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 $\displaystyle g \in G \text{ such that } g \cdot a_i = a_j \text{ where } a_i \in O_i, a_j \in O_j$

for such a g there exists a bijection:

$\displaystyle g \colon O_i \rightarrow O_j$

$\displaystyle g \cdot m_i \mapsto m_j \text{ where } m_i \in O_i, m_j \in O_j$

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 $\displaystyle g\cdot a_i=a_j$ , but what happens with some

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

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

to prove this consider $\displaystyle a_i, b_i \in O_i \text{ such that } a_i \neq b_i$

now $\displaystyle b_i = h_1 \cdot a_i \text{ for some } h_1 \in H$

from my first post $\displaystyle g \cdot b_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j$

to show that the mapping is injective i must show that $\displaystyle g \cdot a_i \neq g \cdot b_i, \text{ that is, } a_j \neq h_2 \cdot a_j$

i will use contradiction to do so. Assume $\displaystyle a_j=h_2 \cdot a_j \Rightarrow a_j=(g h_1 g^{-1}) \cdot a_j \Rightarrow a_j= (g h_1) \cdot a_i \Rightarrow a_j=g \cdot b_i \Rightarrow g^{-1} \cdot a_j = b_i \Rightarrow a_i =b_i$ 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. **