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I am studying elementary Algebraic Topology recently. I have seen that a topological space is identified with a group. We are telling the group as Fundamental Group. So every topological space $X$ and for any $x \epsilon X$ there is a fundamental group $\pi_1 (X,x)$. I have a question about the converse. If we consider any arbitrary group $G$, then does there exist a topological space $X$ and an element $x\epsilon X$ such that $\pi_1 (X,x)=G$?

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That statement is true. First, every free group $F$ is a the fundamental group of a bouquet of circles (having a common base point); each circle represents a generator. Now every group $G$ is a quotient of a free group $F$ by a subgroup (of $F$) generated by the "relations" $f_i\in F$.

Note that each $f_i$ is a word in $F$, that is, a product of the generators (and their inverses) of $F$. We then "kill" the relation $f_i$ by attaching a disk $D^2$ along the circles in $f_i$.

As a (very good) practice problem, try to show the fundamental group of the torus is $$\Bbb Z\times \Bbb Z\cong F(a,b)/\langle aba^{-1}b^{-1}\rangle.$$

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    $\begingroup$ May I ask, why exactly does attaching $2$-cells by the words $f_i$ amount to quotienting the free groups by the subgroup the words generate? $\endgroup$ – Chelsea Dirks Sep 21 '14 at 10:59
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    $\begingroup$ A $2$-cell represents a homotopy of the boundary loop to a point. $\endgroup$ – Quang Hoang Sep 21 '14 at 13:10
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    $\begingroup$ Van Kampen Theorem plus the fact that $\pi_1(S^1) = \mathbb Z$. $\endgroup$ – Justin Young Sep 21 '14 at 14:25

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