If $G$ is a discrete or topological group, $G$ is a closed subgroup of $EG$, and normal iff $G$ is abelian, according to Segal, *Cohomology of topological groups*, Symposia Mathematica IV (1970) (a reference I found from two answers by Chris Schommer-Pries: [Classifying Space of a Group Extension](https://mathoverflow.net/questions/12144/classifying-space-of-a-group-extension) and [Good functorial model for BG](https://mathoverflow.net/questions/42819/good-functorial-model-for-bg)). Hence for $G$ not abelian, $G$ is not normal in $EG$, hence $BG=EG/G$ is not a group.


On the other hand, we also learn from Segal that the reason $EG$ is a group is that the universal bundle functor $E$ is monoidal with respect to the Cartesian monoidal structure. Any monoidal functor takes group objects in one category to group objects in another simply by functoriality. But the classifying space functor $B$ is also monoidal (see [Good functorial model for BG](https://mathoverflow.net/questions/42819/good-functorial-model-for-bg) or Peter May's answer at [Group structure on Eilenberg-MacLane spaces](https://math.stackexchange.com/questions/45556/group-structure-on-eilenberg-maclane-spaces)). Hence $BG$ should be a group?



Why isn't $BG$ a group, for $G$ not abelian?
The classifying space functor may be a monoidal functor out of $\text{Grp}$, but nonabelian groups aren't group objects in $\text{Grp}$. (The group objects in $\text{Grp}$ are precisely the abelian groups. This is also a corollary of the Eckmann-Hilton argument.)


