MHB Stabilizer subgroups - proof verification

kalish1
Messages
79
Reaction score
0
I have a problem that I would like help on. I'm preparing for an exam, and I have provided my work below.

**Problem statement:** Let $G$ act on $X$, and suppose $x,y\in X$ are in the same orbit for this action. How are the stabilizer subgroups $G_x$ and $G_y$ related?

**My attempt:** $G_x = G_y = \{g \in G: gx=x\} = \{g \in G: gy=y\} \forall x,y \in X, \forall g \in G$.

Thanks.
 
Physics news on Phys.org
kalish said:
I have a problem that I would like help on. I'm preparing for an exam, and I have provided my work below.

**Problem statement:** Let $G$ act on $X$, and suppose $x,y\in X$ are in the same orbit for this action. How are the stabilizer subgroups $G_x$ and $G_y$ related?

**My attempt:** $G_x = G_y = \{g \in G: gx=x\} = \{g \in G: gy=y\} \forall x,y \in X, \forall g \in G$.
If $x$ and $y$ are in the same orbit then $y=hx$ for some $h\in G.$ Use that information to express the definition of $G_y$ in terms of $x$ rather than $y$. Then see how that compares with the definition of $G_x$.
 
Opalg's post hints at, but does not explicitly state an important fact about group actions:

If $y = hx$, then $x = h^{-1}y$.

To avoid "symbol-overloading" let's use the following notation:

$G_x = \{g \in G: gx = x\}$

$G_y = \{g' \in G: g'y = y\}$.

What can we say about the relationship of $g'$ to $g$? We can start with the most trivial of observations:

$y = y$.

That is:

$y = g'y$ (for $g' \in G_y$).

Since we are given that $y = hx$ for some $h \in G$, we have:

$hx = y = g'y$

Now for $g \in G_x$, we certainly have:

$x = gx$.

Thus $y = hx = h(gx) = (hg)x$.

Hence $(hgh^{-1})y = (hg)(h^{-1}y) = (hg)(x) = h(gx) = hx = y$.

This shows that $hgh^{-1} \in G_y$, that is: $hG_xh^{-1} \subseteq G_y$.

Can you show the other inclusion (you may find it easier to show that $h^{-1}G_yh \subseteq G_x$)?
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top