i need full answer with explanation please Determine which o

i need full answer with explanation please

Determine which of the following groups are cyclic, justifying your answer in each case. Write down all the generators of each group that is cyclic. (You are NOT asked to prove that any of the following are groups.) G_1 = ({1, 3, 9, 11, 17, 19, 25, 27}, times _32) G_2 = ({1, 5, 7, 11, 13, 17, 19, 23}, times _24) G_3 = ({0, 3, 6, 9, 12, 15, 18, 21}, + _24) Using an appropriate Strategy, determine an isomorphism between two of the groups given in part (a). Determine all the distinct subgroups of the group G_1 given in part (a)(i), justifying your answer.

Solution

a)i) lets find out the elements that can be generated by 3

3 X₃₂ 3 =9 ; 9 X₃₂ 3 = 27 ; 27 X₃₂ 3= 17 ; 17 X₃₂ 3= 19 ; 19 X₃₂ 3= 25 ; 25 X₃₂ 3= 11 ; 11 X₃₂ 3= 1

since 3 generates all the elements of G₁ hence G₁ is cyclic and 3 is the generator of G₁

If order of a group is 8 then total number of generators of group G are equal to positive integers less than 8 and co-prime to 8. The numbers 1,3,5,7 are less than 8 and co-prime to 8, therefore if a is generator of G, then a³,a⁵,a⁷ are also generators of G.

hence 3 ; 3³ = 27 ; 3⁵ = 19 ; 3⁷ = 11 are the generators of G₁.

ii)  lets find out the elements that can be generated by 5.

5 X₂₄ 5 = 1

elements that can be generated by 7:- 7 X₂₄ 7 =1

Also, 11 X₂₄ 11 =  1 ; 13 X₂₄ 13 = 1 ; 17 X₂₄ 17 = 1 ; 19 X₂₄ 19 = 1 ; 23 X₂₄ 23 = 1

since no element generate G_{2 ,}hence G₂ is not cyclic.

iii) the elements that can be generated by 3 :-

3 +₂₄ 3 = 6 ; 6 +₂₄ 3 = 9 ; 9 +₂₄ 3 = 12 ; 12 +₂₄ 3 = 15 ; 15 +₂₄ 3 = 18 ; 18 +₂₄ 3 = 21 ; 21 +₂₄ 3 = 1

hence G₃ is cyclic and 3 is a generator of G₃.

to find other generators, 3³ = 3+3+3 = 9 ;  3⁵ = 3+3+3+3+3 = 15 ;  3⁷ = 21 are the generators of G₃ .

c) in group G₁ , elements 3,11,19,27 have order 8

elements 9 , 25 have order 4

17 has order 2, & 1 has order 1.

any subgroup of group G₁ should have order 1,2,4, or 8.

Also for a finite cyclic group, converse of Lagrange Theorem holds i. e. if G is a finite cyclic group and n is a non negative integer that is a divisor of |G|, then G has a subgroup of order n.

therefore G₁ will have a subgroup of order 1,2,4,8.

subgroup of order 1 = {1}

subgroup of order 2 = {1,17}

subgroup of order 4 = { 1,17,9, 25}

subgroup of order 8 = G₁