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Let $X$ be the Hawaiian earring and let $x_0$ be the point $(0,0)$. Consider the wedge sum $Y:=X\vee X=X\coprod X/x_0\sim x_0$. Then using $\Bbb N=\{1,3,5,\dots\}\cup \{2,4,6,\dots\}$, $X$ is homeomorphic to $Y$. My question is: Do we have $G*G\cong G$ (isomorphic as groups), where $G=\pi_1(X,x_0)$?

If this is not true, but this will be a counterexample of the statement that $\pi_1(X_1\vee X_2)\cong \pi_1(X_1)*\pi_1(X_2)$ for path-connected topological spaces $X_1,X_2$.

(This is true if the wedge point has a neighborhood that deformation retracts onto itself, by the van Kampen theorem).

The group $G$ is quite complicated, and I have no idea whether $G*G$ is isomorphic to $G$ or not.

Edit. It is known that the $H_1(X)$, which is the abelianization of $G$, is $(\prod_{i=1}^\infty \Bbb Z) \oplus (\prod_{i=1}^\infty \Bbb Z/\bigoplus_{i=1}^\infty \Bbb Z)$, so one way to prove that $G*G\not\cong G$ is showing $$ \left( \prod_{i=1}^\infty \Bbb Z \right) \oplus \left( \prod_{i=1}^\infty \Bbb Z/\bigoplus_{i=1}^\infty \Bbb Z \right) \not\cong \left( \prod_{i=1}^\infty \Bbb Z \right) \oplus \left( \prod_{i=1}^\infty \Bbb Z/\bigoplus_{i=1}^\infty \Bbb Z \right) \oplus \left( \prod_{i=1}^\infty \Bbb Z \right) \oplus \left( \prod_{i=1}^\infty \Bbb Z/\bigoplus_{i=1}^\infty \Bbb Z \right).$$

azif00
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user302934
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  • Are you looking for a counterexample to $\pi_1(X_1\vee X_2)\cong \pi_1(X_1)\pi_1(X_2)$ or are interested especially in $G \cong GG$? – Paul Frost Nov 07 '22 at 10:22
  • @PaulFrost Actually the latter one. I've seen a counterexample of the former one: https://math.stackexchange.com/questions/2788471/fundamental-group-of-wedge-sum-of-two-coni-over-hawaiian-earrings – user302934 Nov 07 '22 at 11:13
  • The abelianizations should indeed not be isomorphic as can be observed by taking their duals. – Thorgott Nov 07 '22 at 20:29
  • Is the natural topology of the fundamental group of the earring visible purely from the group structure, i.e. can we define it just using group theoretic language? If so, I imagine there is a topological proof this isomorphism can't hold. – Connor Malin Nov 08 '22 at 01:20

1 Answers1

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There is a negative answer by applying a result stated on math overflow: For every free product decomposition $G = G_1 * G_2$, one of the two factors is free of finite rank. Since $G$ is not free of finite rank, there cannot be an isomorphism $G = G * G$.

Lee Mosher
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