normal subgroup and orbits

**abhishekkgp** · February 11th 2011, 07:27 PM

Let the group G act on the finite set A and let $H \trianglelefteq G$ . Let $O_1, O_2,\ldots,O_r$ be the distinct orbits of H on A.

now read the following:
$g \cdot a_i = a_j$ , where $g \in G, a_i \in O_i, a_j \in O_j$
let $b_i = h_1 \cdot a_i \text{ for some } h_1 \in H \text{ and } a_i \neq b_i$ .
now $g \cdot b_i = g \cdot (h_1 \cdot a_i)=(g h_1) \cdot a_i = (g h_1 g^{-1} g) \cdot a_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j \text{ where }g h_1 g^{-1} = h_2; h_2 \in H$ .
now $h_2 \cdot a_j = c_j \text { for some } c_j \in O_j$ . also it can be shown that $a_j \neq c_j$ very easily.

from the above argument can i say that $|O_i|=|O_j|$

**tonio** · February 12th 2011, 01:16 AM

Originally Posted by abhishekkgp

Let the group G act on the finite set A and let $H \trianglelefteq G$ . Let $O_1, O_2,\ldots,O_r$ be the distinct orbits of H on A.

now read the following:
$g \cdot a_i = a_j$ , where $g \in G, a_i \in O_i, a_j \in O_j$
let $b_i = h_1 \cdot a_i \text{ for some } h_1 \in H \text{ and } a_i \neq b_i$ .
now $g \cdot b_i = g \cdot (h_1 \cdot a_i)=(g h_1) \cdot a_i = (g h_1 g^{-1} g) \cdot a_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j \text{ where }g h_1 g^{-1} = h_2; h_2 \in H$ .
now $h_2 \cdot a_j = c_j \text { for some } c_j \in O_j$ . also it can be shown that $a_j \neq c_j$ very easily.

from the above argument can i say that $|O_i|=|O_j|$

I'm not sure: why won't you directly show a bijection between $O_i\,,\,O_j$ ?

Tonio

**abhishekkgp** · February 12th 2011, 06:54 AM

Originally Posted by tonio

I'm not sure: why won't you directly show a bijection between $O_i\,,\,O_j$ ?

Tonio

Thats what i tried to do. is there another way to show a bijection between the two??

**tonio** · February 12th 2011, 08:01 AM

Originally Posted by abhishekkgp

Thats what i tried to do. is there another way to show a bijection between the two??

"Another" way? Which one has been shown? If you think that your calculations in your first post show

this then do define directly and unmistakenly the bijection, otherwise search for one...

Tonio

**abhishekkgp** · February 13th 2011, 03:34 AM

Originally Posted by tonio

"Another" way? Which one has been shown? If you think that your calculations in your first post show

this then do define directly and unmistakenly the bijection, otherwise search for one...

Tonio

In the question i forgot to mention that G is transitive on A(sorry for that).
with this in mind there exists a $g \in G \text{ such that } g \cdot a_i = a_j \text{ where } a_i \in O_i, a_j \in O_j$
for such a g there exists a bijection:
$g \colon O_i \rightarrow O_j$
$g \cdot m_i \mapsto m_j \text{ where } m_i \in O_i, m_j \in O_j$
to prove this consider $a_i, b_i \in O_i \text{ such that } a_i \neq b_i$
now $b_i = h_1 \cdot a_i \text{ for some } h_1 \in H$
from my first post $g \cdot b_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j$
to show that the mapping is injective i must show that $g \cdot a_i \neq g \cdot b_i, \text{ that is, } a_j \neq h_2 \cdot a_j$
i will use contradiction to do so. Assume $a_j=h_2 \cdot a_j \Rightarrow a_j=(g h_1 g^{-1}) \cdot a_j \Rightarrow a_j= (g h_1) \cdot a_i \Rightarrow a_j=g \cdot b_i \Rightarrow g^{-1} \cdot a_j = b_i \Rightarrow a_i =b_i$ which is a contradiction. so the map is injective.

has the injection been shown correctly? If this is correct then i can attempt to show the surjection too. Please comment.

**tonio** · February 14th 2011, 03:31 AM

Originally Posted by abhishekkgp

In the question i forgot to mention that G is transitive on A(sorry for that).
with this in mind there exists a $g \in G \text{ such that } g \cdot a_i = a_j \text{ where } a_i \in O_i, a_j \in O_j$
for such a g there exists a bijection:
$g \colon O_i \rightarrow O_j$
$g \cdot m_i \mapsto m_j \text{ where } m_i \in O_i, m_j \in O_j$

Uuh?? You define a function between two sets naming it the same as an element in G? And besides this,

how is the function defined? Because we know that $g\cdot a_i=a_j$ , but what happens with some

$a_i\neq x\in O_i??$ How's defined the map you're talking about on x? Can you be sure that $g\cdot x\in O_j$

to begin with at all? So, how do you define the map $g$ ?

to prove this consider $a_i, b_i \in O_i \text{ such that } a_i \neq b_i$
now $b_i = h_1 \cdot a_i \text{ for some } h_1 \in H$

from my first post $g \cdot b_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j$
to show that the mapping is injective i must show that $g \cdot a_i \neq g \cdot b_i, \text{ that is, } a_j \neq h_2 \cdot a_j$
i will use contradiction to do so. Assume $a_j=h_2 \cdot a_j \Rightarrow a_j=(g h_1 g^{-1}) \cdot a_j \Rightarrow a_j= (g h_1) \cdot a_i \Rightarrow a_j=g \cdot b_i \Rightarrow g^{-1} \cdot a_j = b_i \Rightarrow a_i =b_i$ which is a contradiction. so the map is injective.

has the injection been shown correctly? If this is correct then i can attempt to show the surjection too. Please comment.

.

This looks fine, if we address the above questions I wrote.

About surjectivity; Since $|O_i|=|O_j|<\infty$ , a function between them is

injective iff it is surjective.

Tonio

**abhishekkgp** · February 14th 2011, 06:05 AM

I have committed a mistake in defining the bijection, what i should have written is the following:
$g \colon O_i \rightarrow O_j$
$x_i \mapsto g \cdot x_i$

you have asked whether it can be said for sure that $g \cdot x \in O_j \text{ when } x \in O_i, x \neq a_i$ .
I show that it is sure to hold. proof: since $x \in O_i \text{ there exists } h \in H \text{ such that } h \cdot a_i = x$
now $g \cdot x = g \cdot ( h \cdot a_i)$
$\Rightarrow g \cdot x = (g h g^{-1})g \cdot a_i$
but we know that $g \cdot a_i =a_j \text{ and } g h g^{-1} \in H$
$\Rightarrow g \cdot x = h^\prime \cdot a_j$
now since $h^\prime \cdot a_j \in O_j \text{ we have } g \cdot x \in O_j$

Thank you for your help. is the definition of the bijection alright now? it was a blunder on my part

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