Find out $\mathrm{Aut}\ \mathbb{Z}_{p^n}$

Asked Nov 24 '15 at 12:36

Active Nov 24 '15 at 15:09

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Prove that $$\mathrm{Aut}\ \mathbb{Z}_{2^n}=\mathbb{Z}_2\times \mathbb{Z}_{2^{n-2}}$$ $$\mathrm{Aut}\ \mathbb{Z}_{p^n}=\mathbb{Z}_{p^{n-1}(p-1)}$$

I figured out that $\mathrm{Aut}\ \mathbb{Z}_n=U(\mathbb{Z}_n)$ and $\lvert U(\mathbb{Z}_n) \rvert=\varphi(n)$, where $\varphi$ is the Euler's totient function, how should I go on?

edited Nov 24 '15 at 15:09

Andreas Caranti

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asked Nov 24 '15 at 12:36

Idele

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Can you find generators? – Michael Burr Nov 24 '15 at 12:39
Is your question really why are the group of units of $\mathbb{Z}_{p^n}$ cyclic? – Nex Nov 24 '15 at 15:48
@Nex yes that's part of the question when p is odd prime number – Idele Nov 25 '15 at 05:13
@MichaelBurr I have no clue how to find generators:( – Idele Nov 25 '15 at 05:16
See here: http://math.stackexchange.com/questions/73498/is-the-group-of-units-of-a-finite-ring-cyclic – Nex Nov 25 '15 at 08:10

Find out $\mathrm{Aut}\ \mathbb{Z}_{p^n}$

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