Relationship between centralizer and conjugate of an element

**cmj1988** · October 13th 2009, 01:15 PM

Today in class we made a table that dealt with S_8 and then we were given a handout with S_3 and S_4 and their various conjugate classes and centralizers. I was just wondering how one goes about proving the rigorous relationship between S_n where n! is the total elements and n!=centralizer*conjugate class, if there is a rigorous proof. Obviously I noticed, through Lagrange's theorem, that the order of the conjugate and centralizer both divide S_n, but is there an actual rigorous relationship? I just wanted to at least start a proof or be walk through one, kind of for kicks and giggles but also maybe for future preparation in this class.

**tonio** · October 13th 2009, 01:47 PM

Originally Posted by cmj1988

Today in class we made a table that dealt with S_8 and then we were given a handout with S_3 and S_4 and their various conjugate classes and centralizers. I was just wondering how one goes about proving the rigorous relationship between S_n where n! is the total elements and n!=centralizer*conjugate class, if there is a rigorous proof. Obviously I noticed, through Lagrange's theorem, that the order of the conjugate and centralizer both divide S_n, but is there an actual rigorous relationship? I just wanted to at least start a proof or be walk through one, kind of for kicks and giggles but also maybe for future preparation in this class.

Looks like you haven't yet studied actions of groups on sets. After that you'll see these questions can be answered without much difficulty.

Tonio

**cmj1988** · October 13th 2009, 02:20 PM

We have indeed studied group actions and cosets. I'm just not seeing how all this stuff fits together. This is one of those in text (not problem section) excercises. Sort of left to the reader to discover the relationship.

**aliceinwonderland** · October 13th 2009, 07:01 PM

Originally Posted by cmj1988

We have indeed studied group actions and cosets. I'm just not seeing how all this stuff fits together. This is one of those in text (not problem section) excercises. Sort of left to the reader to discover the relationship.

As you said you studied the group action, there is a general theorem about a group action and the cardinal number of orbits.

Theorem. If a group G acts on a set S, then the cardinal number of the orbit of $x \in S$ is the index $|G : G_x|$ , where $G_x$ is the stabilizer of x.

In your case, the group acts on S by conjugation (acting on itsef), and the orbit corresponds to the conjugacy class. So $G_x = \{g \in G | gxg^{-1} = x \}$ , which is the centralizer of x in S.

The cardinal number of the orbit of x = the size of conjugacy class of x = $|G|/|G_x|$ .

I think the most algebra books include the proof of the orbit and stabilizer relationship of groups.

**cmj1988** · October 13th 2009, 07:52 PM

I found this proof:

An introduction to the theory of groups - Google Books

I'm not sure how the whole coset thing gets pulled into it. I do understand that a coset is a subgroup acting on an element of G on a certain side (be it right or left).

**Gamma** · October 13th 2009, 10:42 PM

It's not just any subgroup, its the stabilizer of x. That is really important.

You just literally build the bijection from elements in the orbits under the action into the cosets of the stablizer of x.
$\phi(b\cdot x)=bStab(x)$
Well defined and injective:
$\phi(b\cdot x)=\phi(a\cdot x) \Leftrightarrow bStab(x)=aStab(x) \Leftrightarrow$ $a^{-1}bStab(x)=Stab(x) \Leftrightarrow a^{-1}b\in Stab(x) \Leftrightarrow$ $a^{-1}b\cdot x=x \Leftrightarrow b\cdot x = a \cdot x$

Surjectivity is totally obvious by definition of orbit.

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