Commutativity of the fundamental group of any Lie Group  How do we formally prove that the fundamental group of any Lie group is always commutative?
 A: It is actually true for all topological groups. Topological groups possess a structure which makes them H-spaces and fundamental group of every H-space is abelian. The formulation and the proof is given in Algebraic Topology, Homotopy and Homology, by Switzer Pages 14-16.
A: One-sentence explanation: because the fact that a topological group $G$ is a group object in topological spaces makes its fundamental group $\pi_1(G)$ a group object in groups, and this is an abelian group.
A: As Vahid says, it is true for any topological group.  Here is a proof.  I'm sure there are nicer, more conceptual ones out there, but here goes.
Let $G$ be your topological group.  Take two loops $\sigma$ and $\gamma$ in $G$, based at the identity of $G$, which we will denote by $e$.  Let $\sigma \cdot \gamma$ be the concatenation of the two loops.  This is given by
$$ (\sigma \cdot \gamma) (t) =   
\begin{cases} \sigma(2t) & \quad \text{ if } 0 \le t \le 1/2 \\\
 \gamma(2t-1) &\quad \text{ if } 1/2 \le t \le 1 \end{cases} $$
(Sorry, couldn't manage to format that any better.  Feel free to edit if you know how to put a nice brace bracket to the left of that definition.)
The idea is this.  We will show that $\sigma \cdot \gamma$ is homotopic to to the loop given by the pointwise product of $\sigma$ and $\gamma$.  Let's call that loop $\rho$, so
$$ \rho(t) = \sigma(t)\gamma(t).$$
Now define an auxiliary function $P : [0,1] \times [0,1] \to G$ by
$$  P(s,t) =
\begin{cases} \sigma\left(  \frac{2t}{1+s}  \right) &  \quad \text{ if } 0 \le t \le \frac{1+s}{2} \\\
 e &\quad \text{ if } \frac{1+s}{2} \le t \le 1 \end{cases}$$
At $s=0$, this function does the whole loop $\sigma$ as $t$ goes from $0$ to $1/2$, then sits at $e$.  In other words, at $s=0$ this is the first half of the loop $\sigma \cdot \gamma$.  As $s$ gets larger, $P$ does the whole loop $\sigma$ as $t$ goes from $0$ to $\frac{1+s}{2}$.  At $s=1$, $P$ does the loop $\sigma$ at normal speed.
Then similarly define a function $Q : [0,1] \times [0,1] \to G$ by
$$  Q(s,t) =
\begin{cases} e &  \quad \text{ if } 0 \le t \le \frac{1-s}{2} \\\
 \gamma \left( \frac{2t-1+s}{1+s} \right) &\quad \text{ if } \frac{1-s}{2} \le t \le 1 \end{cases}$$
At $s=0$ this is just the second half of the loop $\sigma\cdot\gamma$, while at $s=1$ it is exactly the loop $\gamma$.
So finally, define
$$ H(s,t) = P(s,t) \cdot Q(s,t). $$
At $s=0$ this is $\sigma \cdot \gamma$, while at $s=1$ it is the pointwise product loop $\rho$.  $H$ is clearly continuous, and $H(s,0) = e = H(s,1)$ for all $s$, so this is a homotopy of loops between $\sigma \cdot \gamma$ and $\rho$.
Now we can redo that process and show that $\rho$ is homotopic to the other concatenation $\gamma \cdot \sigma$.  So this shows that $\pi_1(G)$ is abelian.
A: Geometric proof: A connected Lie group $G$ is homotopy equivalent to a maximal 
compact subgroup, so we may assume $G$ is compact. Being compact, $G$ admits a bi-invariant 
Riemannian metric with respect to which it is a symmetric space, the symmetry $s$ at the 
identity being just the inversion map. Now a homotopy class in $\pi_1(G,1)$ can be represented
by a closed geodesic $\gamma$ (of minimal length in its homotopy class, by a shortening process). Since the differential of $s$ at $1$ is minus identity, $s$ sends $\gamma$ to itself
parametrized backwards. It follows that the homomorphism induced by $s$ on the $\pi_1$-level is inversion. However, the inversion map in a group is a homomorphism if and only if the 
group is Abelian. 
