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Stabilizer subgroups - proof verification

kalish

Member
Oct 7, 2013
99
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
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 $G_x$.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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 $h^{-1}G_yh \subseteq G_x$)?