# Computing the order of n-folded automorphisms

1. Jul 20, 2015

### 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

Last edited: Jul 20, 2015
2. Jul 25, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook