Number of Orbits: Showing Equality of Group G and G_a

[mathmari]

Hey!

Let the finite group $G$ act transitively on the set $\Omega$. Then the action of $G$ and on $\Omega\times\Omega$ is defined as follows $(a,b)\cdot x=(a\cdot x, b\cdot x)$.
Let $a\in \Omega$.
Show that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$. From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. $\forall \omega_1, \omega_2\in \Omega, \exists g\in G$ such that $g*\omega_1=\omega_2$, right? (Wondering) Could you give me some hints how we could show the equality of the two number of orbits? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. $\forall \omega_1, \omega_2\in \Omega, \exists g\in G$ such that $g*\omega_1=\omega_2$, right? (Wondering)

Could you give me some hints how we could show the equality of the two number of orbits? (Wondering)

Right. (Nod)

Have you tried to apply the Orbit-stabilizer theorem and Burnside's lemma? (Wondering)
To be honest, I'm not quite getting the requested result yet, but it's a good exercise.

 

[mathmari]

We have that $G_a=\{g\in G\mid g\cdot a=a\}, a\in \Omega$.
Since $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. $\forall \omega_1, \omega_2\in \Omega, \exists g\in G$ such that $g \cdot \omega_1=\omega_2$.
For $\omega_1=\omega_2=a$, we have that $g\cdot a=a$, right? (Wondering)

We have that $\Omega^g=\{\omega \in \Omega \mid g\omega=\omega\}$.

Does it stand that $|\Omega^g|=|\Omega|$, because of the above, i.e., because $G$ acts transitively on the set $\Omega$ ? (Wondering)

From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}|\Omega^g|$$ right? (Wondering) From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}|(\Omega\times\Omega)^g|$$ right? (Wondering)

 

[I like Serena]

We have that $G_a=\{g\in G\mid g\cdot a=a\}, a\in \Omega$.
Since $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. $\forall \omega_1, \omega_2\in \Omega, \exists g\in G$ such that $g \cdot \omega_1=\omega_2$.

Let's take this step by step. (Thinking)
Yes, we have indeed only one orbit on $\Omega$.

Proof
Let $a \in \Omega$ and suppose there is more than one orbit.
Then there is an element $b\in\Omega$ that is not in the orbit $Ga$.
Then for every $g \in G$ we have that $ga \ne b$, which is a contradiction with the transitivity property.
So indeed, there is only one orbit on $\Omega$. (Nerd)

For $\omega_1=\omega_2=a$, we have that $g\cdot a=a$, right? (Wondering)

We have that $\Omega^g=\{\omega \in \Omega \mid g\omega=\omega\}$.

Does it stand that $|\Omega^g|=|\Omega|$, because of the above, i.e., because $G$ acts transitively on the set $\Omega$ ? (Wondering)

For $\omega_1=\omega_2=a$, we have that there is a $g\in G$ such that $g\cdot a=a$.
This is the case for $g=e$, but I don't think it holds in general for all $g$. (Worried)

From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}|\Omega^g|$$ right?

From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}|(\Omega\times\Omega)^g|$$ right?

Yep. (Nod)

Additionally, the orbit-stabilizer theorem tells us that:
$$|Ga| \cdot |G_a|=|G|$$

 

[mathmari]

For $\omega_1=\omega_2=a$, we have that there is a $g\in G$ such that $g\cdot a=a$.
This is the case for $g=e$, but I don't think it holds in general for all $g$. (Worried)

So, is the definition of $\Omega^g$ that I wrote in post #3 wrong? (Wondering)

 

[I like Serena]

We have that $\Omega^g=\{\omega \in \Omega \mid g\omega=\omega\}$.

Does it stand that $|\Omega^g|=|\Omega|$, because of the above, i.e., because $G$ acts transitively on the set $\Omega$ ? (Wondering)

So, is the definition of $\Omega^g$ that I wrote in post #3 wrong? (Wondering)

Your definition is correct. (Nod)

Let's pick an example.
Suppose we pick $G=S_3=\{(1), (12), ..., (132)\}$ and $\Omega = \{1,2,3\}$.
Does $G$ act transitively on $\Omega$?

Then we have indeed that $\Omega^{(1)} = \Omega$.
What will $\Omega^{(12)}$ be? (Wondering)

 

[mathmari]

Your definition is correct. (Nod)

Let's pick an example.
Suppose we pick $G=S_3=\{(1), (12), ..., (132)\}$ and $\Omega = \{1,2,3\}$.
Does $G$ act transitively on $\Omega$?

Then we have indeed that $\Omega^{(1)} = \Omega$.
What will $\Omega^{(12)}$ be? (Wondering)

We have that $$(12)(1)=(12) \\ (12)(2)=(12) \\ (12)(3)=(12)$$ or not? (Wondering)

 

[I like Serena]

We have that $$(12)(1)=(12) \\ (12)(2)=(12) \\ (12)(3)=(12)$$ or not? (Wondering)

Not quite. (Shake)

That's because $(1) \notin \Omega$, just like $(12) \notin \Omega$.
Note that elements of $G$ are permutations, while elements of $\Omega$ are merely numbers.
The permutation acts on a number and results in a number.

So:
$$(12)1=2 \\ (12)2=1 \\ (12)3=3$$
(Thinking)

While we're at it, can you also tell what $G1$ and $G_1$ are? (Wondering)

 

[mathmari]

Note that elements of $G$ are permutations, while elements of $\Omega$ are merely numbers.
The permutation acts on a number and results in a number.

So:
$$(12)1=2 \\ (12)2=1 \\ (12)3=3$$
(Thinking)

How do we get these numbers? (Wondering)

 

[I like Serena]

How do we get these numbers? (Wondering)

$(12)$ represents the function that maps $1$ to $2$, and that maps $2$ to $1$.
Additionally, it maps $3$ to itself.
If we apply that function to the number $1$, the result is $2$. (Thinking)

 

[mathmari]

So, we have the function $\sigma$ with $\sigma (1)=2$ and $\sigma(2)=1$, right? (Wondering)

And by $(12)1$ do we mean $\sigma (1)$ ? (Wondering)

 

[I like Serena]

So, we have the function $\sigma$ with $\sigma (1)=2$ and $\sigma(2)=1$, right?

And by $(12)1$ do we mean $\sigma (1)$ ? (Wondering)

Yep. That's what we mean when we say that $\sigma$ acts on $1$. (Nod)

 

[mathmari]

Yep. That's what we mean when we say that $\sigma$ acts on $1$. (Nod)

Ah ok... (Nerd)

Let's pick an example.
Suppose we pick $G=S_3=\{(1), (12), ..., (132)\}$ and $\Omega = \{1,2,3\}$.
Does $G$ act transitively on $\Omega$?

Then we have indeed that $\Omega^{(1)} = \Omega$.
What will $\Omega^{(12)}$ be? (Wondering)

So, $\Omega^{(12)}=\Omega$, right? (Wondering)

can you also tell what $G1$ and $G_1$ are? (Wondering)

We have that $G1=\{x\in \Omega : x=g\cdot 1 \text{ for some } g\in G\}=\{g\cdot 1: g\in G\}$ and $G_1=\{g\in G: g\cdot 1=1\}$, right? (Wondering)

 

[I like Serena]

So, $\Omega^{(12)}=\Omega$, right? (Wondering)

Let's see...

We are looking for those elements $\omega \in \Omega$ such that $(12)\omega = \omega$.
Is this true for $\omega=1$? (Wondering)

We have that $G1=\{x\in \Omega : x=g\cdot 1 \text{ for some } g\in G\}=\{g\cdot 1: g\in G\}$ and $G_1=\{g\in G: g\cdot 1=1\}$, right? (Wondering)

Yep... (Sweating)

 

[mathmari]

We are looking for those elements $\omega \in \Omega$ such that $(12)\omega = \omega$.
Is this true for $\omega=1$? (Wondering)

Ahh... So, we have that $\Omega^{(12)}=\{3\}$, right? (Wondering)

 

[I like Serena]

Ahh... So, we have that $\Omega^{(12)}=\{3\}$, right? (Wondering)

Right! (Nod)

 

[mathmari]

Additionally, the orbit-stabilizer theorem tells us that:
$$|Ga| \cdot |G_a|=|G|$$

So, $|G_a|=\frac{|G|}{|Ga|}$.

From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}X(g)=\frac{|Ga|}{|G|}\sum_{g\in G_a}X(g)$$

where $X(g)=|\{\omega \in \Omega \mid g*\omega=\omega\}|$ and $|Ga|=|\{g\cdot a: g\in G\}|$. From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}\tilde{X}(g)$$ where $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid (a,b)\cdot g=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid (a\cdot g, b\cdot g)=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid a\cdot g=a \text{ and } b\cdot g=b\}|$
Is this equal to $|\{a\in \Omega \mid a\cdot g=a\}|\cdot |\{b \in \Omega \mid b\cdot g=b\}|$ or $|\{a\in \Omega \mid a\cdot g=a\}|+ |\{b \in \Omega \mid b\cdot g=b\}|$ ? (Wondering)

Is the formula $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid (a,b)\cdot g=(a,b)\}|$ correct? (Wondering)
Or should it be $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid g\cdot (a,b)=(a,b)\}|$ ? Or are these formulas the same? (Wondering)

 

[I like Serena]

So, $|G_a|=\frac{|G|}{|Ga|}$.

Yep. And we already know from the transitive property that $Ga=\Omega$.

From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}X(g)=\frac{|Ga|}{|G|}\sum_{g\in G_a}X(g)$$

where $X(g)=|\{\omega \in \Omega \mid g*\omega=\omega\}|$ and $|Ga|=|\{g\cdot a: g\in G\}|$.

What does $*$ mean?
Is it conjugation? (Wondering)

From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}\tilde{X}(g)$$ where $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid (a,b)\cdot g=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid (a\cdot g, b\cdot g)=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid a\cdot g=a \text{ and } b\cdot g=b\}|$
Is this equal to $|\{a\in \Omega \mid a\cdot g=a\}|\cdot |\{b \in \Omega \mid b\cdot g=b\}|$ or $|\{a\in \Omega \mid a\cdot g=a\}|+ |\{b \in \Omega \mid b\cdot g=b\}|$ ? (Wondering)

It's the first.
Every combination $(a,b)$ that satisfied $ag=a$ and $bg=b$ counts. (Thinking)

Is the formula $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid (a,b)\cdot g=(a,b)\}|$ correct? (Wondering)
Or should it be $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid g\cdot (a,b)=(a,b)\}|$ ? Or are these formulas the same? (Wondering)

There's a difference between left actions and right actions.
If nothing is specified, we assume that it's a left action, which is more intuitive.
However, the problem statement suggests we're talking about a right action, although it's not explicitly stated.
That means we should always multiply on the right with $g$ to be consistent. (Nerd)

 

[mathmari]

And we already know from the transitive property that $Ga=\Omega$.

Why do we have from the transitive property that $Ga=\Omega$ ? (Wondering)

What does $*$ mean?
Is it conjugation? (Wondering)

It is an action. In my notes there is the following:

$G\rightarrow G$ conjugation
$g*a=a^{g^{-1}}=g^{-1}ag$

It's the first.
Every combination $(a,b)$ that satisfied $ag=a$ and $bg=b$ counts. (Thinking)

Could you explain to me further why it is the first one? (Wondering)

There's a difference between left actions and right actions.
If nothing is specified, we assume that it's a left action, which is more intuitive.
However, the problem statement suggests we're talking about a right action, although it's not explicitly stated.
That means we should always multiply on the right with $g$ to be consistent. (Nerd)

In my notes there is the following formula: $$G \overset{\text{ action }}{\longrightarrow } \Omega \\ X(g)=|\{\omega \in \Omega \mid g*\omega =\omega\}|$$

So, can we not apply the right action of the problem statement? (Wondering)

 

[I like Serena]

Why do we have from the transitive property that $Ga=\Omega$ ? (Wondering)

$Ga$ is the orbit of $a$ and you said that from the transitive property it follows that there is only one orbit.
If there is only one orbit all elements of $\Omega$ (commonly referred to as points) must be in that orbit. (Nerd)

It is an action. In my notes there is the following:

$G\rightarrow G$ conjugation
$g*a=a^{g^{-1}}=g^{-1}ag$

Huh?

A (left) action of $g$ on $x$ is usually written as $g\cdot a$, which means that we apply the permutation $g$ to $a$.
It is actually a function application, so we might also write it as $g(a)$.
Often, it is abbreviated to just $ga$.

The $*$ you mention is a conjugation, which is a possible type of action, but it's not a generic action.
So I don't think we should use it in this particular problem to avoid confusion, unless it's for an example.

Could you explain to me further why it is the first one? (Wondering)

Let's get back to the example we came up with.
That is, $G=S_3$ and $\Omega=\{1,2,3\}$.
Suppose we pick $g=(1)$, then $\Omega^{(1)}=\Omega=\{1,2,3\}$.
Which elements of $(a,b)\in\Omega\times\Omega$ have $(a,b) \cdot g= (a,b)$?
How many are there? Is it $|\Omega|\cdot|\Omega|$ or $|\Omega|+|\Omega|$? (Wondering)

In my notes there is the following formula: $$G \overset{\text{ action }}{\longrightarrow } \Omega \\ X(g)=|\{\omega \in \Omega \mid g*\omega =\omega\}|$$

So, can we not apply the right action of the problem statement? (Wondering)

That's not a right action.
A right action is if we multiply by $g$ on the right.
If $G$ is a right action, we would have for instance $X(g)=|\{\omega \in \Omega \mid \omega * g =\omega\}|$.
And then it's still confusing whether $*$ represents a generic action, or if it represents the specific conjugation action.

 

[mathmari]

$Ga$ is the orbit of $a$ and you said that from the transitive property it follows that there is only one orbit.
If there is only one orbit all elements of $\Omega$ (commonly referred to as points) must be in that orbit. (Nerd)

Ah ok... I see... (Nerd)

A (left) action of $g$ on $x$ is usually written as $g\cdot a$, which means that we apply the permutation $g$ to $a$.
It is actually a function application, so we might also write it as $g(a)$.
Often, it is abbreviated to just $ga$.

The $*$ you mention is a conjugation, which is a possible type of action, but it's not a generic action.
So I don't think we should use it in this particular problem to avoid confusion, unless it's for an example.

$X(g)$ of Burnside's formula is the permutation character, it is equal to the number of elements that $g$ leaves unchanged and the formula that I found in my noyes is $X(g)=|\{\omega \in \Omega \mid g*\omega =\omega\}|$.

So, should that formula be $X(g)=|\{\omega \in \Omega \mid g\cdot \omega =\omega\}|$ ? (Wondering)

Let's get back to the example we came up with.
That is, $G=S_3$ and $\Omega=\{1,2,3\}$.
Suppose we pick $g=(1)$, then $\Omega^{(1)}=\Omega=\{1,2,3\}$.
Which elements of $(a,b)\in\Omega\times\Omega$ have $(a,b) \cdot g= (a,b)$?
How many are there? Is it $|\Omega|\cdot|\Omega|$ or $|\Omega|+|\Omega|$? (Wondering)

For each $a$ there are three possible values for $b$.
Therefore, it is $|\Omega|\cdot|\Omega|$, right? (Wondering)

 

[I like Serena]

$X(g)$ of Burnside's formula is the permutation character, it is equal to the number of elements that $g$ leaves unchanged and the formula that I found in my noyes is $X(g)=|\{\omega \in \Omega \mid g*\omega =\omega\}|$.

So, should that formula be $X(g)=|\{\omega \in \Omega \mid g\cdot \omega =\omega\}|$ ? (Wondering)

For each $a$ there are three possible values for $b$.
Therefore, it is $|\Omega|\cdot|\Omega|$, right? (Wondering)

Yes and yes. (Nod)

 

[mathmari]

Yes and yes. (Nod)

So, we have the following: From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}X(g)=\frac{|Ga|}{|G|}\sum_{g\in G_a}X(g)=\frac{|\Omega|}{|G|}\sum_{g\in G_a}X(g)$$

where $X(g)=|\{\omega \in \Omega \mid g\cdot \omega=\omega\}|$. From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}\tilde{X}(g)$$ where $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid g\cdot (a,b)=(a,b)\}|$.

So, we cannot apply here the operation of the problem statement, can we? (Wondering)

 

[I like Serena]

So, we have the following: From Burnside's formula we have that the number of orbits of $G_a$ on $\Omega$ is equal to $$\frac{1}{|G_a|}\sum_{g\in G_a}X(g)=\frac{|Ga|}{|G|}\sum_{g\in G_a}X(g)=\frac{|\Omega|}{|G|}\sum_{g\in G_a}X(g)$$

where $X(g)=|\{\omega \in \Omega \mid g\cdot \omega=\omega\}|$. From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}\tilde{X}(g)$$ where $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid g\cdot (a,b)=(a,b)\}|$.

Correct.
Additionally, I believe we found that:
$$\tilde X(g) = |(\Omega \times \Omega)^g| = |\Omega^g| \cdot |\Omega^g| = X(g)^2$$

Now we might fill in the example to see how it works out. (Thinking)

 

[mathmari]

Additionally, I believe we found that:
$$\tilde X(g) = |(\Omega \times \Omega)^g| = |\Omega^g| \cdot |\Omega^g| = X(g)^2$$

But when we found that we applied the operation of the problem statement...

From Burnside's formula we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $$\frac{1}{|G|}\sum_{g\in G}\tilde{X}(g)$$ where $\tilde{X}(g)=|\{(a,b) \in \Omega\times\Omega \mid (a,b)\cdot g=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid (a\cdot g, b\cdot g)=(a,b)\}|=|\{(a,b) \in \Omega\times\Omega \mid a\cdot g=a \text{ and } b\cdot g=b\}|$
Is this equal to $|\{a\in \Omega \mid a\cdot g=a\}|\cdot |\{b \in \Omega \mid b\cdot g=b\}|$

(Thinking)

 

[I like Serena]

But when we found that we applied the operation of the problem statement... (Thinking)

We're using how the problem statement defined the operation $\cdot$ on $\Omega\times\Omega$.
It defined that $(a,b)\cdot g = (a\cdot g, b\cdot g)$, so now we can use it.
Btw, implicitly it defined both the action on $\Omega$ and the action on $\Omega\times\Omega$ to be right actions. (Thinking)

If we pick a specific $g$, we can find the set $\Omega^g$.
Each element $a$ in it has $a\cdot g=a$.
As you said, the set $(\Omega \times \Omega)^g$ has exactly those elements $(a,b)$ that has $a\in\Omega^g$ and $b\in\Omega^g$.
Since we can combine each $a\in\Omega^g$ with any of the $b\in\Omega^g$, there are in total $|\Omega^g|\cdot|\Omega^g|$ elements for a specific $g$. (Thinking)

 

[mathmari]

We're using how the problem statement defined the operation $\cdot$ on $\Omega\times\Omega$.
It defined that $(a,b)\cdot g = (a\cdot g, b\cdot g)$, so now we can use it.
Btw, implicitly it defined both the action on $\Omega$ and the action on $\Omega\times\Omega$ to be right actions. (Thinking)

If we pick a specific $g$, we can find the set $\Omega^g$.
Each element $a$ in it has $a\cdot g=a$.
As you said, the set $(\Omega \times \Omega)^g$ has exactly those elements $(a,b)$ that has $a\in\Omega^g$ and $b\in\Omega^g$.
Since we can combine each $a\in\Omega^g$ with any of the $b\in\Omega^g$, there are in total $|\Omega^g|\cdot|\Omega^g|$ elements for a specific $g$. (Thinking)

Ah OK.. Now we have that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to $\frac{1}{|G|}\sum_{g\in G}X(g)^2$ and the number of orbits of $G_a$ on $\Omega$ is $\frac{|\Omega|}{|G|}\sum_{g\in G_a}X(g)$.

How can we show that these two are equal? (Wondering)

 

[mathmari]

We have that $G_a=\{g\in G\mid a\cdot g=a\}$ and $X(g)=|\{\omega \in \Omega \mid \omega \cdot g=\omega\}|$, right? (Wondering)

For $g\in G_a$ we have that $X(g)$ is equal to $1$ since it contains only the elment $a$, or not? (Wondering)

 

[I like Serena]

We have that $G_a=\{g\in G\mid a\cdot g=a\}$ and $X(g)=|\{\omega \in \Omega \mid \omega \cdot g=\omega\}|$, right? (Wondering)

For $g\in G_a$ we have that $X(g)$ is equal to $1$ since it contains only the elment $a$, or not? (Wondering)

Suppose we pick $G=S_3$ and $a=1$.
Then $G_a = G_1 = \{ (1), (23) \}$.

For $g=(1) \in G_a$, we get $X(g) = |\Omega^{(1)}| = |\Omega| = |\{1,2,3\}| = 3$. (Thinking)

 

[mathmari]

Suppose we pick $G=S_3$ and $a=1$.
Then $G_a = G_1 = \{ (1), (23) \}$.

For $g=(1) \in G_a$, we get $X(g) = |\Omega^{(1)}| = |\Omega| = |\{1,2,3\}| = 3$. (Thinking)

Ah ok... I see... (Thinking) Could you maybe give me a hint how we could show the equality of the two number of orbits? (Wondering)
I got stuck right now...

 

[I like Serena]

Could you maybe give me a hint how we could show the equality of the two number of orbits? (Wondering)
I got stuck right now...

Have you tried to apply the Orbit-stabilizer theorem and Burnside's lemma? (Wondering)
To be honest, I'm not quite getting the requested result yet, but it's a good exercise.

Best I can do, is to suggest to work out a couple of examples.
Such as $G=S_3$, $G=A_4$, and $G=S_4$.
That will give some insight in what's happening. (Thinking)

Beyond that, I'm stuck right now... (Worried)

 

[mathmari]

Could we do the following?

Let $(a,b)G=\{(a,b)g: g\in G\}=\{(a\cdot g, b\cdot g):g\in G\}$, $(a,b)\in \Omega\times\Omega$ be an orbit of $G$ on $\Omega\times\Omega$.
We have that $(a,b)\in (a,b)G$ if $a=a\cdot g$ and $b=b\cdot g$, i.e., $g\in G_a=\{g\in G: g\cdot a=a\}$.
Also $(a,c)\in (a,b)G$ if $a=a\cdot g$ and $c=b\cdot g$, i.e., $c\in bG_a$, where $bG_a$ is an orbit of $G_a$ on $\Omega$.
We have that $b\in bG_a$.
That means that $(a,b)$ and $(a,c)$ belong to the same orbit of $G$ on $\Omega\times\Omega$ iff $b$ and $c$ belong to the same orbit of $G_a$ on $\Omega$.
Therefore, the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$.

Is this correct? (Wondering)

But now it depends on $a$.

What could we do? (Wondering)

 

[I like Serena]

Could we do the following?

Let $(a,b)G=\{(a,b)g: g\in G\}=\{(a\cdot g, b\cdot g):g\in G\}$, $(a,b)\in \Omega\times\Omega$ be an orbit of $G$ on $\Omega\times\Omega$.
We have that $(a,b)\in (a,b)G$ if $a=a\cdot g$ and $b=b\cdot g$, i.e., $g\in G_a=\{g\in G: g\cdot a=a\}$.
Also $(a,c)\in (a,b)G$ if $a=a\cdot g$ and $c=b\cdot g$, i.e., $c\in bG_a$, where $bG_a$ is an orbit of $G_a$ on $\Omega$.
We have that $b\in bG_a$.
That means that $(a,b)$ and $(a,c)$ belong to the same orbit of $G$ on $\Omega\times\Omega$ iff $b$ and $c$ belong to the same orbit of $G_a$ on $\Omega$.
Therefore, the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$.

Is this correct? (Wondering)

But now it depends on $a$.

What could we do? (Wondering)

Let's pick an example.
Say $G=S_3$, $\Omega=\{1,2,3\}$, and $a=1$.
Then $G_a = G_1 = \{(1), (23)\}$, which are the elements that leave $1$ unchanged.
And $aG = 1G = \Omega=\{1,2,3\}$. (Nerd)Then the sets of orbits are:
$$\Omega\times\Omega/G = \{1,2,3\}\times\{1,2,3\}/S_3 = \Big\{ \{ (1,1), (2,2), (3,3) \}, \quad\{ (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) \} \Big\}$$
and:
$$\Omega/G_1 = \{1,2,3\}/\{(1), (23)\} = \Big\{ \{ 1 \}, \quad\{ 2, 3 \} \Big\}$$
So indeed, in both cases the number of orbits is $2$.
However, I don't see how considering that $(1,2)$ and $(1,3)$ are in the same orbit helps us if we only consider the permutations that leave $1$ unchanged. We don't even take into account that in this case for instance $(2,1)$ is in the same orbit. (Worried)Alternatively, Burnside tells us that:
$$|\Omega\times\Omega/G| = \frac{1}{|S_3|}\sum_{g\in S_3}|(\Omega\times\Omega)^g|
= \frac 16\left(|(\Omega\times\Omega)^{(1)}| + 3 |(\Omega\times\Omega)^{(23)}| + 2 |(\Omega\times\Omega)^{(123)}|\right)
= \frac 16\left(|\Omega\times\Omega| + 3 | \{ (1,1) \} | + 2 \cdot 0\right)
=2$$
And:
$$|\Omega/G_a| = \frac{1}{|G_1|}\sum_{h\in G_1}|\Omega^h| = \frac{1}{|\{(1), (23)\}|}\sum_{h\in \{(1), (23)\}}|\Omega^h|
= \frac 12\left(|\Omega^{(1)}| + |\Omega^{(23)}| \right)
= \frac 12\left(|\Omega| + | \{ 1 \} |\right)
=2$$

It seems we might have a match between the terms in both cases... (Thinking)

 

[mathmari]

Let's pick an example.
Say $G=S_3$, $\Omega=\{1,2,3\}$, and $a=1$.
Then $G_a = G_1 = \{(1), (23)\}$, which are the elements that leave $1$ unchanged.
And $aG = 1G = \Omega=\{1,2,3\}$. (Nerd)Then the sets of orbits are:
$$\Omega\times\Omega/G = \{1,2,3\}\times\{1,2,3\}/S_3 = \Big\{ \{ (1,1), (2,2), (3,3) \}, \quad\{ (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) \} \Big\}$$
and:
$$\Omega/G_1 = \{1,2,3\}/\{(1), (23)\} = \Big\{ \{ 1 \}, \quad\{ 2, 3 \} \Big\}$$
So indeed, in both cases the number of orbits is $2$.
However, I don't see how considering that $(1,2)$ and $(1,3)$ are in the same orbit helps us if we only consider the permutations that leave $1$ unchanged. We don't even take into account that in this case for instance $(2,1)$ is in the same orbit. (Worried)

From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. for $a\in \Omega$ and $\forall \omega_1\in \Omega, \exists g\in G$ such that $ga=\omega_1$.
So, $\forall \omega_1, \omega_2\in \Omega$ we have that $(\omega_1, \omega_2)=(ga, \omega_2)=(ga, gg^{-1}\omega_2)=g(a, g^{-1}\omega_2)$.

Is this correct? (Wondering)

 

[I like Serena]

From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$, i.e. for $a\in \Omega$ and $\forall \omega_1\in \Omega, \exists g\in G$ such that $ga=\omega_1$.
So, $\forall \omega_1, \omega_2\in \Omega$ we have that $(\omega_1, \omega_2)=(ga, \omega_2)=(ga, gg^{-1}\omega_2)=g(a, g^{-1}\omega_2)$.

Is this correct? (Wondering)

Since we have an (unusual) right action, let's make that:

$\forall \omega_1, \omega_2\in \Omega\, \exists g \in G$ such that $(\omega_1, \omega_2)=(ag, \omega_2)=(ag, \omega_2g^{-1}g)=(a, \omega_2g^{-1})g$