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the problem statement is as follows: Given G an abelian group, I have to show that there exist a simplicial complex, K (finite), such that the $H_{0}(K,\mathbb{Z})=\mathbb{Z}$ and $H_{1}(K,\mathbb{Z})=G$.

The first thing that came to my mind was use the theorem of classification for compact surfaces that let me the sphere, the connected sum of the tori, or the connected sum of the real projective plane the only possibilities. All of them connected, so its $0$-homology is $\mathbb{Z}$ and the $1$-homology of the sphere, the sum of the tori and the sum of the projective planes $0,\mathbb{Z}^{m},\mathbb{Z}^{n}\oplus\mathbb{Z}_{2}$, respectively.

and use the fundamental theorem of finitely generated abelian groups, then $$G=\mathbb{Z}^{m}\oplus(\mathbb{Z}_{a_{i}}\oplus\cdots\oplus\mathbb{Z}_{a_{k}})$$ the problem is the torsion. If G is free, then G is torsion-free and there is no problem but I don't know if this can be proved.

another way I thought was consider the boundery of a 2-simplex this is connected so the condition for the $0$-homology is satisfied. in the other hand there is no 2-simplex so $B_{1}=0$, and $H_{1}(\partial\Delta^{2})=Ker(\partial_{1})$ and considert an arbitrary $1$-chain $$c=a_{1}\sigma+a_{2}\delta+a_{3}\kappa$$ where $a_{i}\in G$ that will give me the $1$-homology required, but then I thought that there was something wrong with this.

Do you have any suggestion that can help me?

Monkey
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  • Search for "Moore space". If you even think in $\pi_1$, Eilenberg-Maclane space. – A.G Sep 24 '16 at 07:46
  • @A.G OP didn't ask for CW_complex. Otherwise the answer would be very easy. – Anubhav Mukherjee Sep 24 '16 at 11:58
  • Wedges on a vertex should get you the sum. The obvious triangle gives you the $\mathbb Z$s. Thus, you just need to find a simplicial complex with first homology equal to $\mathbb Z /n$ for any $n$, though $n= p^k$ a prime power would suffice. Maybe try $n=2, 3$ to get the idea. Think about the CW-complex version and subdivide. – Justin Young Sep 24 '16 at 12:04
  • even what you can prove is that given any finitely presented group, you can have a Simplicial complex (finite) , whose fundamental group is the given group. – Anubhav Mukherjee Sep 24 '16 at 12:08
  • What about this @Anubhav? Let us suppose I have a simplicial complex K (finite) of dimension greater than $1$. I know that $C_{1}(K)$ is a free abelian group finitely generated and $Z_{1}(K)\subset C_{1}(K)$ which is abelian, so $Z_{1}(K)$ is also finitely generated and its quotient ($H_{1}$) is also finitely generated, if dim(K)=1, better! because the classification is easier, else I only have to consider orientable simplicial complex. – Monkey Sep 25 '16 at 07:29

1 Answers1

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Take the direct product of circles and lens spaces $L(n,1)$ for suitable $n$'s. The latter satisfy $H_1(L(n,1), Z)\cong Z_n$.

Moishe Kohan
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