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Let $A$ be a finely generated Abelian group. Prove that there is a topological space $X$ with a basic point $x_0$ such that $\pi_1(X,x_0)\cong A$.

I'm trying to solve this problem but I do not know where to start, here are several posts similar to this that I do not understand: Given a group $G$, the existence of a space such that $\pi_1(X)\simeq G$. Every Group is a Fundamental Group

Is there an easy way to prove my proposition? Thank you very much

user402543
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    It would be helpful to know what you've learned about fundamental groups. For each cyclic group, you might try to find a space whose fundamental group is the chosen cyclic group. Then, if $G$ is finitely generated and abelian, then it is a product of cyclic groups. So, the product of corresponding topological spaces will do the trick. – Robert Bell Mar 19 '18 at 21:09
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    I would add that unless you intend to use the classification of finitely generated abelian groups as Robert Bell indicated, the restriction to finitely generated abelian groups as opposed to all groups does not actually make the problem any easier. (In particular, the conditions "finitely generated" and "abelian" are not actually helpful for solving the problem, except in that when they are combined you can completely classify such groups.) – Eric Wofsey Mar 19 '18 at 21:20
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    If you drop the unnecessary assumptions that the group is finitely generated and abelian then your question is a duplicate of the ones that you cite. Rather than asking for an "easy way", you should study the answers to the other questions and then ask for help with any points that you don't understand. – Rob Arthan Mar 19 '18 at 21:21

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By the fundamental theorem of finitely generated abelian groups there exists an integer $n\geq 0 $ and integers $q_1,\ldots,q_k$ such that $$A\cong \mathbb{Z}^n\times \mathbb{Z}_{q_1}\times\cdots\times \mathbb{Z}_{q_k}.$$

Now recall that if $(X,x_0,Y,y_0)$ are based topological spaces then $$\pi_1(X\times Y,(x_0,y_0))\cong\pi_1(X,x_0)\times \pi_1(Y,y_0).$$

Now $\pi_1(\mathbb{S}^1,z_0)\cong \mathbb{Z}$ for any $z_0\in\mathbb{S}^1,$ so if $T^n$ is the $n$-torus $\mathbb{S}^1\times \cdots\times\mathbb{S}^1$ ($n$-times), then $$ \pi_1(T^n,t_0)\cong \mathbb{Z}^n$$ for any $t_0\in T^n.$

Now consider the space $X=\mathbb{D}^2/\sim$ where $$ (\cos\theta, \sin \theta)\sim (\cos(\theta+\frac{2\pi}{n}),\sin(\theta+\frac{2\pi}{n}))$$ for all $\theta\in [0,2\pi].$ Using Seifert-van Kampen one can show that $\pi_1(X,\overline{x_0})\cong\mathbb{Z}_n$ for any $\overline{x_0}\in X.$

With this in mind define, for $i=1,\ldots,k,$ the space $X_i:=\mathbb{D}^2/\sim$ where $$ (\cos\theta, \sin \theta)\sim (\cos(\theta+\frac{2\pi}{q_i}),\sin(\theta+\frac{2\pi}{q_i}))$$ for all $\theta\in [0,2\pi],$ so that $\pi_1(X_i,\overline{x_i})\cong\mathbb{Z}_{q_i}$ for each $i=1,\ldots,k$ and every $\overline{x_i}\in X_i.$

Therefore, defining $$Y:=T^n\times X_1\times X_2\times\cdots\times X_k,$$ we have $$ \pi_1(Y,(t_0,\overline{x_1},\ldots,\overline{x_k}))\cong \mathbb{Z}^n\times \mathbb{Z}_{q_1}\times\cdots\times \mathbb{Z}_{q_k}\cong A$$ for all $(t_0,\overline{x_1},\ldots,\overline{x_k})\in Y$

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