$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian

Or a one-line proof: "the fundamental group functor preserves products, hence it sends group objects to group objects". Note that the group objects in ${\bf Top}$ are precisely the topological groups, and group objects in ${\bf Grp}$ correspond to abelian groups.

One can get more out of this situation,which especially useful if the topological group $G$ is not connected. For more details, see this paper.

Let $\tilde{G}$ be the set of homotopy classes rel end points of paths in $G$ which start at $e$. The group structure of $G$ induces a group structure on $\tilde{G}$. The final point map defines $t: \tilde{G} \to G$, which under appropriate local conditions is the universal cover of $G$ at $e$. This morphism may also be given the structure of crossed module, using the conjugation operation of $G$ on $\tilde{G}$.

Recall that a crossed module $\mu: M \to P$ is a morphism of groups together with an action of $P$ on $M$ written $(m,p) \mapsto m^p$ with the properties

1. $\mu(m^p)= p^{-1}\mu(m) p$;

2. $m^{-1}nm=n^{\mu m}$

for all $m,n \in M, p \in P$. Such a crossed module determines a $k$-invariant $k \in H^3(Cok\; \mu, Ker\; \mu)$.

For the crossed module coming from a topological group as above, the $k$-invariant is trivial if and only if the topological group $G$ has a universal cover of all components such that the totality can be given the structure of topological group with the covering map a morphism of topological groups.