Why is ##Aut(C_8)## not a cyclic group?

[jimmycricket]

Im doing a question where I have to calculate of composition of automorphisms of a cyclic p-group and something has got me confused. When constructing decompositions of cyclic groups I have gotten used to grouping the direct products of groups with orders of the same prime to a power e.g C_{20}\cong C_4\times C_5.
In this question however I have gotten to a stage where accorging to my lecturer Aut(C_8)\times Aut(C_9)\cong C_2\times C_2 \times C_6 and i don't understand why. I would have expressed it has C_4\times C_6 since there are 4 numbers less than 8 that are coprime to 8 (eulers totient function). Can anyone help clear up my confusion?

 

[jbunniii]

$Aut(C_8)$ is not a cyclic group.

If we write $C_8$ additively, with elements $\{0,1,2,3,4,5,6,7\}$, then there are four possible generators: $1,3,5,7$. Any automorphism $\phi \in Aut(C_8)$ is determined completely by $\phi(1)$, and there are four possibilities: $\phi$ can map $1$ to any of $1,3,5,7$. So $Aut(C_8)$ has four elements.

Expressed in cycle notation, $Aut(C_8)$ consists of these four elements:
$$\phi_1 = \text{identity}$$
$$\phi_2 = (0)(1 3)(2 6)(4)(5 7)$$
$$\phi_3 = (0)(1 5)(2)(3 7)(4)(6)$$
$$\phi_4 = (0)(1 7)(2 6)(3 5)(4)$$
The orders of the non-identity elements are all $2$, so $Aut(C_8)$ cannot be $C_4$. Since $C_2 \times C_2$ is (up to isomorphism) the only other group of order 4, by process of elimination, $Aut(C_8)$ must be $C_2 \times C_2$.