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

Recall that the Hawaiian earring group is an uncountable group isomorphic to the free -product of , denoted .

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

*Proof.* Since is uncountable (we proved this fact in this post), if it were free, then it would be isomorphic to a free group on an uncountable set . But the set of group homomorphisms to the additive group of integers is in bijective correspondence with the (obviously uncountable) set of functions . However, according to Theorem 7 (which we prove below), is isomorphic to the direct sum , which is countable. Therefore, we can conclude that is not a free group. .

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 be the subgroup of consisting of sequences in which the letters do not appear. Formally,

The elements of were also used in Lemma 3 in Part I.

**Lemma 5:** For each , the group is the free product of the subgroup and

* Proof.* Since the sets of letters and 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 can be written uniquely as a product where and (where at most one of or is allowed to be the identity).

Fix an element . By definition, for all . In particular, for , there is a such that for all . Let . Thus the appearances of the letters do not change in the words . Rather, we can only get from by adding letters (where ) in between the already present elements of . So by the uniqueness of reduced words in free groups, there are elements

and

such that . We can demand that at most one of or may be an identity element. Let be the projection when . Then if

we get the desired, unique representation .

**Lemma 6:** Every homomorphism such that for all is the trivial homomorphism.

* Proof.* Suppose for all and that for some . By Lemma 5, we can (for each ) write where and . Let . Basically, is the unique word obtained from by deleting all appearances of the letters . Notice that since , we have and thus for each .

The sequence meets the two conditions necessary to apply Lemma 3 from the previous post. This means there is a self-homomorphism such that . But now is a homomorphism such that for all and this contradicts Lemma 4 (again from the previous post) which says that all but finitely many must be killed by every homomorphism !

**Remark about the proof of Lemma 6:** One might be tempted to skip Lemma 5 altogether and try to construct using the self-homomorphism induced by collapsing the first circles of to the identity and leaving the other circles untouched. Indeed, it is true that , but a priori it is not clear that you get the next step unless you have the free product decomposition!

**Theorem 7:** The natural homomorphism given by is an injection onto the direct sum .

* Proof.* Lemma 6 says precisely that is injective so we are left to identify the image. By Lemma 4 from the previous post, for every homomorphism , there is an such that for all . Thus the image of the homomorphism lies in the direct sum . To see that the image is equal to the direct sum take any element . The natural map collapsing the -th circle, induces a surjection . Now define a homomorphism by . The composition of these to homomorphisms yields the homomorphism for which we have when and otherwise.

Here is a strengthened version of Lemma 6 that relates this stuff to a non-commutative version of* slender groups* and suggests that behaves in many ways like the non-commutative analogue of the Specker group .

**Theorem 8:** Suppose is any group which has as a quotient group. For every homomorphism , there is a such that .

Proof. Since we are assuming the existence of a surjective homomorphism , it suffices to prove the theorem for . Again, we’ll use the fact that there is an such that for all . But notice that the retract of consisting of all but the first circles is homeomorphic to the Hawaiian earring. Moreover, notice that the image of the inclusion is precisely . Since the restriction of to takes to for all . So by Lemma 6, .

I think it is interesting that Theorem 8 applies to all free groups but does not apply when 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.

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