Determine how many elements of $\text{Aut}(\mathbb{Z}_{105})$ have order 6.
My attempt:
$\text{Aut}(\mathbb{Z}_{105}) \cong \text{U}(105) \cong \text{U}(7) \times \text{U}(3) \times \text{U}(5) \cong \mathbb{Z}_6 \times \mathbb{Z}_2 \times \mathbb{Z}_4$
The combinations of orders that will work are:
- (6,1,1) - 2 elements
- (6,1,2) - 2 elements
- (6,2,1) - 2 elements
- (6,2,2) - 2 elements
- (3,1,1) - 2 elements
- (3,1,2) - 2 elements
- (3,2,1) - 2 elements
- (3,2,2) - 2 elements
Total 16 elements.
The answer key says 14 though -- what did I over count?
$Aut(\Bbb Z/105) = \Bbb Z/2 + \Bbb Z/2 + \Bbb Z/3 + \Bbb Z/4$. Elements of order 6 are $ker \, 6 \setminus (ker \, 2 \cup ker \, 3) $, where $n$ is multiplication by $n$ map. $ker \, 6$ is $(*, *, *, 2*)$; kernel of $2$ is $(*, *, 0, 2*)$, kernel of $3$ is (0,0,*,0) and they intersect trivially (as 2 and 3 are coprime). So kernel of $6$ has 24 elements, and we should remove 8+3-1 = 10 elements. You can try to match your counting with mine, but computing orders of multiplicative residues is too much for me.