When is $\pi_0$ a group?

Question

In general, $\pi_0(X)$ is the set of path components of $X$ and does not have a group structure. After all, $S_0$ is just two points and the usual way of multiplying using the equation of a sphere doesn't work. But sometimes it is. For example, if $X$ is an H space we still have a 0-th homotopy group.

What are other cases? In particular, if $G$ is a discrete group (group with the discrete topology), then is $\pi_0(G)=G$ as a set and as a group?

I just realized you mentioned H-spaces in your question. You... realize every group is an H-space, right? So a discrete group trivially has that property. — , Mar 29 '14 at 23:36
I was aware of the statement for H-spaces. But I was under the impression that H-spaces were required to be path-connected spaces. — Josh, Mar 30 '14 at 19:41
(I just realized that sentence sounded rude - that wasn't intended at all! Sorry!) I wasn't under the impression that an H-space needed to be path-connected; in that case, $\pi_0$ is of course the trivial group. (I'm not aware of the proof that an H-space has $\pi_0$ a group. I think you're right that we need something stronger than H-space in that case, if we allow H-spaces to be disconnected.) — , Mar 30 '14 at 19:49

score 20 · Accepted Answer · answered Mar 29 '14 at 23:29

20

When $G$ is a topological group, we always have that $\pi_0(G)$ is a (topological) group!

Lemma: The connected component of the identity in $G$ - let's denote it $C$ - is a normal subgroup of $G$. By continuity of the group operation, $f(C,C)$ is connected. It contains the identity, so $f(C,C) \subset C$ because $C$ is the connected component of the identity. Similarly, $C^{-1} \subset C$ by continuity of the inversion operator. So $C$ is a subgroup.

We need to show normality still: that $xCx^{-1} \subset C$ for $x \in G$. Continuity (again!) shows us that $f(xC, x^{-1})$ is connected, because $xC = f(x,C)$ is connected, so that $xCx^{-1} \subset C$. This proves what we wanted to prove.

Now $G/C$ is the set of connected components of $G$, and is a group. So $\pi_0(G)$ always inherits a group structure from $G$.

answered Mar 29 '14 at 23:29

The map f is the group operation? – Josh Mar 30 '14 at 19:43
Yes, I should have clarified. $f(g,h) = gh$. – Mar 30 '14 at 19:44
So you mean $f(C,C)$ to mean ${gh \in G | g, h \in C}$? – Josh Mar 30 '14 at 19:47
@jd.r Yep (and similarly in other cases, where I make the arguments as sets rather than elements). – Mar 30 '14 at 19:50
That's what I thought, and what made sense in context...just wanted to be sure. – Josh Mar 30 '14 at 20:14
Could you explain what you mean by the "continuity of the group operation?" – Dylan C. Beck Feb 14 '18 at 01:56
@Dylan_Carlo_Beck A topological group is a space equipped with a group operation $m:G \times G \to G$ which is continuous and satisfies the usual group axioms. – Feb 14 '18 at 02:03
You seem to be using the set of all connected components of $G$ rather than the set of all path-components of $G$. – murray Jul 26 '22 at 14:36

When is $\pi_0$ a group?

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