# Group Action, Double costs

• Nov 19th 2011, 09:03 AM
Amer
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
• Nov 19th 2011, 01:04 PM
Drexel28
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 $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$.

EDIT: I noticed that I did the "opposite side" than you wanted, but by symmetry they're the same.
• Nov 19th 2011, 06:43 PM
Amer
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 $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$.

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)
• Nov 19th 2011, 07:30 PM
Drexel28
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 $\displaystyle |HxK|=|H|[K:K\cap xHx^{-1}]$.

• Nov 19th 2011, 08:00 PM
Amer
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 ?
• Nov 19th 2011, 08:02 PM
Drexel28
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 $\displaystyle [HxK]=|K|[H:H\cap xKx^{-1}$.
• Nov 19th 2011, 08:03 PM
Amer
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