# Prove that $\Bbb Z\oplus \Bbb Z_2$ is not isomorphic to $\Bbb Z∗\Bbb Z_2$.

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.

• Do you mean isomorphic as groups? Do you mean to ask whether there's a non-trivial group homomorphism $f:\Bbb Z \oplus \Bbb Z/2 \Bbb Z \to \Bbb Z \ast \Bbb Z /2 \Bbb Z$? – Robert Shore Apr 9 at 20:19
• Well, I'm not really sure, but my teacher put it this way: Prove that the direct sum Z+Z2 is not homeomorphic to the free product Z∗Z2. – Student Apr 9 at 20:24
• In my experience, "homeomorphic" is a term usually applied to topological spaces. That's why I'm confused. Are we talking about anything other than group structure? – Robert Shore Apr 9 at 20:26
• If you want to clarify the question, I suggest editing the question itself. People often don't read comments. – Robert Shore Apr 9 at 20:28
• Okay. You need to prove that the spaces (as topological spaces) are not homeomorphic by showing that their fundamental groups are not isomorphic (as groups). As you've noted, one group is abelian and the other is not, so they can't be isomorphic. Take any two elements of the non-abelian that don't commute, and there's no faithful way to map them to the abelian group. – Robert Shore Apr 9 at 20:46

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.
• Your notation will probably confuse you. Use numbers for $\Bbb Z$ but use $\{e, a \}$ for $\Bbb Z / 2 \Bbb Z$. Then for any $n \neq 0, na \neq an$ because you're taking the free product. – Robert Shore Apr 14 at 8:56