Group Action, Double costs

Let H,K subgroups of G for each x in G define the double cost HK of x

$\displaystyle HxK=\{hxk\mid h\in H ,k\in K\} $

a) Prove that HxK is the union of the left costs $\displaystyle x_1K,x_2K,...,x_nK$ where $\displaystyle \{x_1K,x_2K,...,x_nK\}$ is the orbit containing xK of H acting by left Multiplication on the set of left costs of K

My work

I want to prove that $\displaystyle HxK = \bigcup_i^n x_iK $ it is clear that

$\displaystyle 1xK = xK $ so $\displaystyle \bigcup_i^n x_iK \subseteq HxK$

let $\displaystyle hxK \in HxK $ hx in G so hxK in the union

d) Prove that $\displaystyle \mid HxK \mid = \mid H \mid .\mid H : H\bigcap xKx^{-1}\mid $

my question is about d) I put a) because I think it will help in solving d)

I was thinking about the order of orbit equal to the index group of the stabilizer

$\displaystyle \mid HxK\mid = \mid \bigcup x_iK\mid = \mid O \mid = \mid H: Stab_H(O)\mid $

$\displaystyle Stab_H(O)= \{ h \in H \mid hxK = xK \}$

$\displaystyle Stab_H(O) = \{h \in H \mid x^{-1}hx \in K \Rightarrow h\in xKx^{-1}\} $

$\displaystyle Stab_H(O) = H \bigcap xKx^{-1} $

so we will have

$\displaystyle \mid HxK\mid = \mid H: H \bigcap xKx^{-1}\mid $ which is not like what i want to prove

THanks

Re: Group Action, Double costs

Quote:

Originally Posted by

**Amer** Let H,K subgroups of G for each x in G define the double cost HK of x

$\displaystyle HxK=\{hxk\mid h\in H ,k\in K\} $

a) Prove that HxK is the union of the left costs $\displaystyle x_1K,x_2K,...,x_nK$ where $\displaystyle \{x_1K,x_2K,...,x_nK\}$ is the orbit containing xK of H acting by left Multiplication on the set of left costs of K

My work

I want to prove that $\displaystyle HxK = \bigcup_i^n x_iK $ it is clear that

$\displaystyle 1xK = xK $ so $\displaystyle \bigcup_i^n x_iK \subseteq HxK$

let $\displaystyle hxK \in HxK $ hx in G so hxK in the union

d) Prove that $\displaystyle \mid HxK \mid = \mid H \mid .\mid H : H\bigcap xKx^{-1}\mid $

my question is about d) I put a) because I think it will help in solving d)

I was thinking about the order of orbit equal to the index group of the stabilizer

$\displaystyle \mid HxK\mid = \mid \bigcup x_iK\mid = \mid O \mid = \mid H: Stab_H(O)\mid $

$\displaystyle Stab_H(O)= \{ h \in H \mid hxK = xK \}$

$\displaystyle Stab_H(O) = \{h \in H \mid x^{-1}hx \in K \Rightarrow h\in xKx^{-1}\} $

$\displaystyle Stab_H(O) = H \bigcap xKx^{-1} $

so we will have

$\displaystyle \mid HxK\mid = \mid H: H \bigcap xKx^{-1}\mid $ which is not like what i want to prove

THanks

I think you are going about the last part wrong. You know that $\displaystyle HxK$ should be equal to $\displaystyle \displaystyle \bigsqcup_{gK\in\mathcal{O}_{xK}}|gK|$ where $\displaystyle \mathcal{O}_{xK}$ is orbit of $\displaystyle xK$ under the left $\displaystyle H$-action on $\displaystyle G/K$. But, each $\displaystyle |gK|$ is equal to $\displaystyle |K|$ and so $\displaystyle |HxK|=|K||\mathcal{O}_{xK}|$. But, using the orbit stabilizer theorem, as you indicated, you can show that $\displaystyle |\mathcal{O}_{xK}|=[H:H\cap xKx^{-1}]$ and so $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$.

For more information you can see my blog post here.

EDIT: I noticed that I did the "opposite side" than you wanted, but by symmetry they're the same.

Re: Group Action, Double costs

Quote:

Originally Posted by

**Drexel28** I think you are going about the last part wrong. You know that $\displaystyle HxK$ should be equal to $\displaystyle \displaystyle \bigsqcup_{gK\in\mathcal{O}_{xK}}|gK|$ where $\displaystyle \mathcal{O}_{xK}$ is orbit of $\displaystyle xK$ under the left $\displaystyle H$-action on $\displaystyle G/K$. But, each $\displaystyle |gK|$ is equal to $\displaystyle |K|$ and so $\displaystyle |HxK|=|K||\mathcal{O}_{xK}|$. But, using the orbit stabilizer theorem, as you indicated, you can show that $\displaystyle |\mathcal{O}_{xK}|=[H:H\cap xKx^{-1}]$ and so $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$.

For more information you can see my blog post

here.

EDIT: I noticed that I did the "opposite side" than you wanted, but by symmetry they're the same.

Thanks very much for the direction I did not noticed that but there still a small problem the question show that

$\displaystyle \mid HxK \mid = \mid H \mid . \mid H : H \bigcap xKx^{-1} \mid $

not

$\displaystyle \mid HxK \mid = \mid K \mid . \mid H : H \bigcap xKx^{-1} \mid$

you can check the book David Dammit same source which you gave in your Blog

chapter 4

is it a typo ?

nice blog I like it (Nerd)

Re: Group Action, Double costs

Quote:

Originally Posted by

**Amer** Thanks very much for the direction I did not noticed that but there still a small problem the question show that

$\displaystyle \mid HxK \mid = \mid H \mid . \mid H : H \bigcap xKx^{-1} \mid $

not

$\displaystyle \mid HxK \mid = \mid K \mid . \mid H : H \bigcap xKx^{-1} \mid$

you can check the book David Dammit same source which you gave in your Blog

chapter 4

is it a typo ?

nice blog I like it (Nerd)

What I wrote is correct, namely $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$. Of course, since $\displaystyle |HxK|=|KxH|$ (by the obvious bijection) we also have that $\displaystyle |HxK|=|H|[K:K\cap xHx^{-1}]$.

And, thanks about the blog!

Re: Group Action, Double costs

when i said typo i mean in the book I made an example

let $\displaystyle G=Z_{24}$ , $\displaystyle H = <3> = \{0,3,6,9,...,21\}$ , $\displaystyle K = <4> = \{0,4,8,...,20\}$

$\displaystyle \mid H\mid = 8 , \mid K \mid = 6 $

H acting on xK by left additive

$\displaystyle G/K = \{K, 1+K,2+K,3+k\}$

the orbit of H acting on the left cost will be $\displaystyle \{K, 1+K,2+K,3+k\}$

$\displaystyle \mid HxK \mid = \mid K \mid . \mid H:Stab_H(xK)\mid = 6.4 = 24 $

but if it is $\displaystyle \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ??$ which is not correct since the order should equal to the G order

so there is a typo in book

am I right ?

Re: Group Action, Double costs

Quote:

Originally Posted by

**Amer** when i said typo i mean in the book I made an example

let $\displaystyle G=Z_{24}$ , $\displaystyle H = <3> = \{0,3,6,9,...,21\}$ , $\displaystyle K = <4> = \{0,4,8,...,20\}$

$\displaystyle \mid H\mid = 8 , \mid K \mid = 6 $

H acting on xK by left additive

$\displaystyle G/K = \{K, 1+K,2+K,3+k\}$

the orbit of H acting on the left cost will be $\displaystyle \{K, 1+K,2+K,3+k\}$

$\displaystyle \mid HxK \mid = \mid K \mid . \mid H:Stab_H(xK)\mid = 6.4 = 24 $

but if it is $\displaystyle \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ??$ which is not correct since the order should equal to the G order

so there is a typo in book

am I right ?

Yes, you are correct. There is a typo. It is not $\displaystyle |HxK|=|H|[H:H\cap xKx^{-1}]$ but $\displaystyle [HxK]=|K|[H:H\cap xKx^{-1}$.

Re: Group Action, Double costs

Quote:

Originally Posted by

**Drexel28** Yes, you are correct. There is a typo. It is not $\displaystyle |HxK|=|H|[H:H\cap xKx^{-1}]$ but $\displaystyle [HxK]=|K|[H:H\cap xKx^{-1}$.

Thanks for the fast respond I appreciate that