Suppose Z20 Z20 is an automorphism and 5 5 What are the po

Suppose : Z₂₀ Z₂₀ is an automorphism and (5) = 5. What are the possibilities of (x)?

Solution

First, since Z20 is a cyclic group generated by 1, any automorphism of Z20 is completely determined by where it sends the element 1, since (x) = x(1) for any x Z20. In order to be a valid automorphism, (1) must also be a generator of Z20, so there are 8 possible automorphisms: (1) {1, 3, 7, 9, 11, 13, 17, 19}. We check whether each automorphism maps 5 to itself. If (1) = 1, then (5) = 5(1) = 5, so this is a possibility. If (1) = 3, then (5) = 5(1) = 5·3 = 15 6= 5 mod 20, so this is not a valid possibility. Checking each of the 8 automorphisms, we see that (1) = 1, 9, 13, or 17, so the answer is (x) = x, 9x, 13x or 17x.