# When is $\pi_0$ a group?

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?

– user122283
Mar 29, 2014 at 23:09
• 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.
– user98602
Mar 29, 2014 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, 2014 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.)
– user98602
Mar 30, 2014 at 19:49

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 $G$.

• The map f is the group operation?
– Josh
Mar 30, 2014 at 19:43
• Yes, I should have clarified. $f(g,h) = gh$.
– user98602
Mar 30, 2014 at 19:44
• So you mean $f(C,C)$ to mean $\{gh \in G | g, h \in C\}$?
– Josh
Mar 30, 2014 at 19:47
• @jd.r Yep (and similarly in other cases, where I make the arguments as sets rather than elements).
– user98602
Mar 30, 2014 at 19:50
• That's what I thought, and what made sense in context...just wanted to be sure.
– Josh
Mar 30, 2014 at 20:14