abstract algebra Consider the group GU2 which cuteiets of t
abstract algebra
Solution
group U(28) has all the elements (less than 28)coprime to 28
a)
so a = 5
b = 15
c = 27
order of 5 = 6
5^2 = 25 , 5^3 = 125 = 13 mod 28 , 5^4 = 9 mod 28 ,5^5 = 17 mod 28 5^6 = 1 mod 28
order of 15 = 2
15^2 = 225 = 1 mod 28
order of 27 = 2
27^2 = 1 mod 28
c) H = {1, 13, 15,27} all elements of order 2 and unit element
K = {1,9,25} all elements of order 4 and unit element
H X K will have unit element (1,1) , three elemets of order 2 (13,1) , (15,1) , (27,1), two elements of order 3 (1,9) , (1,25) and the rest of elements of order 6. Hence this is isomorphic to G
d) G is isomorphic to H X K
but H is not isomorphic to Z/4Z since it has an element of order 4 (and hence it is cyclic)
so G is not isomorphic to Z/4Z X Z/3Z
H is isomorphic to Z/2Z X Z/2Z
and K is isomorphic to Z/3Z
hence G is isomorphic to Z/2Z X Z/2Z X Z/3Z
