TITLE: Why isn't $BG$ a group, for $G$ not abelian?
QUESTION [8 upvotes]: 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 and 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 or Peter May's answer at Group structure on Eilenberg-MacLane spaces). Hence $BG$ should be a group?

REPLY [2 votes]: [As a non-homotopy theorist, I am not certain about this answer.]
If your topological group $G$ has a $2$-fold loop space structure, but not an infinite loop space structure, then $BG$ is naturally a group, but there is some $n \geq 2$ for which $B^nG$ is not.  In this case, $G$ is not strictly abelian, despite $BG$ being a group.  In fact, even in the infinite loop space case, you can have groups that are not strictly abelian, like the infinite unitary group $U$.
As other answers have pointed out, this behavior is not possible when $G$ is discrete.  Indeed, 2-fold loop structure automatically strictifies to the abelian property in the discrete case.