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I have seen and studied this class equation for a finite group acting on itself by conjugations. The only applications I know are Cauchys' theorm and Sylow's theorem. Are there more?

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4 Answers 4

http://en.wikipedia.org/wiki/Burnside's_lemma

Burnside's lemma can be easily deduced from the class equation and is useful in combinatorics

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The class equation implies a nontrivial finite $p$-group has a nontrivial center, or more generally that a nontrivial normal subgroup of a nontrivial finite $p$-group $G$ contains a nontrivial element of the center of $G$. (Note: there are infinite groups where every element has $p$-power order and the center is trivial, so the finiteness assumption on the group is important.) This has standard further consequences for finite $p$-groups, although they are no longer direct consequences of the class equation.

The class equation itself is a special case of a more general result: the orbit-stabilizer formula for group actions. That has lots of uses.

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A nice application is Wedderburn's theorem: every finite skewfield is necessarily commutative. Here a skewfield is something which satisfies the same axioms as field, except that multiplication is not required to be commutative; the typical example are quaternions.

To see this, let $F$ be a finite skewfield, $Z$ be its center. It is easy to see that $Z$ is a field, hence it must be $\mathbb{F}_q$ for some prime power $q = p^k$. $F$ will then be a vector space of finite dimension $n$ over $\mathbb{F}_q$, hence it will have $q^n$ elements.

Now write the class equation for the multiplicative group of $F$:

$$q^n - 1 = q - 1 + \sum_i \frac{q^n - 1}{q^{t_i} - 1}$$

Here $q - 1$ appears as the cardinality of the center, while the sum extends over a set of representatives of the non-trivial conjugacy classes.

Note that for $q^{t_i} - 1$ to divide $q^n - 1$, $t_i$ must divide $n$. Indeed the order of $q$ modulo $q^{t_i} - 1$ is $t_i$, and $q^n = 1 \pmod{q^{t_i} - 1}$.

Now let $f_n$ be the $n$-th cyclotomic polynomial. Then $f_n(q)$ divides $q^n - 1$ and it also divides each term in the sum, so $f_n(q)$ divides $q - 1$.

But this is impossible unless $n = 1$ and the sum is empty, in which case $F$ is commutative. Indeed $f_n(q)$ is a product of terms of the form $q - \omega$, where $\omega$ is a root of unity, and this product will have bigger absolute value than $q - 1$ as soon as $n > 1$.

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1  
Ah, true. I remember seeing in Weil, "Basic Number Theory". –  user218 Jul 27 '10 at 21:19
    
This made me super happy when I first learned it. –  BBischof Aug 3 '10 at 3:14
    
Wow, now we have $\LaTeX$!! :-) –  Andrea Ferretti Aug 3 '10 at 12:51

Sorry, I don't have enough reputation to comment on Andrea Ferretti's answer, but that proof of Wedderburn's theorem is also given in detail in Herstein's Topics in Algebra.

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