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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?

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In the discrete case, this is a corollary of the Eckmann-Hilton argument. – Qiaochu Yuan Jun 29 '13 at 18:18
The correct spelling is Chris Schommer-Pries. Sommer-Preis sounds very german. ;) I would have suggested an edit, but an edit must change at least six characters, I was told. – Rasmus Bentmann Jun 29 '13 at 19:17
I just did the edit you suggest, Rasmus, and the software didn't object. – Tom Leinster Jun 29 '13 at 20:51
@Tom: Perhaps the lower limit only applies to suggested edits (I don't have enough rep to edit)? – Rasmus Bentmann Jul 2 '13 at 7:41
Apologies to Chris. – ziggurism Jul 2 '13 at 23:37
up vote 18 down vote accepted

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.)

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Of course! The group objects are abelian groups. Thank you – ziggurism Jun 29 '13 at 18:38
So we have two ways to see that $BG$ is a group when $G$ is abelian, 1. $B$ is monoidal functor, and 2. $BG$ represents $H^1(-;G)$ which has an obvious group structure which descends to $BG$ by Yoneda. I just learned that second one from Andrew Stacey here: – ziggurism Jun 29 '13 at 18:43
@Joe: yes, but in with the latter argument we only get a group structure on the homotopy type of $BG$, whereas with an appropriate construction of $B$ we should get a group structure on an actual topological space. – Qiaochu Yuan Jun 29 '13 at 18:52

A simple answer for why $BG$ is not a group when $G$ is discrete is that $BG=K(G,1)$, so $\pi_1(BG)=G$. However, if $K(G,1)$ were a topological group, then $\pi_1(K(G,1))=G$ must be abelian. This is proved in the usual way: the pointwise product of two loops in the group structure on $BG$ is homotopic to the concatenation in either order, by homotoping the first loop to be the identity for the first half of the interval, and homotoping the second loop to be the identity for the second half of the interval, or vice-versa.

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Your account seems to have bifurcated, @Agol – David Roberts Jun 29 '13 at 23:58

[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.

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I know how to strictify a loopspace into a monoid... but where are you getting strict inverses? – Dylan Wilson Jun 30 '13 at 15:45
@DylanWilson Sorry, I do not have a good answer to your question. – S. Carnahan Jul 22 '13 at 17:30

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