Thread: prove by orbit-stabilizer

1. prove by orbit-stabilizer

i got stuck on part three

2. Prove $C(A)$ is a subgroup.
This should be a straightforward calculation.

Show $\rho: G\times G\mapsto G$ defines an action on itself.
Again a straightforward calculation, just look at definition of group action.

$|G| = |C(G)|+\sum_{x}|G|/|C(\{ x\})|$
This is just the conjugacy class equation.
Here $C(G) = Z(G)$, i.e. the group center. While $[G:C(\{x\})] = |G|/C(\{ x\})$. And this follows because $x$ is a sum which goes through all the non-trivial conjugacy classes, i.e. when the orbit is greater than 1.

Show $|C(G)| > 1$
This result is known as Burnside's Lemma.
Note, $|G| = |C(G)| + \sum_{x} |G|/|C(\{ x\})|$.
Now since $|C(\{ x\})| \not = p^r$ it means $|G|/|C\{ x\}| > 1$. Thus, the sum is divisible by $p$ and $|G| = p^r$ is divisible by $p$. It must mean that $|C(G)|$ is divisible by $p$.

3. please give proof

please give proof of the last two parts...
i m exactly confused with that...

4. Originally Posted by szpengchao
please give proof of the last two parts
Let $G$ be a finite group. Define a relation $x- y$ if and only if " $x$ is conjugate to $y$".
And "conjugate" means there is $g\in G$ so that $x=gyg^{-1}$.
It is easy to check that $x-x$, i.e. " $x$ is conjugate to itself".
Also $x-y$ implies $y-x$.
And finally, if $x-y$ and $y-z$ implies $x-z$.
This tells us that $-$ is an equivalence relation on $G$.

Let $x\in G$ and define $[x] = \{ a\in G | x-a\}$, more simply it is the set of all elements conjugate with $x$.
This is called the "conjugacy class of $x$".

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, $|G| = \sum_{x}| \text{conj}(x)|$ where $\text{conj}(x)$ is the conjugacy class of $x$, and the sum ranges over different conjugacy classes.

Now let us take out all the single element conjugacy classes.
Thus. $|G| = |A| + \sum_{x}|\text{conj}(x)|$ where $x$ is a sum over conjugacy classes with more than one element and $A$ is the set of all single element conjugacy classes.

Let us examine $A$ in more detail. Note that $a$ is conjugate to itself only i.e. its conjugacy class consists only of itself if and only if $gag^{-1} = a$ for all $g$. Thus $A$ is precisely $C(G)$.

Thus, far we have, $|G| = |C(G)| + \sum_{x} |\text{conj}(x)|$.

We are going to state a theorem: there is a bijection between the element of the conjugacy class and the cosets of $C(\{ x\})$.

The proof is not difficult. Define $\phi: \text{conj}(x) \mapsto G/C(\{ x\})$ as $\phi (a) = aC(\{ x\})$. Try that, I think that will work.

Thus, this tells us that $|\text{conj}(x)| = |G/C(\{x\})| = |G|/|C(\{x\})$.
The last equality follows by Lagrange's theorem.

Thus, $|G| = |C(G)| + \sum_{x} |G|/|C(\text{x})|$.