I need to prove that $\Bbb Z\oplus \Bbb Z_2$ is not homeomorphic to the free product $\Bbb Z∗\Bbb Z_2$.
I know that $\Bbb Z\oplus \Bbb Z_2$ is abelian while the free product $\Bbb Z∗\Bbb Z_2$ is not so they can't be isomorphic but I don't know how to prove in detail.
(1) the fundamental group of a manifold is $\pi_1(M)=\langle a,b|bab^{-1} a^{-1},a a \rangle \cong \Bbb Z\oplus \Bbb Z_2$.
(2) the fundamental group of another manifold is $\pi_1(N)=\langle a,b|b^{2}=1 \rangle\cong \Bbb Z*\Bbb Z_2$.
Now I have to prove that (1) is not isomorphic to (2) ($\Bbb Z\oplus \Bbb Z_2$ is not isomorphic to $\Bbb Z*\Bbb Z_2$).
I'd appreciate any help.
If there's anything wrong with my question please let me know so that I can correct it.
Thanks in advance!
Best Answer
If two spaces are homeomorphic, then their fundamental groups are isomorphic. Conversely, if two spaces have non-isomorphic fundamental groups, then they cannot be homeomorphic to each other. So to prove that two spaces are not homeomorphic, it's enough to show that their fundamental groups are not isomorphic to each other.
Assume $G$ is a non-abelian group and $H$ is an abelian group. Assume $f:G \to H$ were an isomorphism. Choose $x, y \in G$ such that $xy \neq yx$. Then because $f$ is a homomorphism, $f(xy)=f(x)f(y)$. Since $H$ is abelian, $f(x)f(y)=f(y)f(x)=f(yx)$. And because $f$ is an isomorphism, $f(xy)=f(yx) \Rightarrow xy=yx$, contradicting our assumption that $xy \neq yx$. Thus, no such isomorphism $f$ can exist.