The main goal of this two-part post will be to study the homomorphisms out of the Hawaiian earring group. Click here to get to Part II. In particular, we’ll end up concluding that the set of homomorphisms to the additive group of integers is countable. This may seem a bit strange considering is an uncountable group. As a direct consequence, we can see that is not isomorphic to any free group on a set .

**Theorem 1 (de Smit [1]):** is not isomorphic to a free group.

The Hawaiian earring group is complicated enough that this should not be completely obvious. In this post I’ll fully hash out the details of Bart de Smits proof in [1]. Working through all these details has certainly helped me understand it better.

Theorem 1 is in contrast with the fact that the fundamental group on a countable wedge of circles (the Hawaiian earring with the CW-topology) is isomorphic to the free group on a countably infinite set of generators.

**The Hawaiian earring group**

In a previous post, I discussed how to begin understanding the algebraic structure of the fundamental group of the Hawaiian earring , where is the circle of radius centered at . The basepoint is the origin , the one point where the shrinking circles meet.

We decided that the fundamental group is an uncountable group that could be understood a subgroup of the inverse limit

where is the free group on the generators and the map identifies the letter to the identity element (or empty word). The inverse limit consists of sequences where is obtained from by removing all occurrences of the letter . The fundamental group corresponds to a certain subgroup of .

**Definition:** Suppose is a reduced word in the letters . *Reduced* means that and for each . The *k-weight* of is

.

Essentially, is the number of times or appears in .

If , then the sequence (for fixed ) is non-decreasing since the projections only delete letters. We wish to consider the elements of the inverse limit where each such sequence is also bounded (and thus eventually constant). Let

be the subgroup of consisting of sequences where every k-weight is eventually constant. These sequences are usually called *locally eventually constant* sequences and the group is often called the free -product of .

In a previous post, we found the following canonical group isomorphism.

**Theorem 2:** .

So to study the properties of the Hawaiian earring group, we can focus our attention on the purely algebraic structure of . For instance, is certainly uncountable and also must be torsion free since it is a subgroup of the torsion free group .

**Some homomorphisms out of **

First let’s exploit the inverse limit structure of to construct some interesting self-homomorphisms of . For simplicity of notation, we’ll just write for the element where the first non-trivial term is in the k-th position. More generally, we could also identify with it’s image under the canonical embedding of the infinite free group.

Consider a sequence of locally eventually constant sequences. The should satisfy the follow two properties:

- for all
- For all and , we have

For instance, we could have something like:

*which corresponds to an infinite word*

*which corresponds to an infinite word*

*which corresponds to an infinite word*

and so on…

*What is important* is that 1) the first terms of are trivial and that 2) the letters don’t show up in .

Condition 1) means precisely that the sequence must converge to the identity element when has the inverse limit topology, i.e. as a subspace of .

**Lemma 3:** There is a self-homomorphism where .

Proof. Define a homomorphism on the free group by (obviously for ). Since

and

the following diagram commutes.

Consequently, we get a self-homomorphism on the inverse limit such that . We check that and then use the restriction of to prove the lemma. If then for each , we have . Now for fixed :

The inequality must be there since when we replace each letter of with , we may have some word reduction to do. Our restriction that for all and means that

Since, by assumption, , we have . Thus

showing that is locally eventually constant.

Now let’s see what happens when we map to the additive group of integers .

**Lemma 4:** If is a homomorphism, then there is an such that for all .

*Proof.* Suppose for and let

By Lemma 3, there is a homomorphism such that for all . Thus for each . We might as well now replace with so from now on let’s assume that

.

Ok, now let’s define a special sequence that will help us arrive at a contradiction. We define the n-th term to be

\begin{cases} 1 & \text{ if }n<j\\ g_j & \text{ if }n=j \\ g_jg_{j+1}^{a_j} & \text{ if }n=j+1 \\ g_j(g_{j+1}g_{j+2}^{a_{j+1}})^{a_j} & \text{ if }n=j+2 \\ g_j(g_{j+1}(g_{j+2}g_{j+3}^{a_{j+2}})^{a_{j+1}})^{a_j} & \text{ if }n=j+3 \\ … & … \end{cases}

So the general form for is

.

Notice that removing from gives and that for (i.e. the letter never appears more than times despite the fact that the appearances get further and further apart). Thus is a well defined element of .

The main feature of this sequence is that so that when we apply , we get

Iterating this formula times for gives

So if we let

and ,

then we see that and thus

Let’s make a few more observations about and :

*Proof:* this one is pretty obvious since .

*Proof:* Since , we have and inductively if , .

*Proof:* Here we also use the fact that for each . Notice

,

,

, and so on.

If we recursively define the increasing sequence , , we get , which does the trick.

**Added Remark:** de Smit’s construction uses , however, the inductive proofs of these three bullet points also seem to work if you only assume . So actually, I believe this slight simplification can be made.

Now, let’s finally finish the proof of Lemma 4 by showing that satisfies way to many modular equations.

In general, suppose where . If , then we must have and if , then we must have .

Consequently, if , then we must have for all since and . But this contradicts the fact that . On the other hand, if , then we must have for all but this contradicts the fact that .

**References.**

[1] B. de Smith, *The fundamental group of the Hawaiian earring is not free*, International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33-37.

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