If X is a G-set, X is partitioned by the G-orbits. So if X is finite,

(" " means you should take one
representative x from each orbit, and sum over *the set of
representatives*. This is different from the notation " " which occurred in the proof of Burnside's
Theorem; there I was actually summing over *the set of
orbits*.)

x is a * fixed point* if for all . In this case, x is an orbit
all by itself. Recall that denotes the set of fixed points of
X. I can pull these one-point orbits out of the sum above to obtain

In this case, will denote the set of orbits which have
more than one point. (I'll call them * nontrivial
orbits* for short.) Thus, " " means
that you should take one representative x from each nontrivial orbit.

* Example.* Here is a set X of 9 points
arranged in a square:

Let the group of symmetries of the square act on X. There is one fixed point: 5. There are two nontrivial orbits: and . The equation above says

* Lemma.* If G is a group, G acts on itself by
conjugation:

* Proof.* for all . And if , then

Therefore, conjugation is an action.

* Definition.* The orbit of an element of G
under the conjugation action is called a * conjugacy
class*.

* Remark.* I proved earlier that if x and y are
two points in the same orbit, there is a such that . It follows that
any two elements in a conjugacy class are conjugate.

* Example.* Let . The conjugacy
classes are

(Note that if and x and y are conjugate, then x and y are either both even or both odd.)

* Definition.* Let G be a group, and let . The * centralizer* of x is

* Lemma.* Let G act on itself by conjugation,
and let . Then , the
centralizer of x.

* Proof.* Fix . When is ? means g fixes x, i.e. . This is equivalent to , which in turn is what it means for x to be in .

* Corollary.* Let G act on itself by
conjugation, and let . Then:

(a) .

(b) The number of elements in each conjugacy class divides the order of G.

* Proof.* For (a), recall that the order of the
orbit equals the index of the isotropy group, and I just showed the
the isotropy group of x is . Then (b) is
immediate, since , i.e. .

* Corollary.* Let G act on itself by
conjugation. Then .

(The horrible notation means the elements of G fixed under the conjugation action. I promise not to do it again.)

* Proof.* The orbit of a fixed point consists
of the point itself, so . This is equivalent to , or . This is the same as saying that
everything commutes with x, i.e. .

I'll apply these results to the equation

In this case, the G-set is . The set of fixed points is the center: . The nontrivial orbits consist of conjugacy classes with more than one element. The order of such a conjugacy class is , where is the centralizer of x. (I must take one such x for each nontrivial orbit.) Therefore,

This is called the * class equation*.

* Example.* Let . The center of
is , and the nontrivial
conjugacy classes are

Take one element from each nontrivial class: say and . Their centralizers are

Thus,

* Example.* Consider the group of quaternions. Here are the
conjugacy classes:

I found these by direct computation. For example, the conjugates of i are:

Hence, is a complete conjugacy class.

In the class equation, the sum is taken over a set of representatives for the nontrivial conjugacy classes. In this example, for instance, I could take or or --- i.e. any three elements, one from each of the (nontrivial) classes , , .

Finally, note that , and the nontrivial classes have 2, 2, and 2 elements, respectively. The class equation says (correctly) that

Before I prove the next result, I'll review some things that are
apparent from the derivation of the class equation. I had G acting on
itself by conjugation. Then an orbit (which does
*not* mean "G times x" in this case!) is the
conjugacy class of x --- i.e. the set of elements conjugate to x.
Assuming that G is finite, the order of the orbit equals the index of
of the isotropy group, so , where is the centralizer of x --- the set of elements which
commute with x.

In the term, I'm summing over
conjugacy classes *with more than one element*. This implies
two things:

(a) , because equals the number of elements in the conjugacy class.

(b) , since , and .

* Definition.* Let p be prime. G is a p-* group* if for all , , the order of g is for some . (n may be different for different elements.)

Finite p-groups are an important class of finite groups. Their
structure is described in some detail by * Sylow
theory*. Sylow theory is also an important tool in determining
the structure of arbitrary finite groups. For example, it provides
tools that helps answer questions like: How many different groups of
order 20 are there?

As applications of the class equation, I'll look at some results that are "preliminaries" to Sylow theory.

The first result says that if p is prime and , then G has an element of order p. Note that if n is not prime, does not imply the existence of an element of order n. An easy counterexample: , which is divisible by 9, but has no elements of order 9.

* Lemma.* Let G be a finite abelian group, and
let p be a prime number which divides . Then G has an
element of order p.

* Proof.* Induct on . Observe that has an element
of order 2 and has an element of order 3 (and
those are the only groups of order 2 and 3, respectively). This gets
the induction started.

Now suppose , and assume the result is true for abelian groups of order less than n. Let , . I want to show that G has an element of order p.

Let , where . If x has order for some k, then has order p. Therefore, assume that the order of x is not divisible by p.

Now , else x has order n, and n is divisible by p by assumption. Now

Since ,

Now since , so is an abelian group of order less than whose order is divisible by p. By the induction hypothesis, contains an element of order p, where .

Suppose y has order m. Then

Since m kills , it follows that m is divisible by the order p. Suppose . Then has order p.

This shows that G has an element of order p and completes the induction.

* Theorem.* (Cauchy) Let G be a finite group,
and let p be a prime number which divides . Then G has an element of order p.

* Proof.* Induct on . Observe that has an element
of order 2 and has an element of order 3 (and
those are the only groups of order 2 and 3, respectively). This gets
the induction started.

Now suppose , and assume the result is true for groups of order less than n. Let , . I want to show that G has an element of order p.

Consider , so the conjugacy class of x contains more than one element. By the remarks preceding the lemma, . By induction, if , then has an element of order p. Since , I'll then have an element of order p in G.

Therefore, assume that , for all . Now

Hence, for all . Look at the class equation

p divides each term in the summation, and p divides by assumption. Therefore, . However, is an abelian group, so the Lemma shows that it has an element of order p. Since , G has an element of order p as well.

* Corollary.* Let p be prime. The order of a
finite p-group is for some .

* Proof.* If G is a finite p-group, every
element has order equal to a power of p. If for some n, then , where q is a
prime number and . By Cauchy's theorem, G has an
element of order q, contradicting the fact that every element of a
p-group has order equal to a power of p. Therefore, for some n.

* Proposition.* Let p be prime. The center of a
nontrivial finite p-group is nontrivial.

* Proof.* Examine the class equation

Since for some , . If , the remarks preceding the lemma above show that . Since , it follows that . Now p divides and every term in the summation, so p divides . Therefore, .

* Proposition.* Let G be a group of order , where p is prime. Then G has a subgroup of order
for .

* Proof.* If , then , which plainly has subgroups of orders and .

Assume the result is true for groups of order , and suppose . The center of G is nontrivial; since , it follows that for some j. By an earlier result, contains an element x of order p.

Now , so . Hence, is normal. Form the quotient group . Now , so by induction contains subgroups of order for . Each such subgroup has the form , where is a subgroup of G containing .

Suppose then that . I have

Thus, , , ... are subgroups of G orders p, , ... . Obviously, G contains a trivial subgroup of order . This completes the induction step, and proves the result.

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Copyright 2012 by Bruce Ikenaga