Number of double cosets $Hbackslash G/K$ where $G=S_{10}$ and $H=K=S_9$ is a subgroup permuting the integers...
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Past year paper question
Suppose $pi$ is the group action of $G$ on a set $X$.
Fix $x_0 in X$ and let $K$ be the stabilizer of $x_0$ in $G$.
We assume that $X=Gx_0$ is a single orbit of $G$.
Let $H$ be a subgroup of $G$.
For $xin X$, let $O(x)=Hx$ denote the $H$-orbit of $x in X$.
Let $mathcal{O}_H$ be the set of $H$-orbits on $X$.
(i) Show that $f: Hbackslash G/K to mathcal{O}_H$ given by $f(HgK)=O(gx_0)$ is a well-defined map, and a bijection. (I have done this part).
(ii) Compute the number of double cosets $Hbackslash G/K$ where $G=S_{10}$ and $H=K=S_9$ is a subgroup permuting the integers ${1,2,...,9}$.
Attempt
I am quite confused as to what the answer is. If we let $X={1,2,...,10}$ and $x_0=10$, then we have that $X$ is a single $G$-orbit. And we have $K$ stabilizes $x_0$. Then the $H$-orbit of $1$ is ${1,2,...,9}$. And if we pick say $g=(1;10) in G$, then the orbit of $Hg$ of $1$ is $10$. So there are $2$ $H$-orbits and hence $2$ double cosets...?
Can I check if my reasoning is correct?
abstract-algebra group-theory
asked Nov 23 at 4:29
eatfood
1827
add a comment |
up vote
1
down vote
favorite
Past year paper question
Suppose $pi$ is the group action of $G$ on a set $X$.
Fix $x_0 in X$ and let $K$ be the stabilizer of $x_0$ in $G$.
We assume that $X=Gx_0$ is a single orbit of $G$.
Let $H$ be a subgroup of $G$.
For $xin X$, let $O(x)=Hx$ denote the $H$-orbit of $x in X$.
Let $mathcal{O}_H$ be the set of $H$-orbits on $X$.
(i) Show that $f: Hbackslash G/K to mathcal{O}_H$ given by $f(HgK)=O(gx_0)$ is a well-defined map, and a bijection. (I have done this part).
(ii) Compute the number of double cosets $Hbackslash G/K$ where $G=S_{10}$ and $H=K=S_9$ is a subgroup permuting the integers ${1,2,...,9}$.
Attempt
I am quite confused as to what the answer is. If we let $X={1,2,...,10}$ and $x_0=10$, then we have that $X$ is a single $G$-orbit. And we have $K$ stabilizes $x_0$. Then the $H$-orbit of $1$ is ${1,2,...,9}$. And if we pick say $g=(1;10) in G$, then the orbit of $Hg$ of $1$ is $10$. So there are $2$ $H$-orbits and hence $2$ double cosets...?
Can I check if my reasoning is correct?
abstract-algebra group-theory
asked Nov 23 at 4:29
eatfood
1827
1
Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Past year paper question
Suppose $pi$ is the group action of $G$ on a set $X$.
Fix $x_0 in X$ and let $K$ be the stabilizer of $x_0$ in $G$.
We assume that $X=Gx_0$ is a single orbit of $G$.
Let $H$ be a subgroup of $G$.
For $xin X$, let $O(x)=Hx$ denote the $H$-orbit of $x in X$.
Let $mathcal{O}_H$ be the set of $H$-orbits on $X$.
(i) Show that $f: Hbackslash G/K to mathcal{O}_H$ given by $f(HgK)=O(gx_0)$ is a well-defined map, and a bijection. (I have done this part).
(ii) Compute the number of double cosets $Hbackslash G/K$ where $G=S_{10}$ and $H=K=S_9$ is a subgroup permuting the integers ${1,2,...,9}$.
Attempt
I am quite confused as to what the answer is. If we let $X={1,2,...,10}$ and $x_0=10$, then we have that $X$ is a single $G$-orbit. And we have $K$ stabilizes $x_0$. Then the $H$-orbit of $1$ is ${1,2,...,9}$. And if we pick say $g=(1;10) in G$, then the orbit of $Hg$ of $1$ is $10$. So there are $2$ $H$-orbits and hence $2$ double cosets...?
Can I check if my reasoning is correct?
abstract-algebra group-theory
asked Nov 23 at 4:29
eatfood
1827
Past year paper question
Suppose $pi$ is the group action of $G$ on a set $X$.
Fix $x_0 in X$ and let $K$ be the stabilizer of $x_0$ in $G$.
We assume that $X=Gx_0$ is a single orbit of $G$.
Let $H$ be a subgroup of $G$.
For $xin X$, let $O(x)=Hx$ denote the $H$-orbit of $x in X$.
Let $mathcal{O}_H$ be the set of $H$-orbits on $X$.
(i) Show that $f: Hbackslash G/K to mathcal{O}_H$ given by $f(HgK)=O(gx_0)$ is a well-defined map, and a bijection. (I have done this part).
(ii) Compute the number of double cosets $Hbackslash G/K$ where $G=S_{10}$ and $H=K=S_9$ is a subgroup permuting the integers ${1,2,...,9}$.
Attempt
I am quite confused as to what the answer is. If we let $X={1,2,...,10}$ and $x_0=10$, then we have that $X$ is a single $G$-orbit. And we have $K$ stabilizes $x_0$. Then the $H$-orbit of $1$ is ${1,2,...,9}$. And if we pick say $g=(1;10) in G$, then the orbit of $Hg$ of $1$ is $10$. So there are $2$ $H$-orbits and hence $2$ double cosets...?
Can I check if my reasoning is correct?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 23 at 4:29
eatfood
1827
asked Nov 23 at 4:29
eatfood
1827
asked Nov 23 at 4:29
eatfood
1827
asked Nov 23 at 4:29
eatfood
1827
asked Nov 23 at 4:29
eatfood
1827
1827
1
Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06
add a comment |
1
Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06
1
1
Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06
add a comment |
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Yes, except that I don't know why you consider orbits of $g1$ instead of directly of elements of $X$. -- Following (i) , you are supposed to count $mathcal O_H$, and the two orbits are ${10}$ and ${1,ldots,9}$
– Hagen von Eitzen
Nov 23 at 6:09
Yes I think that was a mistake, I should have counted the distinct orbits of $10$ cause thats the $x_0$ I chose
– eatfood
Nov 23 at 11:06