \langle G, t | t^{-1}at=\phi(a) \ \forall \ a \in A \rangle \notag \ \ \ \ \ (14)$

The key result relating to these groups is Britton’s lemma, which allows us to develop a normal form for elements of ${G_\phi}$.

Proposition 26 (Britton’s Lemma) Let ${w=g_0t^{e_1}g_1t^{e_2}\cdots g_nt^{e_{n+1}}}$ where ${e_i=\pm1, n\geq1, g_i \mbox{in} G}$. Suppose ${w=1\in G}$. Then it contains a subword of the form

1. ${t^{-1}gt}$ for some ${g \in A}$, or
2. ${tgt^{-1}}$ for some ${g \in B}$

This subword is called a pinch

${G_\phi}$ is a quotient of ${G \ast \langle t \rangle}$. Every element ${w \in G_\phi}$ is an image of

$\displaystyle x=g_1t^{e_1}\cdots g_nt^{e_n}g_{n+1} \in G \ast \langle t \rangle \notag \ \ \ \ \ (15)$

where ${g_i \in G, \epsilon_i=\pm1}$, and ${x}$ contains no pinches. Such an element of ${G \ast \langle t \rangle}$ is said to be of reduced form.

HNN-extensions over finite subgroups play nicely with residual finiteness.

Theorem 27 Suppose ${G}$ is residually finite, and ${|A| < \infty}$. Then ${G_\phi}$ is residually finite.

We sketch the proof of Cohen [Residual finiteness and Britton’s Lemma] Proof: Let ${w \in G_\phi}$ be a non-trivial word, with corresponding reduced form ${x=g_1t^{e_1}\cdots g_nt^{e_n}g_{n+1}}$, as above. The first step is to reduce to the case ${|G|}$ finite. Since ${G}$ is residually finite, and ${|A|, |B| < \infty}$, we can find a finite index normal subgroup ${N \lhd G}$ such that

$\displaystyle N \cap \left( \bigcup_{i=0}^{n+1} g_i(A \cup B) \right) = \{1\} \notag \ \ \ \ \ (16)$

We use here the fact that residually finite implies fully residually finite, meaning, that for any finite set of non-trivial elements $g_1, \ldots, g_n$ there exists a finite index normal subgroup $N \lhd G$ such that $g_i \notin N$.

We take the natural map

$\displaystyle G \rightarrow \overline{G}:=G/N \notag \ \ \ \ \ (17)$

under which ${A \rightarrow \overline{A}}$ and ${B \rightarrow \overline{B}}$. Moreover, if ${g_i \mapsto \overline{g_i}}$, then ${g_i \in A \Rightarrow \overline{g_i} \in \overline{A}}$ and ${g_i \in B \Rightarrow \overline{g_i} \in \overline{B}}$. Hence we have a map

$\displaystyle G_\phi \rightarrow \overline{G}_\phi = \langle \overline{G}, t | t^{-1}\overline{a}t= \phi(\overline{a}) \ \forall \ \overline{a} \in \overline{A} \rangle \notag \ \ \ \ \ (18)$

where ${\phi(\overline{a}) \in \overline{B}}$. Under this map we have ${w \mapsto \overline{w}}$ which has reduced form

$\displaystyle \overline{g} = \overline{g}_1t^{e_1}\cdots \overline{g}_nt^{e_n}\overline{g}_{n+1} \notag \ \ \ \ \ (19)$

We have now reduced to the problem to finding a homomorphism ${\psi:\overline{G}_{\phi}\rightarrow F}$ with ${F}$ finite such that ${\psi(\overline{g})\neq1}$. Since we can take ${N}$ to be of finite index in ${G}$, we have reduced the problem to the case ${|G|<\infty}$.

Given that ${G}$ is finite, we can do something similar to our proof that free groups are residually finite, namely, building a non-trivial permutation of ${G \times \{1, \ldots, n+1\}}$ corresponding to our word ${w}$.

$\Box$

6. More examples

We start by showing that surface groups are residually finite. We follow  simultaneously the proofs of [Hemple] and of [Malestein-Putman].

Theorem 28 Let ${\Sigma}$ be a compact surface. Then ${\pi_1(S)}$ is residually finite.

Proof: We can assume that $\Sigma$ is orientable, as if not we can take its double cover. If ${\pi_1(\Sigma)}$ is abelian it is residually finite, so we assume it is not. Let ${\alpha: S^1 \rightarrow S}$ be a non-trivial curve on ${S}$. Our aim is to prove that there exists a finite sheeted cover ${\widetilde{S}}$ of ${S}$ such that the lift ${\widetilde{\alpha}}$ of ${\alpha}$ is not closed, as per (RF4), or equivalently that there exists a finite quotient ${\phi:\pi_1(\Sigma)\rightarrow H}$ for which ${\phi(\alpha)\neq1}$ (RF2). We will use both definitions in the proof.

We prove this by induction on the singular set ${S(\alpha)}$, that is, the number of self-intersections of ${\alpha}$.

Suppose ${\alpha}$ is an embedding. If ${\alpha}$ is not nullhomologous, then the image of ${\alpha}$ under the abelianisation map, ${\Lambda: \pi_1(\Sigma) \rightarrow H_1(\Sigma;\mathbb{Z})}$ is non-trivial. But ${H_1(\Sigma;\mathbb{Z})}$ is finitely generated abelian and hence residually finite, so we are done.

Suppose then that ${\alpha}$ is nullhomologous. Let ${\Sigma}$ have genus ${g}$ and ${b}$ boundary components. Then we have the following presentation for ${\pi_1(\Sigma)}$

$\displaystyle \pi_1(\Sigma) = \langle x_1, y_1, \ldots, x_g, y_g, z_1, \ldots, z_b | [x_1, y_1]\cdots[x_g, y_g]=z_1\cdots z_b \rangle \notag \ \ \ \ \ (20)$

in which we can write ${\alpha = [x_1, y_1]\cdots[x_r, y_r]}$ for some ${r. We construct a homomorphism ${\phi}$ to the dihedral group of order 8, for which ${\phi(\alpha)\neq 1}$, as follows.

$\displaystyle \phi: \pi_1(\Sigma) \rightarrow D_8=\langle \sigma, \rho:\sigma^2=\rho^4=1, \sigma\rho\sigma=\rho^{-1} \rangle \notag \ \ \ \ \ (21)$

by

$s \mapsto \left\{\begin{array}{ccc}\sigma & ~ & s=x_1, x_g \\ \sigma\rho & ~ & s=y_1, y_g \\1 & ~ & \mbox{else}\end{array}\right.$

It is simple to check that ${\phi([x_1, y_1][x_g, y_g])=1}$, and hence ${\phi}$ is a homomorphism (in fact, a surjective homomorphism).

Now for the induction, suppose that ${S(\alpha)\neq\varnothing}$. Let ${U}$ be a regular neighborhood of ${\alpha}$ in ${\Sigma}$. Then there exists a simple loop ${\beta:(S^1, \ast)\rightarrow(U, \ast)}$ representing a non-trivial element of ${\pi_1(\Sigma)}$. By the previous case there exists a finite sheeted covering

$\displaystyle q:\widetilde{\Sigma}\rightarrow\Sigma \notag \ \ \ \ \ (21)$

such that ${\beta}$ doesn’t lift to a loop in ${\widetilde{\Sigma}}$. The idea is that in this cover we have reduced the number of self-intersections of ${\alpha}$ (by picking out a loop created by the self-intersection, and lifting it to a finite sheeted cover in which it unwinds). Indeed, if ${\alpha}$ lifts to a loop ${\widetilde{\alpha}:(S^1, \ast) \rightarrow (\widetilde{\Sigma}, \ast)}$, then ${S(\widetilde{\alpha})\subseteq S(\alpha)}$. Suppose that ${S(\widetilde{\alpha})= S(\alpha)}$. Then ${q|_{\widetilde{\alpha}}}$ would be an embedding and as such ${q}$ would map a neighborhood of ${\widetilde{\alpha}}$ homomorphically onto ${U}$. But then ${\beta}$ would lift to a loop in ${\widetilde{\Sigma}}$, which is our contradiction. We are thus done by induction.

$\Box$

References

[Hemple] John Hemple – Residual finiteness of surface groups

[Malestein, Putman] On the self intersection of curves deep in the lower central series of a surface group