# Calculating an automorphism

• jimmycricket
In summary: Similarly, ##Aut(C_9)## is also ##C_2 \times C_2##. So we have ##Aut(C_8) \times Aut(C_9) \cong C_2 \times C_2 \times C_2 \times C_2##, which is isomorphic to ##C_2 \times C_2 \times C_6##.

#### 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?

##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##.

• jimmycricket