# Every Group is a Fundamental Group

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 $\pi_1 (X,x)=G$?

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 $$\Bbb Z\times \Bbb Z\cong F(a,b)/\langle aba^{-1}b^{-1}\rangle.$$
• 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? – Chelsea Dirks Sep 21 '14 at 10:59
• A $2$-cell represents a homotopy of the boundary loop to a point. – Quang Hoang Sep 21 '14 at 13:10
• Van Kampen Theorem plus the fact that $\pi_1(S^1) = \mathbb Z$. – Justin Young Sep 21 '14 at 14:25