Product of complete sets of representatives of the cosets
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Suppose we have the subgroups $H subseteq K subseteq G$ (not necessarily normal), all finite groups.
Denote $mathcal{R}^G_H$ be a complete set of representatives of the cosets of $G$ in $H$, and similarly for the other combinations.
Is it then true that:
$$
mathcal{R}^G_H = mathcal{R}^G_K cdot mathcal{R}^K_H,
$$
At least that the product $mathcal{R}^G_K cdot mathcal{R}^K_H$ (i.e. the pairwise product of every pair of elements from those sets) is a complete set of representatives of the cosets of $G$ in $H$?
I didn't manage to prove it and I'm not sure if it's true in the first place.
abstract-algebra group-theory finite-groups
asked Nov 15 at 19:17
Sigurd
473211
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up vote
0
down vote
favorite
Suppose we have the subgroups $H subseteq K subseteq G$ (not necessarily normal), all finite groups.
Denote $mathcal{R}^G_H$ be a complete set of representatives of the cosets of $G$ in $H$, and similarly for the other combinations.
Is it then true that:
$$
mathcal{R}^G_H = mathcal{R}^G_K cdot mathcal{R}^K_H,
$$
At least that the product $mathcal{R}^G_K cdot mathcal{R}^K_H$ (i.e. the pairwise product of every pair of elements from those sets) is a complete set of representatives of the cosets of $G$ in $H$?
I didn't manage to prove it and I'm not sure if it's true in the first place.
abstract-algebra group-theory finite-groups
asked Nov 15 at 19:17
Sigurd
473211
Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have the subgroups $H subseteq K subseteq G$ (not necessarily normal), all finite groups.
Denote $mathcal{R}^G_H$ be a complete set of representatives of the cosets of $G$ in $H$, and similarly for the other combinations.
Is it then true that:
$$
mathcal{R}^G_H = mathcal{R}^G_K cdot mathcal{R}^K_H,
$$
At least that the product $mathcal{R}^G_K cdot mathcal{R}^K_H$ (i.e. the pairwise product of every pair of elements from those sets) is a complete set of representatives of the cosets of $G$ in $H$?
I didn't manage to prove it and I'm not sure if it's true in the first place.
abstract-algebra group-theory finite-groups
asked Nov 15 at 19:17
Sigurd
473211
Suppose we have the subgroups $H subseteq K subseteq G$ (not necessarily normal), all finite groups.
Denote $mathcal{R}^G_H$ be a complete set of representatives of the cosets of $G$ in $H$, and similarly for the other combinations.
Is it then true that:
$$
mathcal{R}^G_H = mathcal{R}^G_K cdot mathcal{R}^K_H,
$$
At least that the product $mathcal{R}^G_K cdot mathcal{R}^K_H$ (i.e. the pairwise product of every pair of elements from those sets) is a complete set of representatives of the cosets of $G$ in $H$?
I didn't manage to prove it and I'm not sure if it's true in the first place.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Nov 15 at 19:17
Sigurd
473211
asked Nov 15 at 19:17
Sigurd
473211
asked Nov 15 at 19:17
Sigurd
473211
asked Nov 15 at 19:17
Sigurd
473211
asked Nov 15 at 19:17
Sigurd
473211
473211
Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12
add a comment |
Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12
Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12
add a comment |
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Yes it's true that the set of products of elements of a right transversal of $K$ in $H$ and of a right transversal of $H$ in $G$ is a right transversal of $H$ in $G$. The proof is routine. But it is not necessarily true that every right transversal of $H$ in $G$ has that form.
– Derek Holt
Nov 16 at 8:15
Thanks! Indeed, the first fact that you mentioned is what I need. I was struggling with the proof, but I think I understand it now. We need to verify two things: (1) Every element $g in G$ can be written as $g = r_1r_2h$ for $r_1 in R^G_K$ and $r_2 in R^K_H$, and (2) if $g = r_1r_2h = r'_1r'_2h'$, then $r'_1 = r_1$ and $r_2 = r'_2$ (again with $r_1,r'_1 in R^G_K$ and $r_2,r'_2 in R^K_H$). Is this correct?
– Sigurd
Nov 16 at 15:11
Yes that's right. First write $g=r_1k$ with $k in K$ and then write $k=r_2h$ with $h in H$, giving $g=r_1r_2h$.
– Derek Holt
Nov 16 at 17:12