The set G 5 {1, 4, 11, 14, 16, 19, 26, 29, 31, 34, 41, 44} is a group under multiplication modulo 45. Write G as an external and an internal direct product of cyclic groups of prime-power order.

G = (19) × (26) × (16)

Step-by-step explanation:

The isomorphism classes of Abelian groups of order 12 are Z₄ ⊕ Z₃ and Z₂ ⊕ Z₂ ⊕ Z₃

SO Let us calculate the orders of some of the elements of G

We have

4² = 16,

4³ = 64

    = 19,

and

4^4 = 19.4

      = 76

      = 31.

furthermore,

4^5 = 31.4

       = 124

       = 34

and

4^6 = 34.4

      = 136

      = 1

Hence, 4 and 34 each have order 6, 16 and 31 each have order 3, and 19 has order 2.

Next, we calculate

11² = 121

    = 31

and

11³ = 11.3

    = 341

    = 26

this is the calculation needed.

26² = 11^6

      = 31³

       = 1

since we already showed that 31 has order 3. This means that 26 has order 2

Since G has two distinct elements of order 2, it cannot be isomorphic to . We conclude

that G = Z₂ ⊕ Z₂ ⊕ Z₃

Finally, we will express as an internal direct product.

The previous calculations show that

(19) = { 1, 19 }

and (26) = { 1, 26 }

are cyclic subgroups of G of order 2 with trivial intersection. We have

(19) × (26) = { 1, 19, 26, 44 }

since

(16) = { 1, 19, 26, 44 }

has trivial intersection with (19) × (26),  conclude that

G = (19) × (26) × (19)

after simplifying it i got: -36.25 as my answer. hope that !

fdvs

step-by-step explanation: vsfdvsfvd