# Some concepts of groups and rings

Not what you're looking for?

Group problems.

1. Let G be a group. Given a E G define the centralizer Z(a) := {b E G : ab = ba}. Prove that Z(a) (less than or equal to) G. For which a E G is Z(a) = G?

2. We say a, b E G are conjugate if there exists g E G such that a = gbg^-1. Recall (HW2.8) that this is an equivalence relation. Let C(a) := {b E G : E g E G, a = gbg^-1} denote the conjugacy class of a E G. Prove that |C(a)| = [G : Z(a)]

3. On HW2 you proved that Aut(Z) is the group with two elements. Now prove that Aut(Z/nZ) is isomorphic to (Z/nZ)^x. (Hint: An automorphism y: Z/nZ --> Z/nZ is determined by y(1). What are the possibilities?) Taking n = 0, we recover the fact that Aut(Z) is approximately Z^x = {(plus or minus)1}

See attached files for additional questions.

##### Purchase this Solution

##### Solution Summary

Some concepts of groups and rings are given. Namely, some properties of the centralizer, conjugacy classes and subgroups of a group are proved. The group of automorphisms of a finite cyclic group is studied. The notion of a homomorphism of rings is introduced. Some properties of rings and fields are proved. In particular, Chinese remainder theorem is proved.

##### Solution Preview

See full solution attached.

1. To show that is a subgroup of it is sufficient to show that the following two conditions are satisfied:

(a) if , , then .

Indeed, since , we have Similarly, Therefore

This implies that .

(b) if , then .

Indeed, since , we have Multiplying this equality by from both left and right, we obtain . Therefore, .

It is obvious that if and only if , for all

2. Let us construct the mapping from the set to the set of right cosets of by . Consider . There exists such that . We send to . This mapping is defined correctly. Indeed, if , then Therefore, . Hence . Therefore , and hence

Let us show that the constructed mapping is injective. Indeed, consider . We have and . Assume that Then . This implies that . Consequently . This implies that .

The surjectivity of the constructed mapping is obvious. Indeed, for any right coset of by we choose its representative , and then take

Since the constructed mapping is both injective and surjective, it is bijective. This implies that

3. Let us construct an isomorphism To this end, let us consider an automorphism . It is clear that is invertible element (since 1 is ...

##### Purchase this Solution

##### Free BrainMass Quizzes

##### Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

##### Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

##### Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

##### Probability Quiz

Some questions on probability

##### Know Your Linear Equations

Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.