# Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

• November 17th 2011, 08:59 AM
Amer
Prove that G acts transitively on normal subgroup orbits on nonempty finite set A
The Question is let $G$ acts transitively on a nonempty finite set A

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

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

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

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

pick $a_1\in O_i \;\;\;\;, b_1\in O_j$
there exist g in G such that $a_1 = g. b_1$ since G acts transitively on A
let $b_s \in O_j \;\;\; b_s = h.b_1$
want to show that $g.b_s$ is an element of $O_i$
$g.b_s = g.h.b_1$
$g.b_s = g.h.g^{-1} .a_1 \;\;\; a_1=g.b_1 \Rightarrow g_{-1}.a_1 = b_1$
$g.b_s = (ghg^{-1}).a_1$
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
$O_i = gO_j$ ends

is it correct ?? is there any shorter way ? any ideas
Thanks
• November 17th 2011, 09:29 AM
Deveno
Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A
is it already given that a normal subgroup H of G induces a well-defined action on the orbits of H? (it's true, but what i'm asking is: have you already proven this earlier?)

aside from that, your proof of the transitivity of this action of G on the orbits of H looks fine to me.
• November 17th 2011, 09:51 AM
Amer
Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A
Quote:

Originally Posted by Deveno
is it already given that a normal subgroup H of G induces a well-defined action on the orbits of H? (it's true, but what i'm asking is: have you already proven this earlier?)

aside from that, your proof of the transitivity of this action of G on the orbits of H looks fine to me.

you are wright it is not the whole question i proved that before, any orbit of H say O, gO still orbit of H

let a,b in orbit O, we have a = h.b for some h in H
if I can show that g.a , g.b is in the same orbit I am done

$ghg^{-1} \in H \;\;\; say \;\;\; ghg^{-1}=h' \Rightarrow h = g^{-1}h'g$

$a = g^{-1}h'g.b$

$g.a = h'.g.b$
so g.a , g.b in same orbit in H
is it Okay ??
• November 17th 2011, 12:07 PM
Deveno
Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A
that's right, and it shows why H has to be normal for this to work.
• November 17th 2011, 12:27 PM
Amer
Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A
in the same question i asked to prove that all orbits has the same cardinality
My work,I will show that the stabalizer of each representative element of the orbits has the same cardinality, let
$O_i , O_j$ orbits of H on A let a,b be the element that present each orbit

$G_a = gG_bg^{-1}\;\;,\text{such that} G_a,G_b \;\; \text{stabalizer of a,b in G}$

Stab of a in H $H_a = G_a \cup H^c$ so $H_a = H_b$

and $\mid O_a\mid = \mid H:H_a\mid$ which end the proof
I did not post another thread because it is one question
is it correct ?
Thanks