Algebra, find orbits and stabilizers

1. Dec 14, 2015

Dassinia

Hello I have the solution of a problem and I don't understand it

1. The problem statement, all variables and given/known data

We know that every subgroup L<S10 acts on [1, 10] := {1, 2,..., 10} by the formula π • i = π(i). Consider L the subgroup of S10 generated by the permutation p = (1, 2, 3, 4)(4, 5)(8, 9, 10).
Find the orbit and stabilizer of 2

2. Relevant equations
Orb(x)={l•x ,l ∈ L }
Stab(2)={ l ∈ L , l•x=x}

3. The attempt at a solution
Here's the solution
L=< { p=(1, 2, 3, 4)(4, 5)(8, 9, 10) = (1, 2, 3, 4,5)(8, 9, 10) } > ⊂ S10 is generated by a 5-cycle and a 3-cycle so |L|=15 because(1, 2, 3, 4,5) &(8, 9, 10) don't commute.
Orb(2) = { 1, 2, 3, 4 , 5 }
Stab(2)={ Identity, p5, p10 }

I don't understand how they found the orbit. By definition Orb(2)= L • 2 := {l • 2; l ∈ L}. l • 2= l(2) and thn I don't know what to do.
For the Stabilizer why they take p5=(8,10,9) and p10=(8,9,10) ? The order of the permutation is 15 so why Stab(2) isn't { Identity, p15} ?
Thanks

2. Dec 14, 2015

andrewkirk

Let $a=(1\ 2\ 3\ 4\ 5)$ and $b=(8\ 9\ 10)$. The subgroup L is $\{a^nb^n\ |\ n\in \mathbb{N}\}$. In calculating the orbit of 2 you can ignore the $b^n$ factor because it doesn't act on 2.
So the orbit of 2 is $\{a^n(2)\ |\ n\in\mathbb{N}\}$ which is $\{a^n(2)\ |\ n\in\{0,1,2,3,4\}\}$ since $a^n=$Identity.

$p^{15}$ is equal to the identity, so will not be listed in any stabilizer group.

From the above we have that $p^5=b^5, p^{10}=b^{10}$ because the $a$ part is the identity, having a cycle of 5. To see why $b^5=$(8 10 9) apply the permutation $b$ five times and see what happens.