The Question is let $\displaystyle G $ acts transitively on a nonempty finite set A

let H be normal subgroup of G,Let Orbits of H on A be $\displaystyle O_1,O_2,...,O_r$

prove that G acts transitively on $\displaystyle O_1,O_2,...,O_r$

My work I want to prove that for any two oribts $\displaystyle O_i,O_j$ there exist $\displaystyle g\in G $ such that $\displaystyle O_i = gO_j$

I will call $\displaystyle O_i$ elements with "a" and $\displaystyle O_j$ with "b"

pick $\displaystyle a_1\in O_i \;\;\;\;, b_1\in O_j $

there exist g in G such that $\displaystyle a_1 = g. b_1 $ since G acts transitively on A

let $\displaystyle b_s \in O_j \;\;\; b_s = h.b_1$

want to show that $\displaystyle g.b_s $ is an element of $\displaystyle O_i$

$\displaystyle g.b_s = g.h.b_1 $

$\displaystyle g.b_s = g.h.g^{-1} .a_1 \;\;\; a_1=g.b_1 \Rightarrow g_{-1}.a_1 = b_1 $

$\displaystyle g.b_s = (ghg^{-1}).a_1 $

but H is normal subgroup so b_s in O_i

b_s is arbitrary so for any b in O_j g.b is an element of O_i

$\displaystyle O_i = gO_j $ ends

is it correct ?? is there any shorter way ? any ideas

Thanks