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I aim to prove that if $p$ is a prime and $p\ne 2$, then $(\Bbb Z/p^n \Bbb Z)^*$ is cyclic.

It is not obvious, since if $n\ne 1$ then we haven't got a field. May I please ask for a proof? Or any reference would be appreciate.

I can find a quick example that if we take $p=2,n=3$, then $(\Bbb Z/p^n \Bbb Z)^*$ is the Klein four group and thus not cyclic. But may I also ask what is wrong for $p=2$?

Thanks in advance!

Y.X.
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  • look up primitive root – Will Jagy May 23 '17 at 04:00
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    You should find another example, at least taking $p=2$, you always get a group that is not cyclic for $n \ge 3$, since $(\Bbb Z/2^n \Bbb Z)^*$ has two distinct subgroups of order 2, namely the subgroups generated by $2^{n-1}-1$ and $2^{n-1}+1$ –  May 23 '17 at 04:33

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Show that $1+p$ has order $p^{n-1}$, therefore $2(1+p)$ has the required order because $2$ has order and divisible by $p-1$, $2^{p-1} \equiv 1 \mod{p} $ (Fermat's little theorem).

to prove the first part use Prove that $1+p$ is an element of order $p^{n-1}$ in $(\Bbb Z/p^n\Bbb Z)^\times$ using the binomial theorem

$|(\mathbb{Z}/p^n\mathbb{Z})^{*}| = (p-1)p^{n-1}$. We have produced an element of the right order..

Appendix: As rightly pointed about by the commentator/questioner propositionx , $2$ is an element of order $(p-1)p^{d}$ (and not p-1 as mistakenly suggested in previous edition), for some $d \in \mathbb{N}$...

Since $p$ is an odd prime $\Rightarrow 2^{p-1} \equiv 1 \mod{p}$, which in turn $\Rightarrow 2^{p^k} \not\equiv 1 \mod{p},\ \forall \ k \in \mathbb{N} $.

Thus $2(1+p)$ must have order exactly equal to $p^{n-1}(p-1)$.

Affineline
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  • May i please ask how can I see that the order of $2$ is coprime to $p$? – Y.X. May 23 '17 at 05:54
  • You are right... edited just now, it's not necessarily coprime to $p$ but divisible by $p-1$ ... no idea why I wrote that. Point is that $2^{p^{n-1}}$ can't be $1$, because $2^d$ is coprime to $p$. – Affineline May 23 '17 at 05:56
  • And thanks a lot for pointing out the error... – Affineline May 23 '17 at 06:25
  • I think the order of $2$ cannot be greater then $p-1$ and divides $p-1$ since the order is the smallest positive number such that $2^{order}=1$ but we already have $2^{p-1}=1$. Could you please give some explaination on why we have the order of $2$ be $(p-1)p^d$? – Y.X. May 23 '17 at 10:17
  • Thanks a lot for the answer, but could you please give some explainition on why $2$ has exactly order $p-1$ and not some divisor of it? – Y.X. May 23 '17 at 10:23
  • sorry that was mistake...I have edited it. And incorporated it in the final version. Hope it is helpful... regards . – Affineline May 23 '17 at 11:23
  • Sorry I cannot see the edit(it is displayed as "edited 7 hours ago")...Maybe there is something wrong with the system... I would really appreciate if you would edit to correct it and tell me how does it work exactly. Thanks so much. – Y.X. May 23 '17 at 13:49