QUESTION 6 Consider the groups U(8) and ZA (1) Determine the identity element in the group U(8) × Z4 (ii) Determine all the elements of order 4 in the group U(8) X Z.. (iii) Determine the subgroup of U(8) × Z4 generated by the element (7,1). Attach File
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- 1.Prove part of Theorem . Theorem 3.4: Properties of Group Elements Let be a group with respect to a binary operation that is written as multiplication. The identity element in is unique. For each, the inverse in is unique. For each . Reverse order law: For any and in ,. Cancellation laws: If and are in , then either of the equations or implies that .Prove part e of Theorem 3.4. Theorem 3.4: Properties of Group Elements Let G be a group with respect to a binary operation that is written as multiplication. The identity element e in G is unique. For each xG, the inverse x1 in G is unique. For each xG,(x1)1=x. Reverse order law: For any x and y in G, (xy)1=y1x1. Cancellation laws: If a,x, and y are in G, then either of the equations ax=ay or xa=ya implies that x=y.Exercises In Section 3.3, the centralizer of an element a in the group G was shown to be the subgroup given by Ca=xGax=xa. Use the multiplication table constructed in Exercise 20 to find the centralizer Ca for each element a of the octic group D4. Construct a multiplication table for the octic group D4 described in Example 12 of this section.
- 31. a. Prove Theorem : The center of a group is an abelian subgroup of. b. Prove Theorem : Let be an element of a group .the centralizer of in is subgroup of.Exercises 1. List all cyclic subgroups of the group in Example of section. Example 3. We shall take and obtain an explicit example of . In order to define an element of , we need to specify , , and . There are three possible choices for . Since is to be bijective, there are two choices for after has been designated, and then only one choice for . Hence there are different mappings in .Consider the additive group of real numbers. Prove or disprove that each of the following mappings : is an automorphism. Equality and addition are defined on in Exercise 52 of section 3.1. a. (x,y)=(y,x) b. (x,y)=(x,y) Sec. 3.1,52 Let G1 and G2 be groups with respect to addition. Define equality and addition in the Cartesian product by G1G2 (a,b)=(a,b) if and only if a=a and b=a (a,b)+(c,d)=(ac,bd) Where indicates the addition in G1 and indicates the addition in G2. Prove that G1G2 is a group with respect to addition. Prove that G1G2 is abelian if both G1 and G2 are abelian. For notational simplicity, write (a,b)+(c,d)=(a+c,b+d) As long as it is understood that the additions in G1 and G2 may not be the same binary operations.
- 3. Consider the group under addition. List all the elements of the subgroup, and state its order.Exercises 8. Find an isomorphism from the group in Example of this section to the multiplicative group . Sec. 16. Prove that each of the following sets is a subgroup of , the general linear group of order over .Lagranges Theorem states that the order of a subgroup of a finite group must divide the order of the group. Prove or disprove its converse: if k divides the order of a finite group G, then there must exist a subgroup of G having order k.
- Exercises 3. Find the order of each element of the group in Example of section. Example 3. We shall take and obtain an explicit example of . In order to define an element of , we need to specify , , and . There are three possible choices for . Since is to be bijective, there are two choices for after has been designated, and then only one choice for . Hence there are different mappings in .Let H be a subgroup of a group G. Prove that gHg1 is a subgroup of G for any gG.We say that gHg1 is a conjugate of H and that H and gHg1 are conjugate subgroups. Prove that H is abelian, then gHg1 is abelian. Prove that if H is cyclic, then gHg1 is cyclic. Prove that H and gHg1 are isomorphic.In Example 3, the group S(A) is nonabelian where A={ 1,2,3 }. Exhibit a set A such that S(A) is abelian. Example 3. We shall take A={ 1,2,3 } and obtain an explicit example of S(A). In order to define an element f of S(A), we need to specify f(1), f(2), and f(3). There are three possible choices for f(1). Since f is to be bijective, there are two choices for f(2) after f(1) has been designated, and then only once choice for f(3). Hence there are 3!=321 different mappings f in S(A).