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# Some concepts of groups and rings

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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}

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##### 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

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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 ...

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