Free product decompositions of the fundamental group of Hawaiian Earrings This is a spin-off of my  question here, separated from the older question following Jeremy's suggestion. 
Definition. Call a group $G$ essentially freely indecomposable if in every free product decomposition $G\cong G_1\star G_2$, one of the free factors $G_1, G_2$ is free of finite rank. 
Question. Let ${\mathbb H}$ denote the Hawaiian Earrings. Is the fundamental group $\pi_1({\mathbb H})$ essentially freely indecomposable? 
The only relevant result I could find in the literature is the theorem  (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $G=\pi_1({\mathbb H})$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall. 
 A: This answer is courtesy of Sam Corson who kindly pointed out the following.
Theorem: The Hawaiian earring group $\pi_1(\mathbb{H})$ is essentially freely indecomposable, i.e. if $\pi_1(\mathbb{H})\cong G_1\ast G_2$, then one of $G_1$ or $G_2$ must be a finitely generated free group. 
The key is to apply a special case of Theorem 1.3 in 

K. Eda, Atomic property of the fundamental groups of the Hawaiian earring and wild locally path-connected spaces, J Math. Soc. Japan 63 (2011), 769-787.

Let $C_n$ be the $n$-th circle of $\mathbb{H}$. Take $\mathbb{H}_{\geq n}=\bigcup_{k\geq n}C_k$ to be the smaller copies of the Hawaiian earring and $\mathbb{H}_{\leq n}=\bigcup_{k=1}^{n}C_k$ to be the union of the first $n$-circles. One of the defining properties of the Hawaiian earring group is that there is a canonical isomorphism $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ for all $n\in\mathbb{N}$.
Suppose that $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism. In the case of the Hawaiian earring, Eda's theorem cited above implies that for any homomorphism $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$, there exists an $n\in\mathbb{N}$, $i\in\{1,2\}$, and $w\in G_1\ast G_2$ such that $\phi(\pi_1(\mathbb{H}_{\geq n+1}))\leq w G_i w^{-1}$. Suppose, without loss of generality, that $i=1$. Now if $\gamma(g)=w^{-1}gw$ is conjugation in $G_1\ast G_2$, then $\gamma\circ \phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism mapping $\pi_1(\mathbb{H}_{\geq n+1})$ into $G_1$. Let $\psi:G_1\ast G_2\to G_2$ be the projection, which is surjective. Then $\psi\circ\gamma\circ\phi:\pi_1(\mathbb{H})\to G_2$ is a surjection and since $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ and $\psi\circ\gamma\circ\phi(\pi_1(\mathbb{H}_{\geq n+1}))=1$, we must have that $\psi\circ\gamma\circ\phi$ maps $\pi_1(\mathbb{H}_{\leq n})$ onto $G_2$. Therefore, $G_2$ is finitely generated. Moreover, since $G_2$ is isomorphic to a subgroup of the locally free group $\pi_1(\mathbb{H})$, it follows that $G_2$ is free.
