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I have asked this question in MSE but have not got any satisfactory answer, so I am asking it here. Any idea on how to approach this problem will be highly appreciated.

Consider the Hawaiian earring. Suppose $f_n$'s are the loops representing the circles of radius $1/n$ centered at $(1/n, 0)$ for $n=1,2\cdots$. Suppose the following holds in the fundamental group

$$\langle [f_1, f_2][f_3, f_4]\cdots\rangle = \langle [g_1, g_2]\cdots [g_{2k-1},g_{2k}]\rangle $$ for some loops $g_1,g_2,\ldots , g_{2k}$, where $\langle\cdot\rangle$ denotes the homotopy class and $[a,b]=aba^{-1}b^{-1}$ denotes the commutator. Then can we say that $$\langle [f_1, f_2]\cdots [f_{2n-1}, f_{2n}]\rangle = \langle [g_1, g_2]\cdots [g_{2k-1},g_{2k}]\rangle$$ for all sufficiently large $n$, or for some $n>k$ ?

I had this question while I was reading the paper here (see page 76, last paragraph), where the above has been mentioned (in some other equivalent form) without proof.

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This question seems to assume that the infinite product loop $[f_1,f_2][f_3,f_4][f_5,f_6]\dots$ is null-homologous in the Hawaiian earring, which is false. In fact, it is false by Katusya Eda's $0$-form Lemma, which says that a loop $\alpha$ in the Hawaiian earring is null-homologous if and only if the reduced representative of $\alpha$ factors into a finite concatenation $\prod_{i=1}^{2n}\alpha_i$ where there is an inverse pairing among the factors. Formally, this means $\{1,2,\dots ,2n\}$ splits into the disjoint union of two $n$-element sets $A,B$ and there is a bijection $\phi:A\to B$ such that for each $i\in A$, $\alpha_i$ is a reparameterization of the reverse of $\alpha_{\phi(i)}$.

The $0$-form Lemma can be found as Lemma 3.6 in Singular homology groups of one-dimensional Peano continua. Another nice generalization for the Hawaiian earring is Lemma 4.3 in Cotorsion-free groups from a topological point of view by Eda and Fischer. The double induction of this proof is rather subtle and is a thing of beauty.

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I think your question is interesting and deserves to be solved. But, I cannot find the corresponding part in Higman's paper. For $c$ in $G'$, let $r(c)$ be the minimal number $n$ such that $c = [u_1,v_1]\dots[u_n,v_n]$. Let $e_i$ be the generator of the $i$-th factor of $F$. Then, $r([e_1,e_2]...[e_{2n-1},e_{2n}]) = n$, which is known but is not a straightforward fact. Therefore, $[e_1,e_2]\dots[e_{2n-1},e_{2n}]\dots$ does not belong to the commutator subgroup.

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    $\begingroup$ Welcome to MathOverflow! $\endgroup$ Commented Jan 26, 2020 at 22:34
  • $\begingroup$ Indeed by this argument it's not even in the commutator subgroup of the larger group, projective limit of free groups. Hence, the argument uses 2 ingredients (a) the canonical map from $\pi_1$*(Hawaiian earring)* to $\mathrm{limproj}F_n$ is injective (b) in $F_{2n}$ the product $\prod_{1}^n[x_{2i-1},x_{2i}]$ has commutator length $c_n$ tending to infinity. $\endgroup$
    – YCor
    Commented Jan 27, 2020 at 3:57
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I believe this sort of question is addressed in the beautiful series of papers by Cannon and Conner (e.g., the first one:

MR1775709 (2001g:20020) Reviewed Cannon, J. W.(1-BYU); Conner, G. R.(1-BYU) The combinatorial structure of the Hawaiian earring group. (English summary) Topology Appl. 106 (2000), no. 3, 225–271.

I am pretty sure your question or some variant is covered in Section 4 of this.

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