Stabilizer subgroups - proof verification

In summary, the stabilizer subgroups $G_x$ and $G_y$ are related by $hG_xh^{-1} \subseteq G_y$ and $h^{-1}G_yh \subseteq G_x$, where $x$ and $y$ are in the same orbit and $h$ is any element in $G$.
  • #1
kalish1
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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.
 
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  • #2
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 $G_x$.
 
  • #3
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 $h^{-1}G_yh \subseteq G_x$)?
 

Related to Stabilizer subgroups - proof verification

1. What is a stabilizer subgroup?

A stabilizer subgroup is a subgroup of a larger group that fixes a particular element or set of elements in that group. It is also known as a stabilizer or point stabilizer, and it is a fundamental concept in group theory and abstract algebra.

2. Why is proof verification important for stabilizer subgroups?

Proof verification is important for stabilizer subgroups because it ensures that the properties and equations used to describe them are accurate and valid. This is crucial for making accurate conclusions and predictions based on the stabilizer subgroup.

3. How do you prove the existence of a stabilizer subgroup?

To prove the existence of a stabilizer subgroup, one must show that the subgroup satisfies the defining properties of a stabilizer subgroup. This includes showing that it fixes the specified element or set of elements and that it is a subgroup of the larger group.

4. Can stabilizer subgroups exist in non-abelian groups?

Yes, stabilizer subgroups can exist in non-abelian groups. In fact, stabilizer subgroups are a common concept in non-abelian groups, and they play an important role in understanding the structure and properties of these types of groups.

5. How are stabilizer subgroups used in applications?

Stabilizer subgroups have various applications in mathematics and other fields. In group theory, they are used to study the structure and properties of groups. In physics, stabilizer subgroups are used to study symmetries and conservation laws. They also have applications in computer science, cryptography, and coding theory.

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