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Definition 7.2.1.
Let G be a group, and let x,y be elements of G.
Then y is said to be a
conjugate
of x if there exists an element a in G such that y = axa-1.
If H and K are subgroups of G,
then K is said to be a
conjugate subgroup of H
if there exists an element
a in G such that
K = aHa-1.
(b) Conjugacy of subgroups defines an equivalence relation on the set of all subgroups of G.
Definition 7.2.3.
Let G be a group. For any element x in G, the set
{ a in G | axa-1 = x }
is called the centralizer of x in G, denoted by C(x).{ a in G | aHa-1 = H }
is called the normalizer of H in G, denoted by N(H).
Proposition 7.2.4.
Let G be a group and let x be an element of G.
Then C(x) is a subgroup of G.
Proposition 7.2.5.
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. 7.2.3.
Two permutations are conjugate in
Sn
if and only if they have the same shape
(i.e., the same number of disjoint cycles, of the same lengths).
Thus, in particular, cycles of the same length are always conjugate.
Theorem 7.2.6. [Conjugacy class Equation]
Let G be a finite group. Then
| G | = | Z(G) | +
[ g : C(x) ]
Definition 7.2.7.
A group of order pn,
with p a prime number and
n
1,
is called a
p-group.
Theorem 7.2.8. [Burnside]
Let p be a prime number. The center of any p-group is nontrivial.
Corollary 7.2.9.
Any group of order p2 (where p is prime) is abelian.
Theorem 7.2.10. [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.
15.
Prove that if the center of the group G has index n,
then every conjugacy class of G has at most n elements.
Solution
16. Find all finite groups that have exactly two conjugacy classes. Solution
17. Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b. Let H = { 1, a3, b, a3b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H. Solution
18. Write out the class equation for the dihedral group Dn. Note that you will need two cases: when n is even, and when n is odd. Solution
19.
Show that for all n
4,
the centralizer of the element (1,2)(3,4) in Sn
has order 8 · (n-4)!.
Determine the elements in the centralizer of ((1,2)(3,4)).
Solution
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