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I am trying to show that the multiplicative group of $(\mathbb Z/p^r)^\times$ is cyclic. I have established that the order of this group is $p^{r-1}(p-1)$. So, to show that it is cyclic, it suffices to product an element of order $p^{r-1}$ and an element of order $p-1$, for then it must be a product of cyclic groups of these two orders, and since $p^{r-1}$ and $p-1$ are relatively prime, it follows that it is cyclic. But, I am having trouble finding elements with these orders. I tried computing the order of several elements using the binomial formula, but it got pretty messy. Any suggestions for which elements to try and how to prove that they have the desired orders, or for another way to do the proof?

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As Serkan noted, you have to assume that $p$ is odd. –  M Turgeon Sep 25 '12 at 15:41
    
Are you familiar with the $p$-adic numbers? The $p$-adic logarithm makes this problem fairly easy. –  Hurkyl Sep 25 '12 at 16:41

2 Answers 2

up vote 1 down vote accepted

Here is an outline of a possible proof (provided $p$ is odd):

  • Let $x,y\in \left(\mathbb Z/N\mathbb Z\right)^\times$ be of respective order $n$ and $m$, and such that $\gcd(x,y)=1$. Then $xy$ has order $nm$ (modulo $N$).
  • From the previous result, and the fact that $\left(\mathbb Z/p\mathbb Z\right)^\times$ is cyclic, prove that $\left(\mathbb Z/p^2\mathbb Z\right)^\times$ is cyclic.
  • Use induction.
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The usual argument for $p$ odd is to prove first for $r=1$ and then prove that if $g$ is a generator for $r=1$ then either $g$ or $g+p$ is a generator for all $r\ge 2$.

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