4
$\begingroup$

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?

$\endgroup$
9
$\begingroup$

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 $n > 1$.

$\endgroup$
  • 1
    $\begingroup$ Ah, true. I remember seeing in Weil, "Basic Number Theory". $\endgroup$ – user218 Jul 27 '10 at 21:19
  • $\begingroup$ This made me super happy when I first learned it. $\endgroup$ – BBischof Aug 3 '10 at 3:14
  • $\begingroup$ Wow, now we have $\LaTeX$!! :-) $\endgroup$ – Andrea Ferretti Aug 3 '10 at 12:51
  • $\begingroup$ Note for people like me who were confused for a moment: $q^t_i -1$ divides $q^n -1$ because we proving the claim by induction. The centerlizer is a skew-field so a field by induction, and thus we can view $F$ as a vector space over it which implies the division thing. $\endgroup$ – Andy Sep 16 '17 at 22:48
  • $\begingroup$ This is a way to see it, but more generally, these are just the terms appearing in the class equation. The cardinality of the orbit, which is an integer, is the cardinality of the group over the cardinality of the stabilizer. Each stabilizer is a vector space over F^q $\endgroup$ – Andrea Ferretti Sep 17 '17 at 11:04
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy