Transitive Group Action
Solution
GROUP THEORY PROBLEM SOLUTION
See defining a permutation representation of 
it is equivalent to find/define a well-defined GROUP ACTION from 
.
So Let us first define the Group Action.
We will try to give an example and then develop the action.
Take an element of , say (abc). If you just apply it on an element of X, see what happens(say 0 and 1)
(abc):0->0
(abc):1->9
What I am doing is “ Just apply the Permutation of (abc) on an element of X and see to which element of X it goes)
Formally,
“Apply the permutation on an element of X and See the new partition, which is the image of the action”.
THIS DEFINITION OF THIS ACTION IS THE MAIN PART OF THE SOLUTION OF THE PROBLEMS BELOW.
Assume 
FAITHFULLNESS:
See, It is easy to check that no single permutation of can take every element of X to the same element [So Taking every partition to the same partition is equivalent to the statement]
TRANSITIVITY:
See that from every partition to another partition there is a Permutation of Z involved.
So for all x,y of X, there exists g of G such that gx=y.
Note: Take 2 distinct elements from X: say x and y.
See that In simple terms x ->y ,by just a simple transposition on the elements of X.[() denotes transposition]
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |
| 0 | Nothing | (cd) | (ce) | (cf) | (bd) | (be) | (bf) | (af) | (ae) | (ad) |
The pointwise stabilizer of abc in 
is
{a,b,c,Sym(d,e,f)} [where it is the permutation of { a,b,c,d,e,f}.
It is Isomorphic to
So,its order in is 3!=6
- The Setwise Stabilizer of abc in is
acts transitively on X. So the orbit of is 1 [G.x=X]. So Order of the Stabilizer is equal to the (Order of the group acting)/|X|= 6!/10= 72=
.[ By Order-Stabilizer Theorem][|X|| |=|G|]
If the set {a,b,c,d,e,f}->{ Permute(a,b,c,),Permute(d,e,f)}/
{Permute(d,e,f),Permute(a,b,c)}]:
So In Total 36+36=72 elements.
1.9.2
| Action of | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | ||
| Effect of action of (abc) on X | (abc) | 0 | 9 | 8 | 7 | 1 | 2 | 3 | 6 | 5 | 4 | |
| Effect of action of (def) on X | (def) | 0 | 2 | 3 | 1 | 5 | 6 | 4 | 9 | 7 | 8 |
is 2-transtive.
We will show that is 2-transitive under this action.[As they are equivalent as before]
We use the Lemma 1.3.6.
We will show 
is transitive on X-{0}.
We may compute 
:= the Stabilizer of both 1 and x where x is nonzero and is an element of X.
See that for x=1.
={Permutations such that ab goes to ab(2) and ef to ef (2)/ ab goes to ef(2) and ef goes to ab(2) and (cd)(1)}
