Show that for an abelian countable group $G$ there exists a compact path connected subspace $K ⊆ \Bbb R^4$ such that $H_1(K)$ isomorphic to $G$

Question

" Given an abelian countable group G thus there exists a compact path connected subspace $K ⊆\Bbb R^4$ such that $H_1(K) ≅ G$ ", where $H_1$ is the first singular homology.

Can I prove it using the Theorem:" For every countable group G there exists a compact path connected subspace $K ⊆ \Bbb R^4$ such that $π_1(K) ≅ G$."?

If i removed abelian, is it still right to be proved or not?

Answer 1

Yes, that's enough. There is a theorem which says (among other things) that $H_1(X)$ is isomorphic to the abelienization of $\pi_1(X)$ for any path connected space $X$.

If $G$ is already abelian, then we have $H_1(X)\cong \pi_1(X)$.

Answer 2

To answer your last question:

No, if you remove abelian it will not work, since all singular homology groups are abelian, and the property of being abelian is preserved through isomorphisms.