Prove that $n$ devides $\phi(p^n-1)$ ($\phi(x)$ being the totient.)
I could not find anything about this particular question on the web, so I will share my argument here.
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See this question or this question(and probably many others on this site). – Jyrki Lahtonen Jun 12 '14 at 10:15
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oh well, I guess my searching skills are weak. – rollover Jun 12 '14 at 10:16
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Don't worry about that @foaly. It helped me a lot that I remembered this question being asked many times. It is easier to find something, when you know it is there. The local search function doesn't satisfactorily recognize TeX-commands, so it is often useless, if you don't know extra keywords. – Jyrki Lahtonen Jun 12 '14 at 10:23
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First, observe that $\phi(p^n-1)$ is the order of the group of automorphisms of a cyclic group of order $p^n-1$. I.e. $$\mid\Bbb{F}_{p^n}^*\mid=p^n-1$$ $$\mid Aut(\Bbb{F}_{p^n}^*)\mid=\phi(p^n-1)$$ $$Aut(\Bbb{F}_{p^n}^*)\ge Aut(\Bbb{F}_{p^n}/\Bbb{F}_p)=Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)$$ $$\because Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)\cong \Bbb{Z}/n\Bbb{Z}$$ $$\therefore n \mid \phi(p^n-1)$$
rollover
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Hmm. Something went wrong here. The ring $\Bbb{Z}{p^n}$ has only $p^{n-1}(p-1)$ units. Did you mean to use the field of $p^n$ elements? Denoted either $\Bbb{F}{p^n}$ or $GF(p^n)$. That, indeed, has $n$ field automorphisms that are thus also automorphism of its multiplicative group, which is cyclic of order $p^n-1$. Consequently you do get the claim from this. – Jyrki Lahtonen Jun 12 '14 at 10:22
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Ok. I'll check back. This is IMHO a creative way of doing it (if a bit overkill), so I will add an upvote after you fix it. Need to go for a quick bite now :-) – Jyrki Lahtonen Jun 12 '14 at 10:27
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Thanks for checking on my notation again. I'm mixing things up sometimes. – rollover Jun 12 '14 at 10:39