I've recently become familiar with the notion of topological group and the theorem that the binary operation on topolgical group and the one on it's fundamental group are the same.
However I questioned whether all topological groups must be path connected, whether it makes actually sense to talk about loops on it generally? And the answer was no, there are topological groups that are not path connected.
My question is: is it that there is just a silent assumption in the theorem (the one stating the binary operations are the same), that the topological group is path connected? or am I missing some important point here?
edit: I mean why there is no assumption on path connectedness in the theorem itself?
You don't need to assume that a topological group is path connected. Normally in the definition of the fundamental group you have to pick a base point. However the fundamental group of a topological group does not depend on the choice of the base point. That is because for any $a, b\in G$ there is a homeomorphism $f:G\to G$ that maps $a\mapsto b$ namely $f(x)=a^{-1}x$. This homeomorphism induces isomorphism of $\pi(G, a)\simeq \pi(G, b)$. Actually the same is true for any space that has this property (i.e. the existance of such homeomorphism for any two points). So that's why the notion $\pi(G)$ makes sense.
Also it is convenient to always take $\pi(G, e)$ where $e$ is the neutral element of $G$. That's because the path connected component of $e$ is a closed normal subgroup of $G$. So from the other point of view you can always assume that $G$ is path connected.