I have posted an identical question in MSE few days ago, but maybe this site is a better adress to discuss this problem:

Let $G$ be a finite group and $K, H \leq G$ two subgroups. Then the right quotients $G/H $ and $G/K$ become naturally left $G$-sets via $\rho: G \times G/H \to G/H, (g, fH) \mapsto gfH$ (we will work only with left $G$-sets and omit the word left).
The product $G/H \times G/K$ (or more generally any finite product $\prod_i G/H_i$) can be naturally endowed with structure of a $G$-set by diagonal $G$-action $(g,(aH, bK)) \mapsto (gaH, gbK)$.

In Tammo tom Dieck's book 'Transformation Groups' it is stated (Prop 1., page 19) that there are bijections between the following sets:

1. $G$-orbits of $G/H \times G/K$.
2. $H$-orbits of $G/K$ with left $H$-action.
3. Double cosets $HgK$, g \in G$.

In the proof the $G$-orbit $G \cdot (eH, gK)$ of $G/H \times G/K$ corresponds to the $H$-orbit $H \cdot gK$ of $G/K$ and to the double coset $HgK$.
This implies that we can decompose $G/H \times G/K$ as set in $G$-orbits as follows

$$ G/H \times G/K = \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK)$$

where $e \in G$ is the identity element and the disjoint union runs over all representatives $\overline{g}$ of double cosets $HgK \in H \backslash G / K$.

But we can say even more. Since $G$ acts on any orbit $G \cdot (eH, gK)$ transitively and the stabilizer of $(eH,gK)$ is $H \cap gKg^{-1} = H \cap K^{g^{-1}}$, the orbit $G \cdot (eH, gK)$ is isomorphic to $G/(H \cap K^{g^{-1}})$ and therefore the decomposition in $G$-orbits of $G/H \times G/K$ is isomorphic to

$$ \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK) G/(H \cap K^{g^{-1}}).$$

Question: Is there a natural similar decomposition of the fiber product $F:= G/H \times_{G/B} G/K$, which sits in diagram

$$ \require{AMScd} \begin{CD} G/H \times_{G/B} G/K @>{} >> G/K \\ @VVV @VVR_y: \ (gK \mapsto gyB)V \\ G/H @>{R_x: \ (gH \mapsto gxB)}>> G/B \end{CD} $$

where the $x, y \in G$ satisfying $x^{-1}Hx \subset B$, $y^{-1}Ky \subset B$ define the $G$-maps $R_x: G/H \to G/B$, $gH \to gxB$ and $R_y: G/K \to G/B$, $gK \to gyB$.

(Note that up to isomorphism every $G$-map $f: G/A \to G/B$ is given by a $R_x: G/A \to G/B$, $gA \mapsto gxB$ where $x$ must satisfy $x^{-1}Ax \subset B$. (Note that $x$ and $x' \in G$ define the same $R_x$ iff $x'x^{-1} \in B$.))

What we know: set theoretically we have $F= G/H \times_{G/B} G/K \subset G/H \times G/K$ and $F$ consists of pairs $(aH, bK)$ with $axB= byB$. Since $R_x$ and $R_y$ are $G$-maps, if a $(aH,bK) \in F$, then the complete $G$-orbit $G \cdot (aH,bK)$ is contained in $F$. Therefore we can assume $(aH,bK)=(eH, gK) \in F$. Then $xB= gyB$ implies $x^{-1}gy \in B$. That appears to be our neccessary and sufficent condition.

So the claim is that $F = G/H \times_{G/B} G/K$ decomposes in $G$-orbits as

$$F = G/H \times_{G/B} G/K = \dot\bigcup_{HgK \in H \backslash G / K, \ x^{-1}gy \in B} G/(H \cap K^{g^{-1}}).$$

Problem: the parametrizing set $HgK \in H \backslash G / K, \ x^{-1}gy \in B$ looks not really nice, especially I would like to get rid of the $x^{-1}gy \in B$ condition. Can the fiber product $F = G/H \times_{G/B} G/K$ be decomposed in orbits over a more amenable parametrizing set without this $x^{-1}gy \in B$ condition?

Maybe there exist a parametrization over double coset $H^x \backslash B / K^y$? Can the later be somehow related to the double coset $\{HgK \in H \backslash G / K \ \vert \ x^{-1}gy \in B \}$?

I have posted an identical question in MSE few days ago, but maybe this site is a better adress to discuss this problem:

Let $G$ be a finite group and $K, H \leq G$ two subgroups. Then the right quotients $G/H $ and $G/K$ become naturally left $G$-sets via $\rho: G \times G/H \to G/H, (g, fH) \mapsto gfH$ (we will work only with left $G$-sets and omit the word left).
The product $G/H \times G/K$ (or more generally any finite product $\prod_i G/H_i$) can be naturally endowed with structure of a $G$-set by diagonal $G$-action $(g,(aH, bK)) \mapsto (gaH, gbK)$.

In Tammo tom Dieck's book 'Transformation Groups' it is stated (Prop 1., page 19) that there are bijections between the following sets:

1. $G$-orbits of $G/H \times G/K$.
2. $H$-orbits of $G/K$ with left $H$-action.
3. Double cosets $HgK$, g \in G$.

In the proof the $G$-orbit $G \cdot (eH, gK)$ of $G/H \times G/K$ corresponds to the $H$-orbit $H \cdot gK$ of $G/K$ and to the double coset $HgK$.
This implies that we can decompose $G/H \times G/K$ as set in $G$-orbits as follows

$$ G/H \times G/K = \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK)$$

where $e \in G$ is the identity element and the disjoint union runs over all representatives $\overline{g}$ of double cosets $HgK \in H \backslash G / K$.

But we can say even more. Since $G$ acts on any orbit $G \cdot (eH, gK)$ transitively and the stabilizer of $(eH,gK)$ is $H \cap gKg^{-1} = H \cap K^{g^{-1}}$, the orbit $G \cdot (eH, gK)$ is isomorphic to $G/(H \cap K^{g^{-1}})$ and therefore the decomposition in $G$-orbits of $G/H \times G/K$ is isomorphic to

$$ \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK) G/(H \cap K^{g^{-1}}).$$

Question: Is there a natural similar decomposition in $G$-orbits of the fiber product $F:= G/H \times_{G/B} G/K$, which sits in diagram

$$ \require{AMScd} \begin{CD} G/H \times_{G/B} G/K @>{} >> G/K \\ @VVV @VVR_y: \ (gK \mapsto gyB)V \\ G/H @>{R_x: \ (gH \mapsto gxB)}>> G/B \end{CD} $$

where the $x, y \in G$ satisfying $x^{-1}Hx \subset B$, $y^{-1}Ky \subset B$ define the $G$-maps $R_x: G/H \to G/B$, $gH \to gxB$ and $R_y: G/K \to G/B$, $gK \to gyB$.

(Note that up to isomorphism every $G$-map $f: G/A \to G/B$ is given by a $R_x: G/A \to G/B$, $gA \mapsto gxB$ where $x$ must satisfy $x^{-1}Ax \subset B$. (Note that $x$ and $x' \in G$ define the same $R_x$ iff $x'x^{-1} \in B$.))

What we know: set theoretically we have $F= G/H \times_{G/B} G/K \subset G/H \times G/K$ and $F$ consists of pairs $(aH, bK)$ with $axB= byB$. Since $R_x$ and $R_y$ are $G$-maps, if a $(aH,bK) \in F$, then the complete $G$-orbit $G \cdot (aH,bK)$ is contained in $F$. Therefore we can assume $(aH,bK)=(eH, gK) \in F$. Then $xB= gyB$ implies $x^{-1}gy \in B$. That appears to be our neccessary and sufficent condition.

So the claim is that $F = G/H \times_{G/B} G/K$ decomposes in $G$-orbits as

$$F = G/H \times_{G/B} G/K = \dot\bigcup_{HgK \in H \backslash G / K, \ x^{-1}gy \in B} G/(H \cap K^{g^{-1}}).$$

Problem: the parametrizing set $HgK \in H \backslash G / K, \ x^{-1}gy \in B$ looks not really nice, especially I would like to get rid of the $x^{-1}gy \in B$ condition. Can the fiber product $F = G/H \times_{G/B} G/K$ be decomposed in orbits over a more amenable parametrizing set without this $x^{-1}gy \in B$ condition?

Maybe there exist a parametrization over double coset $H^x \backslash B / K^y$? Can the later be somehow related to the double coset $\{HgK \in H \backslash G / K \ \vert \ x^{-1}gy \in B \}$?

I have posted an identical question in MSE few days ago, but maybe this site is a better adress to discuss this problem:

Let $G$ be a finite group and $K, H \leq G$ two subgroups. Then the right quotients $G/H $ and $G/K$ become naturally left $G$-sets via $\rho: G \times G/H \to G/H, (g, fH) \mapsto gfH$ (we will work only with left $G$-sets and omit the word left.)
The product $G/H \times G/K$ (or more ginerally any finite product $\prod_i G/H_i$) can be naturally endowed with structure of a $G$-set by diagonal $G$-action $(g,(aH, bK)) \mapsto (gaH, gbK)$.

In Tammo tom Dieck's book 'Transformation Groups' is stated (Prop 1., page 19) that there are bijections between following sets

(i) $G$-orbits of $G/H \times G/K$
(ii) $H$-orbits of $G/K$ with left $H$-action
(iii) Double cosets $HgK, g \in G$

In the proof the $G$-orbit $G \cdot (eH, gK)$ of $G/H \times G/K$ corresponds to the $H$-orbit $H \cdot gK$ of $G/K$ and to the double coset $HgK$.
This implies that we can decompose $G/H \times G/K$ as set in $G$-orbits as follows

$$ G/H \times G/K = \dot\bigcup_{HgK \in H \setminus G / K} G \cdot (eH, gK)$$

where $e \in G$ is the identity element and the disjoint union runs over all representators $\overline{g}$ of double cosets $HgK \in H \setminus G / K$.

But we can say even more. Since $G$ acts on any orbit $G \cdot (eH, gK)$ transitively and the stabilizer of $(eH,gK)$ is $H \cap gKg^{-1} = H \cap K^{g^{-1}}$, the orbit $G \cdot (eH, gK)$ is isomorphic to $G/(H \cap K^{g^{-1}})$ and therefore the decomposition in $G$-orbits of $G/H \times G/K$ is isomorphic to

$$ \dot\bigcup_{HgK \in H \setminus G / K} G \cdot (eH, gK) G/(H \cap K^{g^{-1}})$$

Question: Is there a natural similar decomposition of the fiber product $F:= G/H \times_{G/B} G/K$ possible, which sits in diagram

$$ \require{AMScd} \begin{CD} G/H \times_{G/B} G/K @>{} >> G/K \\ @VVV @VVR_y: \ (gK \mapsto gyB)V \\ G/H @>{R_x: \ (gH \mapsto gxB)}>> G/B \end{CD} $$

where the $x, y \in G$ satisfying $x^{-1}Hx \subset B, y^{-1}Ky \subset B$ define the $G$-maps $R_x: G/H \to G/B, gH \to gxB$ and $R_y: G/K \to G/B, gK \to gyB$.

(Note that up to isomorphism every $G$-map $f: G/A \to G/B$ is given by a $R_x: G/A \to G/B, gA \mapsto gxB$ where $x$ must satisfy $x^{-1}Ax \subset B$. (note that $x$ and $x' \in G$ define same $R_x$ iff $x'x^{-1} \in B$.)

What we know: set theoretically we have $F= G/H \times_{G/B} G/K \subset G/H \times G/K$ and $F$ consists of pairs $(aH, bK)$ with $axB= byB$. Since $R_x$ and $R_y$ are $G$-maps, if a $(aH,bK) \in F$, then the complete $G$-orbit $G \cdot (aH,bK)$ is contained in $F$. Therefore we can assume $(aH,bK)=(eH, gK) \in F$. Then $xB= gyB$ implies $x^{-1}gy \in B$. That appears to be our neccessary and sufficent condition.

So the claim is that $F = G/H \times_{G/B} G/K$ decomposes in $G$-orbits as

$$F = G/H \times_{G/B} G/K = \dot\bigcup_{HgK \in H \setminus G / K, \ x^{-1}gy \in B} G/(H \cap K^{g^{-1}})$$

Problem: the parametrizing set $HgK \in H \setminus G / K, \ x^{-1}gy \in B$ looks not really nice, especially I would like to get rid of the $x^{-1}gy \in B$ condition. Can the fiber product $F = G/H \times_{G/B} G/K$ be decomposed in orbits over a more amenable parametrizing set without this $x^{-1}gy \in B$ condition?

Maybe there exist a parametrization over double coset $H^x \setminus B / K^y$? Can the later be somehow related to the double coset $\{HgK \in H \setminus G / K \ \vert \ x^{-1}gy \in B \}$?

I have posted an identical question in MSE few days ago, but maybe this site is a better adress to discuss this problem:

Let $G$ be a finite group and $K, H \leq G$ two subgroups. Then the right quotients $G/H $ and $G/K$ become naturally left $G$-sets via $\rho: G \times G/H \to G/H, (g, fH) \mapsto gfH$ (we will work only with left $G$-sets and omit the word left).
The product $G/H \times G/K$ (or more generally any finite product $\prod_i G/H_i$) can be naturally endowed with structure of a $G$-set by diagonal $G$-action $(g,(aH, bK)) \mapsto (gaH, gbK)$.

In Tammo tom Dieck's book 'Transformation Groups' it is stated (Prop 1., page 19) that there are bijections between the following sets:

1. $G$-orbits of $G/H \times G/K$.
2. $H$-orbits of $G/K$ with left $H$-action.
3. Double cosets $HgK$, g \in G$.

In the proof the $G$-orbit $G \cdot (eH, gK)$ of $G/H \times G/K$ corresponds to the $H$-orbit $H \cdot gK$ of $G/K$ and to the double coset $HgK$.
This implies that we can decompose $G/H \times G/K$ as set in $G$-orbits as follows

$$ G/H \times G/K = \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK)$$

where $e \in G$ is the identity element and the disjoint union runs over all representatives $\overline{g}$ of double cosets $HgK \in H \backslash G / K$.

But we can say even more. Since $G$ acts on any orbit $G \cdot (eH, gK)$ transitively and the stabilizer of $(eH,gK)$ is $H \cap gKg^{-1} = H \cap K^{g^{-1}}$, the orbit $G \cdot (eH, gK)$ is isomorphic to $G/(H \cap K^{g^{-1}})$ and therefore the decomposition in $G$-orbits of $G/H \times G/K$ is isomorphic to

$$ \dot\bigcup_{HgK \in H \backslash G / K} G \cdot (eH, gK) G/(H \cap K^{g^{-1}}).$$

Question: Is there a natural similar decomposition of the fiber product $F:= G/H \times_{G/B} G/K$, which sits in diagram

$$ \require{AMScd} \begin{CD} G/H \times_{G/B} G/K @>{} >> G/K \\ @VVV @VVR_y: \ (gK \mapsto gyB)V \\ G/H @>{R_x: \ (gH \mapsto gxB)}>> G/B \end{CD} $$

where the $x, y \in G$ satisfying $x^{-1}Hx \subset B$, $y^{-1}Ky \subset B$ define the $G$-maps $R_x: G/H \to G/B$, $gH \to gxB$ and $R_y: G/K \to G/B$, $gK \to gyB$.

(Note that up to isomorphism every $G$-map $f: G/A \to G/B$ is given by a $R_x: G/A \to G/B$, $gA \mapsto gxB$ where $x$ must satisfy $x^{-1}Ax \subset B$. (Note that $x$ and $x' \in G$ define the same $R_x$ iff $x'x^{-1} \in B$.))

What we know: set theoretically we have $F= G/H \times_{G/B} G/K \subset G/H \times G/K$ and $F$ consists of pairs $(aH, bK)$ with $axB= byB$. Since $R_x$ and $R_y$ are $G$-maps, if a $(aH,bK) \in F$, then the complete $G$-orbit $G \cdot (aH,bK)$ is contained in $F$. Therefore we can assume $(aH,bK)=(eH, gK) \in F$. Then $xB= gyB$ implies $x^{-1}gy \in B$. That appears to be our neccessary and sufficent condition.

So the claim is that $F = G/H \times_{G/B} G/K$ decomposes in $G$-orbits as

$$F = G/H \times_{G/B} G/K = \dot\bigcup_{HgK \in H \backslash G / K, \ x^{-1}gy \in B} G/(H \cap K^{g^{-1}}).$$

Problem: the parametrizing set $HgK \in H \backslash G / K, \ x^{-1}gy \in B$ looks not really nice, especially I would like to get rid of the $x^{-1}gy \in B$ condition. Can the fiber product $F = G/H \times_{G/B} G/K$ be decomposed in orbits over a more amenable parametrizing set without this $x^{-1}gy \in B$ condition?

Maybe there exist a parametrization over double coset $H^x \backslash B / K^y$? Can the later be somehow related to the double coset $\{HgK \in H \backslash G / K \ \vert \ x^{-1}gy \in B \}$?