One of the the surprising things about the Hawaiian earring group (the fundamental group of the Hawaiian earring space ) is that the group of homomorphisms to the additive group of integers is countable (see this post for details) even though is torsion-free and uncountable. In this post we’ll see that the situation is completely different when we replace with a finite group!
The Hawaiian earring is the space where is the circle of radius centered at . The basepoint is the origin , which is the only point in the intersection of all of the shrinking circles. Let be the loop traversing once in the counterclockwise direction and be its homotopy class.
In this post, we’ll work through the proof of the following result.
Theorem 1: Let be any non-trivial finite group. There are uncountably many surjective group homomorphisms mapping to the identity element for every .
Theorem 1 could be considered “surprising” since every homomorphism from to a free group which kills all classes must be the trivial homomorphism. An obvious consequence is the following corollary.
Corollary 2: For any non-trivial finite group , the set* of group homomorphism is uncountable.
*since the finite group need not be abelian, the hom set need not have natural group structure.
An anomalous aside
The only place I know of Theorem 1 being proven in the literature is Conner and Spencer’s paper  entitled “The anomalous behavior of the Hawaiian earring group.” This is a nice, short paper which clarified some things about the Hawaiian earring group. Conner and Spencer state in the introduction of  that “The aim of the current article is to understand some of the anomalous behavior of homomorphisms from the fundamental group of the Hawaiian earring into finite groups.” According to Theorem 1, the hom-set 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 Hawaiian 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 , we’ll see that is uncountable essentially because is uncountable. Similarly, Theorem 1 follows directly from the existence of uncountably many epimorphisms 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 .
The infinite product of n-dimensional projective space () exhibits the same sort of “anomalous” behavior since . So yes, the Hawaiian earring is quite anomalous, however, this particular anomaly is not unique to .
Ultrafilters and finite groups
Definition 3: A filter on a set is a non-empty set of subsets of satisfying the properties:
- If , then ,
- If , and , then .
A filter is an ultrafilter if for every subset , either or . An ultrafilter is a non-principal ultrafilter if does not contain any finite subsets of .
We’re mainly going to be using non-principal ultrafilters on the natural numbers . The existence of a non-principal ultrafilters on infinite sets is equivalent to the axiom of choice. In fact the remainder of the Stone-Cech compactification is in bijective correspondence with the non-principal ultrafilters on . In particular, there are uncountably many non-principal ultrafilters on .
Lemma 4: If is a non-principal ultrafilter on and is a finite partition of , then there is a unique such that .
Proof. Suppose for all . Then for each and thus which is false. Thus for some . Suppose for . Then which cannot be. Thus there is a unique such that .
Let be a finite group with trivial element and be the infinite direct product whose elements are sequences where .
Lemma 5: Let be the subgroup of consisting of sequences such that for all but finitely many . There exists uncountably many epimorphisms such that .
Proof. Let be a non-principal ultrafilter on . We define a homomorphism . To define , fix a sequence . For each , let . Since is finite is a finite partition of . According to Lemma 4, there is a unique such that . Thus we have a well-defined function when we set where .
claim 1: is a homomorphism. Suppose is another sequence and where . Consequently,
and therefore . It follows from our definition of that .
claim 2: is surjective. Let and pick any set . If is the sequence where then clearly . Thus is surjective.
claim 3: . Let be a sequence such that for all but finitely many . Since is finite when , we must have . By definition of , .
Claims 1-3 guarantee the existence of a desired epimorphism for each non-principal ultrafilter . We check uniqueness last.
claim 4: if . If is a non-principal ultrafilter, distinct from , there is an infinite set such that and . Pick and define a sequence by . Then and . Thus .
Since there are uncountably many non-principal ultrafilters on , we have proved the lemma.
Remark: In Lemma 5 we basically defined a canonical injection
Corollary 6: For any non-trivial finite group , the hom-sets and are uncountable.
Completing the proof of Theorem 1
Proof of Theorem 1. Suppose has generating set and let be the free group on these generators. Let be the canonical epimorphism induced by the inclusion .
Let be the first wedge of m-circles, be the next m-circles, and so on. Thus . Since each of these spaces is a wedge of m-circles, we have for each . The retraction , which collapses all other circles to the basepoint, induces a surjection on fundamental groups. Together, these surjections induce a canonical surjection , . The argument that is a surjection is essentially the same as the one we used to show is uncountable. We’re going to work with the surjective composition:
Pre-composition with defines a function . In detail, .
Notice that is injective since it is precisely the function induced by the hom-functor from the category of groups to the category of sets. Since is an epimorphism in , is a monomorphism in …or you could just check directly that is injective using the fact that and are surjective. Since we’ve embedded an uncountable set into , 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 :
Fact I: If is surjective, then so is .
Fact II: For fixed , the image can only be non-trivial if for a unique .
Recall from Lemma 5 that there are uncountably many epimorphisms such that . Fact II implies that, for each , we have since this sequence is non-trivial in at most one component. Thus for each of the uncountably many surjections constructed in Lemma 5, we have a unique (since is injective) surjection (by Fact I) such that
for each . This completes the proof of Theorem 1.
 G. Conner, K. Spencer, Anomalous behavior of the Hawaiian earring group, Journal of Group Theory 8 (2005) 223-227.