The earring group is not free (Part II)

This post is Part II in an explanation of why the fundamental group of the earring space is not a free group. I’ll be referencing the notation and results that we worked through in Part I.

Recall that the earring group $\pi_1(\mathbb{E})$ is an uncountable group isomorphic to the free $\sigma$ -product of $\mathbb{Z}$ , denoted $\#_{\mathbb{N}}\mathbb{Z}$ .

Theorem 1 (de Smit [1]): $\pi_1(\mathbb{E})=\#_{\mathbb{N}}\mathbb{Z}$ is not isomorphic to a free group.

Proof. Since $\#_{\mathbb{N}}\mathbb{Z}$ is uncountable (we proved this fact in this post), if it were free, then it would be isomorphic to a free group $F(X)$ on an uncountable set $X$ . But the set of group homomorphisms $Hom(F(X),\mathbb{Z})$ to the additive group of integers is in bijective correspondence with the (obviously uncountable) set $\mathbb{Z}^{X}$ of functions $X\to\mathbb{Z}$ . However, according to Theorem 7 (which we prove below), $Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})$ is isomorphic to the direct sum $\bigoplus_{j=1}^{\infty}\mathbb{Z}$ , which is countable. Therefore, we can conclude that $\#_{\mathbb{N}}\mathbb{Z}$ is not a free group. $\square$ .

To fill in the remaining gaps (namely a proof of Theorem 7), we’ll need a couple of lemmas to break it up into more manageable bits.

Let $G_{\geq j}$ be the subgroup of $\#_{\mathbb{N}}\mathbb{Z}$ consisting of sequences $(w_n)$ in which the letters $g_1,...,g_{j-1}$ do not appear. Formally,

$G_{\geq j}=\{(w_n)|\#_{k}(w_n)=0\text{ for all }n\geq 1\text{ and }k<j\}$

The elements of $G_{\geq j}$ were also used in Lemma 3 in Part I.

Lemma 5: For each $j>1$ , the group $\#_{\mathbb{N}}\mathbb{Z}$ is the free product of the subgroup $F_{j-1}$ and $G_{\geq j}$

Proof. Since the sets of letters $\{g_1,...,g_{j-1}\}$ and $\{g_j,g_{j+1},...\}$ used in the two groups are mutually exclusive we can use the uniqueness of reduced words in free groups to prove this lemma. In particular, we need to show that any element of $\#_{\mathbb{N}}\mathbb{Z}$ can be written uniquely as a product $a_1b_1a_2...a_pb_p$ where $a_i\in F_{j-1}$ and $b_i\in G_{\geq j}$ (where at most one of $a_1$ or $b_p$ is allowed to be the identity).

Fix an element $(w_n)\in\#_{\mathbb{N}}\mathbb{Z}$ . By definition, $\lim_{n\to\infty}\#_{k}(w_n)=M_k<\infty$ for all $k\geq 1$ . In particular, for $k=1,...,j-1$ , there is a $n_{k}$ such that $\#_{k}(w_{n})=M_k$ for all $n\geq n_j$ . Let $N=\max\{n_1,...,n_{j-1}\}$ . Thus the appearances of the letters $g_1,...,g_{j-1}$ do not change in the words $w_{N},w_{N+1},...$ . Rather, we can only get $w_{N+q+1}$ from $w_{N+q}$ by adding letters $g_{k}$ (where $k>j$ ) in between the already present elements of $F_{j-1}$ . So by the uniqueness of reduced words in free groups, there are elements

$a_{1},...,a_{p}\in F_{j-1}$ and $b_{1}^{(N+q)},...,b_{p}^{(N+q)}\in F(g_{j},...,g_{N+q})$

such that $w_{N+q}=a_1b_{1}^{(N+q)}...a_{p}b_{p}^{(N+q)}$ . We can demand that at most one of $a_1$ or $b_{p}^{(N+q)}$ may be an identity element. Let $r_n:F_{N}\to F_n$ be the projection when $n<N$ . Then if

$b_i=(r_1(b_{i}^{(N)}),r_2(b_{i}^{(N)}),...,r_{N-1}(b_{i}^{(N)}),b_{i}^{(N)},b_{i}^{(N+1)},...)\in\#_{\mathbb{N}}\mathbb{Z}$

we get the desired, unique representation $(w_n)=a_1b_1...a_pb_p$ . $\square$

Lemma 6: Every homomorphism $f:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ such that $f(g_j)=0$ for all $j\geq 1$ is the trivial homomorphism.

Proof. Suppose $f(g_j)=0$ for all $j\geq 1$ and that $f(x)\neq 0$ for some $x\in\#_{\mathbb{N}}\mathbb{Z}$ . By Lemma 5, we can (for each $j>1$ ) write $x=a_1b_1\cdots a_pb_p$ where $a_i\in F_{j-1}$ and $b_i\in G_{\geq j}$ . Let $s_j=b_1b_2\cdots b_p$ . Basically, $s_j$ is the unique word obtained from $x$ by deleting all appearances of the letters $g_1,...,g_{j-1}$ . Notice that since $f(g_j)=0$ , we have $f(a_i)=0$ and thus $f(s_j)=f(x)$ for each $j\geq 1$ .

The sequence $s_j$ meets the two conditions necessary to apply Lemma 3 from the previous post. This means there is a self-homomorphism $h:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z}$ such that $h(g_j)=s_j$ . But now $f\circ h:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ is a homomorphism such that $f\circ h(g_j)=f(s_j)=f(x)\neq 0$ for all $j> 1$ and this contradicts Lemma 4 (again from the previous post) which says that all but finitely many $g_j$ must be killed by every homomorphism $\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ ! $\square$

Remark about the proof of Lemma 6: One might be tempted to skip Lemma 5 altogether and try to construct $s_j$ using the self-homomorphism $q_j:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z}$ induced by collapsing the first $j-1$ circles of $\mathbb{E}$ to the identity and leaving the other circles untouched. Indeed, it is true that $s_j=q_j(x)$ , but a priori it is not clear that you get the next step $f(x)=f(s_j)$ unless you have the free product decomposition!

Theorem 7: The natural homomorphism $Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z}$ given by $f\mapsto (f(g_1),f(g_2),f(g_3),...)$ is an injection onto the direct sum $\bigoplus_{j=1}^{\infty}\mathbb{Z}$ .

Proof. Lemma 6 says precisely that $Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z}$ is injective so we are left to identify the image. By Lemma 4 from the previous post, for every homomorphism $f:\#_{\mathbb{N}}\mathbb{Z}\to \mathbb{Z}$ , there is an $N\geq 1$ such that $f(g_n)=0$ for all $n\geq N$ . Thus the image of the homomorphism $Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z}$ lies in the direct sum $\bigoplus_{j=1}^{\infty}\mathbb{Z}$ . To see that the image is equal to the direct sum take any element $(z_1,z_2,...,z_n,0,0,...)\in\bigoplus_{j=1}^{\infty}\mathbb{Z}$ . The natural map $\mathbb{E}\to \bigcup_{j=1}^{n}C_j$ collapsing the $j$ -th circle, $j>n$ induces a surjection $\#_{\mathbb{N}}\mathbb{Z}\to F_n$ . Now define a homomorphism $F_n\to\mathbb{Z}$ by $g_j\mapsto z_j$ . The composition of these to homomorphisms yields the homomorphism $f:\#_{\mathbb{N}}\mathbb{Z}\to \mathbb{Z}$ for which we have $f(g_j)=z_j$ when $j\leq n$ and $f(g_j)=0$ otherwise. $\square$

Here is a strengthened version of Lemma 6 that relates this stuff to a non-commutative version of slender groups and suggests that $\#_{\mathbb{N}}\mathbb{Z}$ behaves in many ways like the non-commutative analogue of the Specker group $\prod_{n=1}^{\infty}\mathbb{Z}$ .

Theorem 8: Suppose $S$ is any group which has $\mathbb{Z}$ as a quotient group. For every homomorphism $f:\#_{\mathbb{N}}\mathbb{Z}\to S$ , there is a $n\geq 1$ such that $f(G_{\geq n})=1$ .

Proof. Since we are assuming the existence of a surjective homomorphism $S\to\mathbb{Z}$ , it suffices to prove the theorem for $S=\mathbb{Z}$ . Again, we’ll use the fact that there is an $N\geq 1$ such that $f(g_n)=0$ for all $n\geq N$ . But notice that the retract $\mathbb{E}_{\geq N}=\bigcup_{n\geq N}C_n$ of $\mathbb{E}$ consisting of all but the first $N-1$ circles is homeomorphic to the earring space. Moreover, notice that the image of the inclusion $\mathbb{E}_{\geq N}\to\mathbb{E}$ is precisely $G_{\geq N}$ . Since the restriction of $f$ to $G_{\geq N}=\pi_1(\mathbb{E}_{\geq N})\cong\pi_1(\mathbb{E})$ takes $g_k$ to $0$ for all $k\geq N$ . So by Lemma 6, $f(G_{\geq N})=f(\pi_1(\mathbb{E}_{\geq N}))=0$ . $\square$

I think it is interesting that Theorem 8 applies to all free groups $S=F(X)$ but does not apply when $S$ is any finite group.

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.

5 Responses to The earring group is not free (Part II)

Pingback: The harmonic archipelago group is not free | Wild Topology
Pingback: The Griffiths Twin Cone | Wild Topology
Pingback: The Baer-Specker Group | Wild Topology
Pingback: The Hawaiian Earring Group is not Free (Part I) | Wild Topology
Pingback: The earring group is not free (Part I) | Wild Topology