i got stuck on part three
This should be a straightforward calculation.Prove is a subgroup.
Again a straightforward calculation, just look at definition of group action.Show defines an action on itself.
This is just the conjugacy class equation.
Here , i.e. the group center. While . And this follows because is a sum which goes through all the non-trivial conjugacy classes, i.e. when the orbit is greater than 1.
This result is known as Burnside's Lemma.Show
Note, .
Now since it means . Thus, the sum is divisible by and is divisible by . It must mean that is divisible by .
Let be a finite group. Define a relation if and only if " is conjugate to ".
And "conjugate" means there is so that .
It is easy to check that , i.e. " is conjugate to itself".
Also implies .
And finally, if and implies .
This tells us that is an equivalence relation on .
Let and define , more simply it is the set of all elements conjugate with .
This is called the "conjugacy class of ".
It should be known to you that equivalence relations gives rise to partitions.
Meaning every element is in some conjugacy class and that distinct conjugacy classes are disjoint.
Therefore, where is the conjugacy class of , and the sum ranges over different conjugacy classes.
Now let us take out all the single element conjugacy classes.
Thus. where is a sum over conjugacy classes with more than one element and is the set of all single element conjugacy classes.
Let us examine in more detail. Note that is conjugate to itself only i.e. its conjugacy class consists only of itself if and only if for all . Thus is precisely .
Thus, far we have, .
We are going to state a theorem: there is a bijection between the element of the conjugacy class and the cosets of .
The proof is not difficult. Define as . Try that, I think that will work.
Thus, this tells us that .
The last equality follows by Lagrange's theorem.
Thus, .