The Hawaiian Earring Group is not Free (Part I)

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 Hom(\pi_1(\mathbb{H}),\mathbb{Z}) to the additive group of integers is countable.  This may seem a bit strange considering \pi_1(\mathbb{H}) is an uncountable group. As a direct consequence, we can see that \pi_1(\mathbb{H}) is not isomorphic to any free group F(X) on a set X.

Theorem 1 (de Smit [1]): \pi_1(\mathbb{H}) 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 \pi_1\left(\bigvee_{n=1}^{\infty}S^1\right) on a countable wedge of circles (the Hawaiian earring with the CW-topology) is isomorphic to the free group F(g_1,g_2,...) 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 \pi_1(\mathbb{H}) of the Hawaiian earring \mathbb{H}=\bigcup_{n=1}^{\infty}C_n\subset \mathbb{R}^2, where C_n is the circle of radius \frac{1}{n} centered at \left(\frac{1}{n},0\right). The basepoint is the origin x_0=(0,0), the one point where the shrinking circles meet.


We decided that the fundamental group \pi_1(\mathbb{H}) is an uncountable group that could be understood a subgroup of the inverse limit

\varprojlim\left(\cdots\to F_{n+1}\to F_n\to \cdots \to F_2 \to F_1\right)

where F_n is the free group on the generators g_1,...,g_n and the map p_{n+1}:F_{n+1}\to F_n identifies the letter g_{n+1} to the identity element (or empty word). The inverse limit \varprojlim_{n}F_n consists of sequences (w_n) where w_n \in F_n is obtained from w_{n+1} by removing all occurrences of the letter g_{n+1}. The fundamental group \pi_1(\mathbb{H}) corresponds to a certain subgroup of \varprojlim_{n}F_n.

Definition: Suppose w=g_{k_1}^{\epsilon_1}g_{k_2}^{\epsilon_2}...g_{k_m}^{\epsilon_m}\in F_n is a reduced word in the letters g_1,...,g_n. Reduced means that k_i\neq k_{i+1} and \epsilon_i\neq 0 for each i. The k-weight of w is


Essentially, \#_{k}(w) is the number of times g_{k} or g_{k}^{-1} appears in w.


If (w_n)\in\varprojlim_{n}F_n, then the sequence \#_{k}(w_n) (for fixed k) is non-decreasing since the projections p_n 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

\#_{\mathbb{N}}\mathbb{Z}=\left\{(w_n)\in\varprojlim_{n}F_n\Big|\lim_{n\to\infty}\#_{k}(w_n)<\infty\text{ for all }k\geq 1\right\}

be the subgroup of \varprojlim_{n}F_n consisting of sequences where every k-weight is eventually constant. These sequences are usually called locally eventually constant sequences and the group \#_{\mathbb{N}}\mathbb{Z} is often called the free \sigma-product of \mathbb{Z}.

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

Theorem 2: \pi_1(\mathbb{H})\cong\#_{\mathbb{N}}\mathbb{Z}.

So to study the properties of the Hawaiian earring group, we can focus our attention on the purely algebraic structure of \#_{\mathbb{N}}\mathbb{Z}. For instance, \#_{\mathbb{N}}\mathbb{Z} is certainly uncountable and also must be torsion free since it is a subgroup of the torsion free group \varprojlim_{n}F_n.

Some homomorphisms out of \#_{\mathbb{N}}\mathbb{Z}

First let’s exploit the inverse limit structure of \varprojlim_{n}F_n to construct some interesting self-homomorphisms of \#_{\mathbb{N}}\mathbb{Z}. For simplicity of notation, we’ll just write g_k for the element (1,1,...,1,g_k,g_k,...)\in\varprojlim_{n}F_n where the first non-trivial term is in the k-th position. More generally, we could also identify w\in F_n with it’s image under the canonical embedding F(g_1,g_2,...)\to\#_{\mathbb{N}}\mathbb{Z} of the infinite free group.

Consider a sequence s_j=(w_{n}^{(j)})\in\#_{\mathbb{N}}\mathbb{Z} of locally eventually constant sequences. The s_j should satisfy the follow two properties:

  • w_{n}^{(j)}=1 for all n<j
  • For all n\geq 1 and k<j, we have \#_{k}(w_{n}^{(j)})=0

For instance, we could have something like:

s_1=(g_1, g_1g_2, g_1g_2g_3,...)   which corresponds to an infinite word g_1g_2g_3g_4...

s_2=(1, g_2, g_2g_3, g_2g_3g_4, ...)   which corresponds to an infinite word g_2g_3g_4g_5...

s_3=(1, 1, g_3, g_3g_4, g_3g_4g_5, ...)   which corresponds to an infinite word g_3g_4g_5g_6...

and so on…

What is important is that  1) the first j-1 terms of s_j are trivial and that 2) the letters g_1,...,g_{j-1} don’t show up in s_j.

Condition 1) means precisely that the sequence s_j must converge to the identity element (1,1,1,...) when \varprojlim_{n}F_n has the inverse limit topology, i.e. as a subspace of \prod_{n}F_n.

Lemma 3: There is a self-homomorphism h:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z} where h(g_j)=s_j.

Proof. Define a homomorphism h_n:F_n\to F_n on the free group by h_n(g_j)=w_{n}^{(j)} (obviously for j\leq n). Since

p_n(h_n(g_j))=p_n(w_{n}^{(j)})=\begin{cases} w_{n-1}^{(j-1)} & \text{ if }j<n\\ 1 &\text{ if }j=n \end{cases}


h_{n-1}(p_n(g_j))=\begin{cases} h_{n-1}(g_j)=w_{n}^{(j-1)} & \text{ if }j<n\\ 1 &\text{ if }j=n \end{cases},

the following diagram commutes.



Consequently, we get a self-homomorphism h:\varprojlim_{n}F_n\to\varprojlim_{n}F_n on the inverse limit such that h(g_j)=s_j. We check that h(\#_{\mathbb{N}}\mathbb{Z})\subseteq\#_{\mathbb{N}}\mathbb{Z} and then use the restriction of h to prove the lemma. If v=(v_n)\in\#_{\mathbb{N}}\mathbb{Z} then for each k\geq 1, we have \lim_{n\to\infty}\#_{k}(v_n)=M_k<\infty. Now for fixed k\geq 1:


The inequality must be there since when we replace each letter g_j of v_n with h_n(g_j), we may have some word reduction to do. Our restriction that \#_{k}(w_{n}^{(j)})=0 for all n\geq 1 and k<j means that


Since, by assumption,  w_{n}^{(j)}\in\#_{\mathbb{N}}\mathbb{Z}, we have \lim_{n\to\infty}\#_{k}(w_{n}^{(j)})=N_k<\infty. Thus

\lim_{n\to\infty}\#_{k}(h(v)_n)\leq\sum_{j=1}^{k}M_k\cdot N_k<\infty

showing that h(v)\in\#_{\mathbb{N}}\mathbb{Z} is locally eventually constant. \square

Now let’s see what happens when we map \#_{\mathbb{N}}\mathbb{Z} to the additive group of integers \mathbb{Z}.

Lemma 4: If f:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z} is a homomorphism, then there is an N\geq 1 such that f(g_n)=0 for all n\geq N.

Proof. Suppose f(g_{n_j})\neq 0 for n_1<n_2<n_3<\cdots and let

s_j=\begin{cases} g_{n_j}^{3} & \text{ if }f(g_{n_j})>0\\ g_{n_j}^{-3} & \text{ if }f(g_{n_j})<0 \end{cases}

By Lemma 3, there is a homomorphism h:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z} such that h(g_j)=s_j for all j\geq 1. Thus f\circ h(g_j)=3|f(g_{n_j})|\geq 3 for each j\geq 1. We might as well now replace f with f\circ h so from now on let’s assume that

a_j=f(g_j)\geq 3.

Ok, now let’s define a special sequence z^{(j)}\in\#_{\mathbb{N}}\mathbb{Z} that will help us arrive at a contradiction. We define the n-th term z^{(j)}_{n}\in F_n 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 n>j is


Notice that removing g_k from z^{(j)}_{k} gives z^{(j)}_{k-1} and that \lim_{n\to\infty}\#_{j+k}(z^{(j)}_{n})=a_{j+k-1}<\infty for k\geq 1 (i.e. the letter g_{j+k} never appears more than a_{j+k-1} times despite the fact that the appearances get further and further apart). Thus z^{(j)}_{n} is a well defined element of \#_{\mathbb{N}}\mathbb{Z}.

The main feature of this sequence is that z^{(j)}=g_j(z^{(j+1)})^{a_j} so that when we apply f, we get


Iterating this formula j-1 times for f(z^{(1)}) gives


So if we let

b_j=a_1+a_1a_2+a_1a_2a_3+\cdots+a_1a_2...a_{j-1}    and    c_j=a_1a_2...a_j,

then we see that f(z^{(1)})=b_j+c_j+c_{j}f(z^{(1)}) and thus

f(z^{(1)})=b_j\text{ mod}(c_j).

Let’s make a few more observations about b_j and c_j:

  • b_j\to\infty

Proof: this one is pretty obvious since a_j\geq 3. \square

  • b_j<c_j

Proof: Since a_j\geq 3, we have a_1<a_1a_2 and inductively if b_{j-1}<c_{j-1}, b_{j}=b_{j-1}+c_{j-1}<c_{j-1}+c_{j-1}=2c_{j-1}<a_{j}c_{j-1}=c_{j}\square

  • c_j-b_j\to\infty

Proof: Here we also use the fact that a_j\geq 3 for each j. Notice

c_2-b_2=a_1a_2-a_1=a_1(a_2-1)\geq 2a_1,

c_3-b_3=a_1(a_2(a_3-1)-1)\geq a_1(2a_2-1)\geq 5a_1,

c_4-b_4=a_1(a_2(a_3(a_4-1)-1)-1)\geq 14a_1, and so on.

If we recursively define the increasing sequence p_2=2, p_{j+1}=3p_{j-1}-1, we get c_j-b_j\geq p_ja_1, which does the trick. \square

Added Remark: de Smit’s construction uses a_j\geq 3, however, the inductive proofs of these three bullet points also seem to work if you only assume a_j\geq 2. So actually, I believe this slight simplification can be made.

Now, let’s finally finish the proof of Lemma 4 by showing that f(z^{(1)}) satisfies way to many modular equations.

In general, suppose y=b\text{ mod}(c) where b<c. If y\geq 0, then we must have y\geq b and if y<0, then we must have y\leq b-c.

Consequently, if f(z^{(1)})\geq 0, then we must have f(z^{(1)})\geq b_j for all j\geq 1 since f(z^{(1)})=b_j\text{ mod}(c_j) and b_j<c_j. But this contradicts the fact that b_j\to \infty. On the other hand, if f(z^{(1)})<0, then we must have f(z^{(1)})\leq b_j-c_j for all j \geq 1 but this contradicts the fact that b_j-c_j\to-\infty. \square


[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.

This entry was posted in Free groups, Fundamental group, Group homomorphisms, Hawaiian earring and tagged , , , , , . Bookmark the permalink.

4 Responses to The Hawaiian Earring Group is not Free (Part I)

  1. Pingback: The Hawaiian Earring Group is not Free (Part II) | Wild Topology

  2. Pingback: The harmonic archipelago group is not free | Wild Topology

  3. Pingback: Homomorphisms from the Hawaiian earring group to finite groups | Wild Topology

  4. Pingback: The Baer-Specker Group | Wild Topology

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