# Some concepts of groups and rings

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.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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 invertible), and hence it belongs to . We send to .

Let us show that it is bijective. If are automorphisms such that , then it is clear that (since any element of is of the form ). Therefore the constructed mapping is injective.

Let us show that the constructed mapping is surjective. Consider any element from , and its representative < . Consider the following mapping : we send an element of to . Let us show that this mapping is constructed correctly. Indeed, if , then = for some integer . Then , and hence = . Let us show that is an automorphism. If , then , for some integer Hence . Since ( )=1, then and . Therefore is injective. Let . We have for some integers . Therefore . Hence .

It is obvious that . Therefore the image of under the constructed mapping is Therefore this mapping is surjective.

Since the constructed mapping is both injective and surjective, it is bijective. Since it is also homomorphism, it is an isomorphism.

Let . Then is isomorphic to . But . Therefore, from the above-proved assertion we obtain, that Aut( ) is isomorphic to .

4. (a) Let be a subgroup of and be a normal subgroup of . Let us show that is a subgroup of . To this end let us verify the following two conditions:

(i) if and , then .

Indeed, we have

, for some .

Here we used the associativity of the multiplication, fact that H is normal, and the fact that both and are closed under multiplication.

(ii) the inverse of any element of the form ( ) with and lies in

Indeed, we have

.

Here we used the fact that is normal and is a subgroup.

(b) Let be abelian and . Let us construct an isomorphism . We send the pair ( ) to the product This mapping obviously is a homomorphism. Indeed,

Moreover,

.

If , then . But and . Since , and . Therefore ( )=( ). This implies that the constructed mapping is injective. It is clear that it is surjective. Therefore it is an isomorphism.

5. A mapping of rings is called a (ring) homomorphism if the following conditions are satisfied:

(i) for any , we have and ;

(ii) .

Let : be an isomorphism of rings. Let be the following mappings: , for . It is clear that it is defined correctly (since if is invertible, is also invertible). The injectivity of follows from the injectivity of . Let , and let and . Then . But is an isomorphism. This implies that . Therefore . Hence is surjective, and therefore is an isomorphism.

6. (a) We have . Hence . Then . This implies that . Therefore . Applying the associativity condition for the addition, we conclude that . Therefore . We obtain . Similarly one can prove that .

(b) Consider (according to the part (a)). On the other hand

(here we again use the part (a)). Therefore both and are equal to , and hence are equal to each other.

7. The injectivity of the constructed mapping is obvious. To show its surjectivity let us consider representative of a coset from and a representative of a coset from . We have , for some integers . Consider the element . We have

=

Similarly,

=

Therefore,

.

Thus is surjective, and hence is an isomorphism.

8. Let . Let us show that the constructed map is injective. Indeed, if , then . Therefore , since is an integral domain. Hence . This implies that and its image under the constructed map have the same cardinality . Since is finite, this implies that they coincide, and hence the map is surjective. Therefore there exists an element of whose image is 1. Thus and is invertible.

The ring of integer numbers is an integral domain, but is not a field.

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