If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)=\pi_0(G)$, as desired.

If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)\cong\pi_0(G)$, as desired. I would even bet that the above isomorphism respects the natural topologies that are present on both sides.

If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. It is well known that the fundamental group of the Hawaiian earrings is a completely wild object, about which not much interesting can be said.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)=\pi_0(G)$, as desired.

If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)=\pi_0(G)$, as desired.