# is the group of rational numbers the fundamental group of some space?

Which path connected space has fundamental group isomorphic to the group of rationals? More generally, is every group the fundamental group of a space?

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See the Wikipedia page on Eilenberg-Mac Lane spaces for an even better statement: For every group $G$ there is a $CW$-complex $K(G,1)$ (unique up to homotopy equivalence) such that $\pi_1(K(G,1)) \cong G$ and $\pi_{n}(K(G,1)) = 0$ for all $n \neq 1$. This is also true for every other value of $1$ (to quote Mariano Suárez-Alvarez) and abelian $G$ and proofs of these statements can be found in almost all books on algebraic topology.

A nice and and rather explicit example for a space with fundamental group $\mathbb{Q}$ can be constructed using the theory of graphs of groups, see exercise 6 on page 96 of Hatcher's book.

In 1988, Shelah proved that there is no "nice" compact space with fundamental group $\mathbb{Q}$, where nice means metric, compact (hence separable) path connected and locally path connected. Indeed, Shelah has shown the fundamental group of a nice compact space is either finitely generated or has the cardinality of the continuum.

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Every group is the fundamental group of a space; a relatively easy choice of such a space is the presentation complex associated to a presentation. First, every group $G$ has a presentation with some generators $g_i$ indexed by some set $I$ and some relations $r_j$ indexed by some set $J$. Let $X$ be the wedge of $|I|$ circles; by Seifert-van Kampen we know that $\pi_1(X) \cong F_{|I|}$.

Now we will add some $2$-cells corresponding to the relations. First, note that every relation $r_j$ determines a homotopy class of paths in $X$, hence a subspace isomorphic to $S^1$ of $X$. Associated to such a subspace is an attaching map $b_j : S^1 \to B^2$, and we can form the adjunction space $X \cup_{b_j} B^2$ in order to attach the appropriate $2$-cell and kill the relation $r_j$ (proven by a second application of Seifert-van Kampen). Doing this for all relations gives the appropriate space.

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Nice! But, if the group is not finitely generated, is it so easy to give a presentation? For example, how is a presentation of rationals? – MBL May 3 '11 at 20:10
@MBL: given a group $G$ there is a canonical map $F_{|G|} \to G$ from the free group on the elements of $G$ to $G$. We can take the set of relations to be the kernel of this map. (I didn't say the presentation had to be nice!) – Qiaochu Yuan May 3 '11 at 20:15