## Homomorphisms from the earring group to finite groups

One of the the surprising things about the earring group (the fundamental group of the earring space $\mathbb{E}$) is that the group of homomorphisms $Hom(\pi_1(\mathbb{E}),\mathbb{Z})$ to the additive group of integers is countable (see this post for details) even though $\pi_1(\mathbb{E})$ is torsion-free and uncountable. In this post we’ll see that the situation is completely different when we replace $\mathbb{Z}$ with a finite group!

The earring space is the planar subspace $\mathbb{E}=\bigcup_{n=1}^{\infty}C_n$  where $C_n\subseteq\mathbb{R}^2$ is the circle of radius $\frac{1}{n}$ centered at $\left(\frac{1}{n},0\right)$. The basepoint is the origin $(0,0)$, which is the only point in the intersection of all of the shrinking circles. Let $\ell_n:S^1\to C_n$ be the loop traversing $C_n$ once in the counterclockwise direction and $[\ell_n]\in\pi_1(\mathbb{E})$ be its homotopy class.

In this post, we’ll work through the proof of the following result.

Theorem 1: Let $G$ be any non-trivial finite group. There are uncountably many surjective group homomorphisms $\pi_1(\mathbb{E})\to G$ mapping $[\ell_n]$ to the identity element for every $n\geq 1$.

Theorem 1 could be considered “surprising” since every homomorphism from $\pi_1(\mathbb{E})$ to a free group which kills all classes $[\ell_n]$ must be the trivial homomorphism. An obvious consequence is the following corollary.

Corollary 2: For any non-trivial finite group $G$, the set* of group homomorphism $Hom(\pi_1(\mathbb{E}),G)$ is uncountable.

### An anomalous aside

The only place I know of Theorem 1 being proven in the literature is Conner and Spencer’s paper [1] entitled “The anomalous behavior of the Hawaiian earring group.” This is a nice, short paper which clarified some things about the earring group. Conner and Spencer state in the introduction of [1] that “The aim of the current article is to understand some of the anomalous behavior of homomorphisms from the fundamental group of the earring group into finite groups.” According to Theorem 1, the hom-set $Hom(\pi_1(\mathbb{E}),G)$ certainly is large and complicated, however, I’m going to try to convince you in this post that this result really doesn’t have a lot to do with the intricate structure of the earring group. Instead, this craziness is due to the generally complicated behavior of morphisms out of infinite categorical limits (specifically products).

As we break apart the argument in [1], we’ll see that $Hom(\pi_1(\mathbb{E}),G)$ is uncountable essentially because $Hom(\prod_{k=1}^{\infty}G,G)$ is uncountable. Similarly, Theorem 1 follows directly from the existence of uncountably many epimorphisms $\prod_{k=1}^{\infty}G\to G$ which vanish on the subgroup of eventually trivial sequences. These algebraic statements about finite groups and their products do not depend on the group structure of $\pi_1(\mathbb{E})$.

The infinite product $\prod_{k=1}^{\infty}\mathbb{RP}^n$ of n-dimensional projective space ($n>1$) exhibits the same sort of “anomalous” behavior since $\pi_1\left(\prod_{k=1}^{\infty}\mathbb{RP}^n\right)=\prod_{k=1}^{\infty}\pi_1(\mathbb{RP}^n)=\prod_{k=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$. So yes, the earring group is quite anomalous, however, this particular anomaly is not unique to $\mathbb{E}$.

### Ultrafilters and finite groups

Definition 3: A filter on a set $X$ is a non-empty set $\mathscr{F}$ of subsets of $X$ satisfying the properties:

1. If $A,B\in\mathscr{F}$, then $A\cap B\in\mathscr{F}$,
2. $\emptyset\notin\mathscr{F}$,
3. If $A\in\mathscr{F}$, and $A\subseteq B$, then $B\in\mathscr{F}$.

A filter $\mathscr{F}$ is an ultrafilter if for every subset $A\subseteq X$, either $A\in\mathscr{F}$ or $X\backslash A\in\mathscr{F}$. An ultrafilter $\mathscr{F}$ is a non-principal ultrafilter if $\mathscr{F}$ does not contain any finite subsets of $X$.

We’re mainly going to be using non-principal ultrafilters on the natural numbers $\mathbb{N}=\{1,2,...\}$. The existence of non-principal ultrafilters on infinite sets is equivalent to the axiom of choice. In fact the remainder of the Stone-Cech compactification $\beta\mathbb{N}\backslash\mathbb{N}$ is in bijective correspondence with the non-principal ultrafilters on $\mathbb{N}$. In particular, there are uncountably many non-principal ultrafilters on $\mathbb{N}$.

Lemma 4: If $\mathscr{U}$ is a non-principal ultrafilter on $X$ and $\{A_1,...,A_p\}$ is a finite partition of $X$, then there is a unique $i$ such that $A_i\in\mathscr{U}$.

Proof. Suppose $A_i\notin\mathscr{U}$ for all $i$. Then $X\backslash A_i\in\mathscr{U}$ for each $i$ and thus $\bigcap_{i=1}^{p}X\backslash A_i=X\backslash\bigcup_{i=1}^{p}A_i=\emptyset\in\mathscr{U}$ which is false. Thus $A_i\in\mathscr{U}$ for some $i$. Suppose $A_i,A_j\in\mathscr{U}$ for $i\neq j$. Then $(X\backslash A_i)\cup (X\backslash A_j)=X\backslash(A_i\cap A_j)=X\notin\mathscr{U}$ which cannot be. Thus there is a unique $i$ such that $A_i\in\mathscr{U}$. $\square$

Let $G$ be a finite group with trivial element $1$ and $\prod_{k=1}^{\infty}G=G\times G\times\cdots$ be the infinite direct product whose elements are sequences $(g_1,g_2,...)$ where $g_k\in G$.

Lemma 5: Let $K$ be the subgroup of $\prod_{k=1}^{\infty}G$ consisting of sequences $(g_k)$ such that $g_k=1$ for all but finitely many $k$. There exists uncountably many epimorphisms $F:\prod_{k=1}^{\infty}G\to G$ such that $F(K)=1$.

Proof. Let $\mathscr{U}$ be a non-principal ultrafilter on $\mathbb{N}$. We define a homomorphism $F:\prod_{k=1}^{\infty}G\to G$. To define $F=F_{\mathscr{U}}$, fix a sequence $s=(g_1,g_2,...)\in\prod_{k=1}^{\infty}G$. For each $g\in G$, let $A_{g}^{s}=\{k\in\mathbb{N}|g_k=g\}$. Since $G$ is finite $\{A_{g}^{s}|g\in G\}$ is a finite partition of $\mathbb{N}$. According to Lemma 4, there is a unique $g\in G$ such that $A_{g}^{s}\in\mathscr{U}$. Thus we have a well-defined function $F$ when we set $F(s)=g$ where $A_{g}^{s}\in\mathscr{U}$.

claim 1: $F$ is a homomorphism. Suppose $t=(h_1,h_2,...)$ is another sequence and $F(t)=h$ where $A_{h}^{t}=\{k\in\mathbb{N}|h_k=h\}\in\mathscr{U}$. Consequently,

$A_{g}^{s}\cap A_{h}^{t}=\{k\in\mathbb{N}|g_k=g\text{ and }h_k=h\}\in\mathscr{U}.$

But $st=(g_1h_1,g_2h_2,...)$ and

$A_{g}^{s}\cap A_{h}^{t}\subseteq A^{st}_{gh}=\{k\in\mathbb{N}|g_{k}h_{k}=gh\}.$

and therefore $A^{st}_{gh}\in\mathscr{U}$. It follows from our definition of $F$ that $F(st)=gh=F(s)F(t)$.

claim 2: $F$ is surjective. Let $g\in G$ and pick any set $A\in\mathscr{U}$. If $s=(g_k)$ is the sequence where $g_k=\begin{cases} g & \text{ if }k\in A\\ 1 & \text{ if }k\notin A \end{cases}$ then clearly $F(s)=g$. Thus $F$ is surjective.

claim 3: $F(K)=1$. Let $s=(g_k)$ be a sequence such that $g_k=1$ for all but finitely many $k\in\mathbb{N}$. Since $A_{g}^{s}$ is finite when $g\neq 1$, we must have $A_{1}^{s}\in\mathscr{U}$. By definition of $F$, $F(s)=1$.

Claims 1-3 guarantee the existence of a desired epimorphism $F_{\mathscr{U}}$ for each non-principal ultrafilter $\mathscr{U}$. We check uniqueness last.

claim 4: $F_{\mathscr{U}}\neq F_{\mathscr{V}}$ if $\mathscr{U}\neq\mathscr{V}$. If $\mathscr{V}$ is a non-principal ultrafilter, distinct from $\mathscr{U}$, there is an infinite set $A\subset X$ such that $A\in\mathscr{U}$ and $A\notin\mathscr{V}$. Pick $g\neq 1$ and define a sequence $s=(g_k)$ by $g_k=\begin{cases} 1 & \text{ if }k\in A\\ g & \text{ if }k\notin A \end{cases}$. Then $F_{\mathscr{U}}(s)=1$ and $F_{\mathscr{V}}(s)=g$. Thus $F_{\mathscr{U}}\neq F_{\mathscr{V}}$.

Since there are uncountably many non-principal ultrafilters on $\mathbb{N}$, we have proved the lemma. $\square$

Remark: In Lemma 5 we basically defined a canonical injection

$\displaystyle\beta\mathbb{N}\backslash\mathbb{N}\to Hom\left(\prod_{k=1}^{\infty}G/K,G\right)$.

Corollary 6: For any non-trivial finite group $G$, the hom-sets $\displaystyle Hom\left(\prod_{k=1}^{\infty}G,G\right)$ and $\displaystyle Hom\left(\prod_{k=1}^{\infty}G/K,G\right)$ are uncountable.

### Completing the proof of Theorem 1

Proof of Theorem 1. Suppose $G$ has generating set $S=\{g_1,...,g_m\}$ and let $F_m=F(g_1,...,g_m)$ be the free group on these generators. Let $q:F_m\to G$ be the canonical epimorphism induced by the inclusion $S\to G$.

Let $D_1=\bigcup_{1\leq n\leq m}C_n$ be the first wedge of m-circles, $D_2=\bigcup_{m+1\leq n\leq 2m}C_n$ be the next m-circles, and so on. Thus $D_k=\bigcup_{(k-1)m+1\leq n\leq km}C_n$. Since each of these spaces is a wedge of m-circles, we have $\pi_1(D_k)=F_m$ for each $k\geq 1$. The retraction $\mathbb{E}\to D_k$, which collapses all other circles to the basepoint, induces a surjection $f_k:\pi_1(\mathbb{E})\to F_m$ on fundamental groups. Together, these surjections induce a canonical surjection $f:\pi_1(\mathbb{E})\to \prod_{k=1}^{\infty}F_m$, $f([\alpha])=(f_1([\alpha]),f_2([\alpha]),...)$. The argument that $f$ is a surjection is essentially the same as the one we used to show $\pi_1(\mathbb{E})$ is uncountable. We’re going to work with the surjective composition:

Pre-composition with $\left(\prod_{k=1}^{\infty}q \right)\circ f$ defines a function $\eta: Hom(\prod_{k=1}^{\infty}G,G)\to Hom(\pi_1(\mathbb{E}),G)$. In detail, $\eta(F)([\alpha])=F(q(f_1([\alpha]),q(f_2([\alpha]),...)$.

Notice that $\eta$ is injective since it is precisely the function $Hom\left(\left(\prod_{k=1}^{\infty}q\right)\circ f,G\right)$ induced by the hom-functor $Hom(-,G):\mathbf{Grp}\to\mathbf{Set}$ from the category of groups to the category of sets. Since $\prod_{k=1}^{\infty}q \circ f$ is an epimorphism in $\mathbf{Grp}$, $\eta=Hom\left(\left(\prod_{k=1}^{\infty}q\right)\circ f,G\right)$ is a monomorphism in $\mathbf{Set}$…or you could just check directly that $\eta$ is injective using the fact that $\prod_{k=1}^{\infty}q$ and $f$ are surjective. Since we’ve embedded an uncountable set into $Hom(\mathbb{E},G)$, this hom-set must be uncountable (this proves Corollary 2). We still need to finish the stronger statement of Theorem 1.

We’ll use two other facts that are easy observations about $\eta$:

Fact I: If $F$ is surjective, then so is $\eta(F)=F\circ(\prod_{k=1}^{\infty}q)\circ f$.

Fact II: For fixed $n\geq 1$, the image $q(f_k([\ell_n]))$ can only be non-trivial if $(m-1)k+1\leq n\leq mk$ for a unique $k$.

Recall from Lemma 5 that there are uncountably many epimorphisms $F:\prod_{k=1}^{\infty}G\to G$ such that $F(K)=1$. Fact II implies that, for each $n\geq 1$, we have $(\prod_{k=1}^{\infty}q)\circ f([\ell_n])\in K$ since this sequence is non-trivial in at most one component. Thus for each of the uncountably many surjections $F=F_{\mathscr{U}}$ constructed in Lemma 5, we have a unique (since $\eta$ is injective) surjection (by Fact I) $\eta(F):\pi_1(\mathbb{E})\to G$ such that

$\eta(F)([\ell_n])=F\circ(\prod_{k=1}^{\infty}q)\circ f([\ell_n])\in F(K)=1$

for each $n\geq 1$. This completes the proof of Theorem 1. $\square$

References.

[1] G. Conner, K. Spencer, Anomalous behavior of the Hawaiian earring group, Journal of Group Theory 8 (2005) 223-227.