# Possible mistake in Boothby's Manifolds book

This question is from section 7, chapter 3 of An introduction to Differentiable Manifolds By William M. Boothby.

$G_{x_0}$ is the stabilizer of $x_0$. The relation that gives sense to the quotient $G/G_{x_0}$ is defined in $G$ by $$g_1\sim g_2\quad\text{iff}\quad g_2=hg_1$$ for some $h\in G_{x_0}$. So $G/G_{x_0}$ is defined as $G/\sim$ as usual.

The condition $\tilde{F}=F\circ\pi$ implies that if $g_1\sim g_2$ then $[g_1]=[g_2]$ and then \begin{align*} \pi(g_1) &=\pi(g_2), &&\text{taking F both sides gives}\\ \tilde{F}(g_1)&= \tilde{F}(g_2)\\ g_1x_0 &= g_2x_0\\ x_0 &= g_1^{-1}g_2x_0, \end{align*} that is, $g_1^{-1}g_2\in G_{x_0}$ and this implies $g_1^{-1}\sim g_2^{-1}$.

So a necessary condition for this $F$ to exist is that $G$ must satisfy $$g_1\sim g_2\quad\text{implies}\quad g_1^{-1}\sim g_2^{-1}.\tag{1}\label{c}$$

There are groups $G$ so that \ref{c} is not true. For such a $G$, $F$ can't exist.

My question is is it the exercise wrong? Do we really need additional conditions on $G$.

I don't see how to use that $G$ acts transitively on $X$ to solve the problem.

EDIT The convention regarding groups and subgroups acting on them is established in the previous example (7.11). Subgroups are supposed to act on the right by translations in which case everything follows as in the Ralth answer. I'll leave the question.

• The question deals with left cosets, so your equivalence relation is really $g_1 \sim g_2$ iff $g_2 = g_1h$ for some $h$.
– user27126
Commented Dec 9, 2012 at 23:19
• As the action is tansitive, this is essentially the orbit-stabilizer theorem. Maybe google that one (proof is fairly easy). Commented Dec 10, 2012 at 0:07

I do not believe the book is wrong, you just need to rethink the equivalence relation you are using.

(i) is basically the orbit stabilizer theorem for group actions. Let $G$ be a group that acts on a set $X$, not necessarily transitively. Then fix a point $x_0 \in X$. The theorem states that there is a bijection between the left cosets of $G_{x_0}$ and $G\cdot x_0$, the orbit of $x_0$ (all the elements one can get by action on $x_0$).

Define the map $F:G/G_{x_0} \to G\cdot x$ by $F(gG_{x_0})=g\cdot x_0$ just as they do in the question. This is well defined since $(gh)\cdot x_0 =g\cdot(h\cdot x_0)=g\cdot x_0$ where $h\in G_{x_0}$. It is clearly surjective, so suppose $F(gG_{x_0}) = F(kG_{x_0})$ for some $g,k \in G$. Then $k^{-1}g\cdot x_0=x_0$, i.e. $k^{-1}g \in G_{x_0}$. Thus $g=kh$ for some $h\in H$.

Here is where I think you are having trouble. It is true that two elements $g_1, g_2$ are in the same left coset of a subgroup $H$ if and only if $g_1=g_2h$ (NOTE THE ORDER) for some $h\in H$. This follows since if $g_1=gh_1$ and $g_2=gh_2$ (that is $g_1, g_2\in gH$), then there exists and $h_3\in H$ such that $h_1h_3=h_2$ so that $g_1h_3=g_2$. Conversely if $g_1=g_2h$, then they belong to the same left coset since multiplication on the right by an element of $H$ is not going to change the coset.

Composing $F$ with $\pi$ gives $\tilde F=(F\circ \pi)(g_1)=F(g_1G_{x_0})=g_1\cdot x_0$ as required.

• I did it with that relation because it is as he define in a previous pages. With your equivalence relation the class of $g_2$ is $[g_2]=g_2G_{x_0}$ which seems like a right coset.
– leo
Commented Dec 10, 2012 at 0:37
• My definition of a left coset is $gH=\{gh | h\in H\}$. Maybe we having a right left issue? Commented Dec 10, 2012 at 0:39
• You are right. I was wrong. The thing is that one page before he calls left coset to something that is right coset in a context where there is no reason for they to be the same thing. That was what confused me.
– leo
Commented Dec 10, 2012 at 0:56