TITLE: Fundamental groups of topological groups.
QUESTION [20 upvotes]: 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.

REPLY [2 votes]: [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.