$G$ 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:

1. 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 $\ast$.

2. 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

   $$ a \circ b = b \ast a $$

3. 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?

Solution 1:

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.

Solution 2:

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.