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.
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, 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.
Here is another construction, which I think shows that it isn't surprising that such groups exist. First, we note the fact that for any abelian group $G$, there is a model of $BG$ that is an abelian topological group (there are many ways to see this; Todd's answer mentions one). Now start with any nontrivial abelian group $G_0$, and let $G_1$ be the underlying (discrete) group of $BG_0$. Similarly, let $G_2$ be the underlying group of $BG_1$, and so on. Now let $G=G_0\times\prod BG_n$. Then $\pi_1(G)=\prod G_n$ is isomorphic to $G$ as an abstract group. If you start with $G_0$ that is not just a discrete group but a simply connected topological group (e.g., $\mathbb{R}$), this gives you an example where $G$ is connected.
(Technically, this construction may require you to work in CGHaus rather than Top--it works in any category of spaces in which for any abelian $G$, you can construct an abelian group object whose fundamental group is $G$. I don't know if this is possible in Top.)
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.
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.
[I originally entered the answer below in response to another question. Later other people pointed out that that question was a duplicate of this one, so I joined them in closing it.]
Let $V$ be a vector space over $\mathbb{Q}$ of dimension $2^{\aleph_0}$ (e.g. $V=\mathbb{R}$) and put $G=BV$. Here we use the simplicially defined classifying space functor $B$, so $G\times G=BV\times BV=B(V\times V)$, so we can apply $B$ to the addition map $V\times V\to V$ to get a topological group structure on $BV$. By construction we have $\pi_1(G)\simeq V$, but also $G$ is a vector space over $\mathbb{Q}$ of dimension $2^{\aleph_0}$, so $G\simeq V$ as abstract groups. I don't think that you can do anything much smaller than this.
