MHB Number of Orbits: Showing Equality of Group G and G_a

[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 $(\omega_1, \omega_2)=(ag, \omega_2)=(ag, \omega_2g^{-1}g)=(a, \omega_2g^{-1})g$