I have to find the number of elements in $\operatorname{Aut}(\mathbb Z_{720})$ with order $6$. Please suggest how to go about it.
1) $\operatorname{Aut}(\mathbb Z_{720})$ isomorphic to $U(720)$ (multiplicative group of units).
2 ) I am using the fundamental theorem of abelian group that a finite abelian group is isomorphic to the direct products of cyclic groups $\mathbb Z_n$.
In this case, $720 = 16 \times 9 \times 5$.
Therefore, $\operatorname{Aut}(\mathbb Z_{720})\cong U(720) \cong \mathbb Z_2 \times \mathbb Z_4 \times \mathbb Z_4 \times \mathbb Z_6$.
Now, the possible orders of elements in $\mathbb Z_2: 1, 2$; $\mathbb Z_4: 1,2,4$; $\mathbb Z_6: 1, 2, 3, 6$.
Using the result defining the order of an element in external direct products:
If $6= \operatorname{Order}(a, b, c, d) = \operatorname{lcm} ( \operatorname{Order} (a), \operatorname{Order} (b), \operatorname{Order} (c), \operatorname{Order} (d))$ then:
Case 1 : If $\operatorname{Order} (d) =6$ then $\operatorname{lcm} ( \operatorname{Order}(a), \operatorname{Order}(b), \operatorname{Order}(c))= 1$ or $2$.
Using the the result for cyclic groups:
for every divisor $d$ of the order of a cyclic group G, there exists $\phi(d)$ elements in G with order $d$.
It seems there are $16$ elements. I am not sure though.
Is this the correct way and how to proceed further? Please suggest.
Case 2: If order of d = 3, then we need that lcm ( O(a), O(b), O(c))= 2
It can happen in three ways:
(a) O(a) = 2, O(b) = 1 or 2, O(c)= 1 or 2. (b) O(b) =2, O(a) = 1, O(c)= 1 or 2. (c) O(c)=2, O(a) = 1, O(b)= 1.
According to this, in each case, there can be 14 elements in all. Total no. of elements 16 + 14 = 30.