Question

Find the automorphism group of Z16 and express it as a product of cyclic groups. A map φ : Z16 → Z16 is a group homomorphism if and only if φ([k]) = [tk], where t ∈ {1, 3, 5, 7, 9, 11, 13, 15} = U(Z16). Composition of automorphisms corresponds to multiplication mod 16 in U(Z16). U(Z16) is an abelian group of order 8. So possible factorizations of U(Z16) as a product of cyclic groups are Z8, Z4 × Z2 or Z2 × Z2 × Z2. Direct computation shows that 1 = 9^2 = 7^2 and 1 = 3^4 = 11^4 = 5^4 = 13^4 in U(Z16). Since U(Z16) does not have elements of order 8 and U(Z16) has an element of order 4, we get U(Z16) ≅ Z4 × Z2.

          Find the automorphism group of Z16 and express it as a product of cyclic groups.

A map φ : Z16 → Z16 is a group homomorphism if and only if φ([k]) = [tk], where t ∈ {1, 3, 5, 7, 9, 11, 13, 15} = U(Z16). Composition of automorphisms corresponds to multiplication mod 16 in U(Z16).

U(Z16) is an abelian group of order 8.

So possible factorizations of U(Z16) as a product of cyclic groups are Z8, Z4 × Z2 or Z2 × Z2 × Z2.

Direct computation shows that 1 = 9^2 = 7^2 and 1 = 3^4 = 11^4 = 5^4 = 13^4 in U(Z16). Since U(Z16) does not have elements of order 8 and U(Z16) has an element of order 4, we get U(Z16) ≅ Z4 × Z2.

Added by Aaron M.

Question

Please give Ace some feedback