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.



Fundamental groups of topological groups.
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: 


$$\textbf{Set} \stackrel{K}{\to} \textbf{Cat} \stackrel{\text{nerve}}{\to} \textbf{Set}^{\Delta^{op}} \stackrel{R}{\to} \textbf{CGHaus}$$ 


($\textbf{CGHaus}$ 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 $\mathbb{F}\_2$-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 $\mathbb{F}\_2$ in $\textbf{CGHaus}$. It can also be considered an internal vector space over $\mathbb{F}\_2$ in $\textbf{Top}$, the category of ordinary topological spaces, because a finite power $X^n$ in $\textbf{Top}$ of a CW-complex $X$ has the same topology as $X^n$ does in $\textbf{CGHaus}$ provided that $X$ has only countably many cells, which is certainly the case for $S^\infty$ (see [Hatcher's book](http://www.math.cornell.edu/~hatcher/AT/ATapp.pdf), Theorem A.6). Thus $S^\infty$ can be considered as an honest commutative topological group of exponent 2. 


The underlying group of $S^\infty$ (in $\textbf{Set}$) 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. 


