number theory - Multiplicative group of $(mathbb Z/p^r)^times$

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?

Answer

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.