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Let $G$ be a topological group, and $\pi_1(G,e)$ its fundamental group at the identity. If $G$ is the trivial group then $G \cong \pi_1(G,e)$ as abstract groups. My question is: If $G$ is a non-trivial topological group can $G \cong \pi_1(G,e)$ as abstract groups? About all I know now is that $G$ would have to be abelian. |
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Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s. The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph_0}$. This is not hard: the total space $S^\infty$ of the universal $\mathbb{Z}_2$-bundle is obtained by applying a composite of functors to the group structure $\mathbb{Z}_2$ in the category of sets: $$Set \stackrel{K}{\to} Cat \stackrel{nerve}{\to} Set^{\Delta^{op}} \stackrel{R}{\to} Top$$ ($Top$ here is the category of compactly generated Hausdorff spaces and continuous maps). Here $K$ is the right adjoint to the "underlying set of objects" functor; it takes a set to the category whose objects are the elements of the set and there is exactly one morphism between any two objects. The functor $R$ is of course geometric realization. Each of these functors is product-preserving, and since the concept of group can be formulated in any category with finite products, a product-preserving functor will map a group object in the domain category to one in the codomain category. Even more: the concept of a $\mathbb{Z}_2$-vector space makes sense in any category with finite products since we merely need to add the equation $\forall_x x^2 = 1$ to the axioms for groups, which can be expressed by a simple commutative diagram. Thus $S^\infty$ is an internal vector space over $\mathbb{Z}_2$ in $Top$. Its underlying group (in $Set$) is clearly a vector space of dimension $2^{\aleph_0}$. We make take this vector space to be the countable product $\mathbb{Z}_2^{\mathbb{N}}$. Modding out by $\mathbb{Z}_2$ (modding out by a 1-dimensional subspace), the space $\mathbb{RP}^\infty$ is also, as an abstract group, isomorphic to this. And so is a countably infinite product $(\mathbb{RP}^\infty)^{\mathbb{N}}$ of copies of $\mathbb{RP}^\infty$. Finally, the functor $\pi_1$ is product-preserving, and so $$\pi_1((\mathbb{RP}^\infty)^{\mathbb{N}}) \cong \mathbb{Z}_{2}^{\mathbb{N}}$$ and we are done. |
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This question occurred as Advanced Problem 5889 in the Amer. Math. Monthly 80 (1973), no. 1, 82. It was listed as still unsolved five years later, in vol. 85, no. 10, p. 834, of the Monthly; however, my recollection is that it mysteriously vanished from the Monthly's "unsolved" list the next time this got updated, but without a solution having appeared in the interim. |
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First, $\pi_1(G;e)=\pi_1(G_e)$, where $G_e$ is the connected component of $e$, therefore we may assume that $G$ is connected. If $G$ is a finite dimensional compact connected manifold, then it is an $n$-torus (because it is abelian), and its fundamental group is discrete (${\mathbb Z}^n$). In this case, $\pi_1(G)$ is very different from $G$. I suspect that in general $\pi_1$ of an abelian group is either trivial or non-compact. It is hard to see how it could not be discrete (if it is discrete, then $G$ has to be trivial). If $\pi_1(G)$ is not discrete, there are arbitrarily small loops that are not homotopic to a point; thus $G$ is full of holes! Definitely a strange beast. |
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