Page 133 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes 11.2 Summary

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
Conjugacy of elements defines an equivalence relation on any group G.

Conjugacy of subgroups defines an equivalence relation on the set of all subgroups of G.

Let G be a group. For any element x in G, the set

{ a in G | axa = x }
-1
is called the centralizer of x in G, denoted by C(x).
For any subgroup H of G, the set
{ a in G | aHa = H }
-1
is called the normalizer of H in G, denoted by N(H).
Let G be a group and let x be an element of G. Then C(x) is a subgroup of G.

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.
11.3 Keywords

Conjugate Element: 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
Centralizer: Let G be a group. For any element x in G, the set
{ a in G | axa = x }
-1
is called the centralizer of x in G, denoted by C(x).

11.4 Review Questions

1. Compute the G-equivalence classes of X for each of the G-sets X = {1, –2, 24, 5, 6} and
G = {(1), (1, 2) (3, 4, 5) ; (1 2) (3 4 5), (1 2) (3 8 4)} for each x  X verify |G| = |O | |G |.
x
x
2. Write the class equation for S5 and for |G |
x
3. Let P be prime. Show that the number of different abelian groups of order P is the same
n
as the number of conjugacy class in S .
n
4. Let a  G, show that for any g  G, gc(a)g = c(gag ).
-1
-1
5. Let |G| = p and suppose that |Z(G)| = p for p prime. Prove that G is abelian.
n-1
n
6. Let G be a group with order p , where p is prime and X a finite G-set. If X = {x  X : gx = x
n
G
for all g  G} is the set of elements in X fixed by the group actions, then prove that
|X| = |X | (mod ).
G
p
Answers: Self Assessment
1. (c) 2. (b) 3. (d) 4. (b) 5. (d)

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