Page 131 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Proposition 2: Let G be a group and let x be an element of G. Then C(x) is a subgroup of G.
Proposition 3: Let x be an element of the group G. Then the elements of the conjugacy class of x
are in one-to-one correspondence with the left cosets of the centralizer C(x) of x in G.

Example: Two permutations are conjugate in S if and only if they have the same shape
n
(i.e., the same number of disjoint cycles, of the same lengths). Thus, in particular, cycles of the
same length are always conjugate.
Theorem 1: [Conjugacy class Equation] Let G be a finite group. Then
| G | = | Z(G) | + [ g : C(x) ]
where the sum ranges over one element x from each nontrivial conjugacy class.

Definition: A group of order p , with p a prime number and n  1, is called a p-group.
n
Theorem 2: [Burnside] Let p be a prime number. The center of any p-group is nontrivial.
Corollary 1: Any group of order p (where p is prime) is abelian.
2
Theorem 3: [Cauchy] If G is a finite group and p is a prime divisor of the order of G, then G
contains an element of order p.

Example: Prove that if the center of the group G has index n, then every conjugacy class of
G has at most n elements.
Solution: The conjugacy class of an element a in G has [G : C(a)] elements. Since the center Z(G)
is contained in C(a), we have [G : C(a)] [G : Z(G)] = n. (In fact, [G : C(a)] must be a divisor of n.)

Example: Find all finite groups that have exactly two conjugacy classes.
Solution: Suppose that |G| = n. The identity element forms one conjugacy class, so the second
conjugacy class must have n-1 elements. But the number of elements in any conjugacy class is a
divisor of |G|, so the only way that n-1 is a divisor of n is if n = 2.

Example: Let G = D , given by generators a, b with |a|=6, |b|=2, and ba=a b. Let H =
-1
12
{ 1, a , b, a b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H.
3
3
Solution: The normalizer of H is a subgroup containing H, so since H has index 3, either N (H)
G
= H or N (H) = G. Choose any element not in H to do the first conjugation.
G
aHa = { 1, a(a )a , aba , a(a b)a } = { 1, a , a b, a b }
5
5
5
3
2
3
3
5
-1
This computation shows that a is not in the normalizer, so N (H) = H. Conjugating by any
G
element in the same left coset aH = { a, a , ab, a b } will give the same subgroup. Therefore, it
4
4
makes sense to choose a to do the next computation.
2
a Ha = { 1, a , a ba , a (a b)a } = { 1, a , a b, ab }
3
4
4
3
2
3
-2
2
2
4
Comment: It is interesting to note that an earlier problem shows that b, a b, and a b form one
2
4
conjugacy class, while ab, a b, and a b form a second conjugacy class. In the above computations,
5
3
notice how the orbits of individual elements combine to give the orbit of a subgroup.
Example: Write out the class equation for the dihedral group D . Note that you will need
n
two cases: when n is even, and when n is odd.
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