- #1
jackmell
- 1,807
- 54
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
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
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