$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian Hypothesis: Let $G$ be a topological group with identity element $e$.  Let $\mu$ denote the multiplication mapping in $G$.
Goal: Show that $\pi_1(G,e) = \pi(G)$ is an abelian group via the hint below.
Hint: There are two products on $\pi(G)$.  The usual product $\circ$ defined for the fundamental group and the product $\ast$ induced by
$$
\ast: \pi(G) \times \pi(G) \cong \pi(G \times G) \overset{\pi(\mu)}{\rightarrow} \pi(G).
$$
Show that there is a common two sided unit, $u \in \pi(G)$ for both products and there is a distributive law
$$
(f \ast g) \circ (a \ast b) = (f \circ a) \ast (g \circ b)
$$
Attempt:


*

*Suppose the distributive law in the hint holds.  Suppose further that $1$ -- the identity element in $\pi(G)$ with respect to $\circ$ -- serves also as the identity element in $\pi(G)$ with respect to $\ast$.

*Then for $a,b \in \pi(G)$, we would have the following relationship:
$$
(1 \ast a) \circ (b \ast 1) = (1 \circ b) \ast (a \circ 1)
$$
so that
$$
a \circ b = b \ast a
$$

*Then if we can show that for all $f,g \in \pi(G)$ that $f \ast g \iff f \circ g$, we could complete the above relation to
$$
a \circ b = b \ast a = b \circ a
$$
so that $\pi(G)$ is abelian as desired.
Question: Am I on the right track?
 A: 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


*

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

*$m^{-1}nm=n^{\mu m}$
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. 
A: 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.
