Given $(G,\cdot)$ a topological group with identity $e$, is it always true that $\pi_1(G,e)$ is abelian?
For what it's worth: I have already shown that if $f,g\in\Omega(X,e)$, then $[f \times g]=[f\cdot g]$, where "$\times$" is the ordinary path multiplication (left to right), "$\cdot$" is the group multiplication (that's to say, $f\cdot g(t)=f(t)g(t)$) and $[\cdot]$ is the homotopy class. So in view of this property, if $G$ is abelian itself then $\pi_1(X,e)$ is trivially abelian too. However, I have no clue how to proceed when $G$ is nonabelian.
