Computing the order of n-folded automorphisms

[jackmell]

I'd like to find ##\operatorname{aut}^2 \mathbb{Z}_n^*## and would first like to just compute it's order. However, I'm pretty sure in general ##\operatorname{aut}\mathbb{Z}_n^*## is non-abelian so can't apply or extend the formula for computing ##\big|\operatorname{aut}\mathbb{Z}_n^*\big|## to the twice-folded case.

Is there a formula for computing ##\big|\operatorname{aut}^2 \mathbb{Z}_n^*\big|##?

For example, empirically I find ##\big|\operatorname{aut}^2\mathbb{Z}_{15}^*\big|=8##. I was wondering if someone could help me prove this algebraically and then perhaps help me extend it to the first 500 groups? I should point out if ##\operatorname{aut}\mathbb{Z}_{15}^*## was abelian, then it would be isomorphic to ##\mathbb{Z}_{2}\times \mathbb{Z}_{2^2}## and then show it's order is 8 but I don't think ##\operatorname{aut}\mathbb{Z}_{15}^*## is abelian.

Ok thanks for reading,
Jack

 

[jvicens]

Dear Jack,

Your question is quite interesting and the answer is not simple. In general, there is no known formula for computing the order of ##\operatorname{aut}^2 \mathbb{Z}_n^*##. However, there are some specific cases where it is possible to find the order.

Firstly, let's define ##\operatorname{aut}^2 \mathbb{Z}_n^*##. This is the set of automorphisms of the group ##\mathbb{Z}_n^*## composed with itself. In other words, it consists of all possible mappings from ##\mathbb{Z}_n^*## to itself, where the composition of two mappings is also a mapping from ##\mathbb{Z}_n^*## to itself.

Now, let's consider the case of ##\mathbb{Z}_n^*## being cyclic. In this case, ##\operatorname{aut} \mathbb{Z}_n^*## is isomorphic to ##\mathbb{Z}_n^*## itself, and therefore ##\operatorname{aut}^2 \mathbb{Z}_n^*## is isomorphic to ##\mathbb{Z}_n^* \times \mathbb{Z}_n^*##. The order of this group is ##\varphi(n)^2##, where ##\varphi(n)## is the Euler totient function.

However, in the case where ##\mathbb{Z}_n^*## is not cyclic, finding the order of ##\operatorname{aut}^2 \mathbb{Z}_n^*## becomes much more challenging. In general, it is necessary to analyze the structure of the group and find all possible automorphisms to determine the order. This is a difficult task and there is no known formula for it.

In your example of ##\mathbb{Z}_{15}^*##, it is possible to show that ##\operatorname{aut} \mathbb{Z}_{15}^*## is isomorphic to ##\mathbb{Z}_2 \times \mathbb{Z}_4## and therefore ##\operatorname{aut}^2 \mathbb{Z}_{15}^*## is isomorphic to ##\mathbb{Z}_2 \times \mathbb{Z}_4 \