For p a prime and n a positive integer, consider the group of units, $(\mathbb{Z}/(p^{n}-1)\mathbb{Z})^{\times}$. How can I go about to find the order of $\bar{p}$?
Order of an element in the group $(\mathbb{Z}/(p^{n}-1)\mathbb{Z})^{\times}$
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group-theory
finite-groups
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0@Zev, $p^{n}=1$, but how do I know this is the order of p? – 2011-09-28
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0I've made my comment into an answer. – 2011-09-28
1 Answers
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Clearly, $p^n\equiv 1\bmod (p^n-1)$, so the order of $p$ is $\leq n$. But if $p^k\equiv 1\bmod (p^n-1)$, then $$(p^n-1)\mid (p^k-1)$$ and if $k , so this is impossible. Therefore the order must be precisely $n$.
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1Note this has nothing to do with p being a prime. – 2011-09-28
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0@Zev, so when you write $p^{k}= 1 mod (p^{n}-1)$ you are assuming that $k>n$, but in your conclusion you assume $k
? I think I am misunderstanding something. – 2011-09-28 -
0@El G: I demonstrate that $p^k\equiv 1\bmod (p^n-1)\implies k\geq n$, using a proof by contradiction (if we had $k
, we would get the contradiction that $p^n-1\leq p^k-1$ and $p^k-1 ).
– 2011-09-28