We know that the automorphism group of the cyclic group 𝐶37=⟨𝑥∣𝑥37=1⟩ is isomorphic to a direct product of cyclic groups.If 𝜙𝑟:𝐶37→𝐶37 is the homomorphism 𝑥↦𝑥𝑟 find a minimal list of generators for Aut(𝐶37)=⟨𝜙𝑟1,…,𝜙𝑟𝑡⟩. (Your answer should be a set of one or more integers. For example if Aut(𝐶37)=⟨𝜙3,𝜙7⟩ then enter {3,7}.)

The automorphism group of a cyclic group of order n, Aut(C_n), is isomorphic to the group of units mod n, (Z/nZ)*. The group of units mod n consists of all integers less than n and relatively prime to n.

In this case, we are looking for the automorphism group of C_37, Aut(C_37). Since 37 is a prime number, all integers less than 37 are relatively prime to 37. Therefore, the group of units mod 37, (Z/37Z)*, consists of all integers from 1 to 36.

However, we are looking for a minimal list of generators for Aut(C_37). A generator of a cyclic group is an element of the group such that every other element of the group can be obtained by repeatedly applying the group operation to this element or its inverse.

The group (Z/37Z)* is a cyclic group of order 36. The generators of a cyclic group of order n are the elements relatively prime to n. Therefore, the generators of (Z/37Z)* are the integers less than 36 and relatively prime to 36.

The integers less than 36 and relatively prime to 36 are 1, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 35.

Therefore, a minimal list of generators for Aut(C_37) is {1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 35}.