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The group of units of the rings $\text{GF}(p^r)$ and $\mathbb{Z}/p^r\mathbb{Z}$ are both cyclic (except for the exception of prime powers are not cyclic when $p=2$ and $r\ge 3$). This is a strong result which I have been using a lot but I don't understand it properly.

I would like to know more alternative proofs of these results. If there are any generalizations (e.g. can both cases be treated together?) and specific cases (e.g. is there something special about the intersection $\text{GF}(p) = \mathbb{Z}/p\mathbb{Z}$.

I hope that question doesn't seem vague, it's more like a couple of questions together - I will greatly appreciate anything on this topic so thanks very much!


Update

Inspired by lhf's answer to a question about Wilson's theorem is there any way to prove that if the product of all elements of a group are $-1$ then the units group is cyclic?

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    any finite multiplicative subgroup of a field is cyclic2011-03-08
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    @yoyo: that doesn't tell us anything directly about the units of $\mathbb{Z}/p^r\mathbb{Z}$.2011-03-08
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    @quanta: The group of units of $\mathbb{Z}/p^r\mathbb{Z}$ is not cyclic for $p=2$, $r \ge 3$.2011-03-08
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    @Chris, thanks. I keep forgetting that.2011-03-08
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    @quanta: The fact that the group of units of $\mathbb{Z}/p^r\mathbb{Z}$ is cyclic is a consequence of the existence of primitive roots, which in turn can be seen as a consequence of Hensel's Lemma; would that be what the kind of observation you are hoping for?2011-03-08
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    @Arturo Magidin, Neat! I can use the fact that Z/pZ is cyclic to prove Z/p^rZ using Hensel. That's really a slick proof that I didn't know!2011-03-08
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    @quanta: You have to work a bit at it, as I recall; you actually show that there is at least one lift to $\mathbb{Z}/p^2\mathbb{Z}$ that works, and then for odd prime you can show that there will always be a lift above $\mathbb{Z}/p^2\mathbb{Z}$ that works. (In other words, as I recall you have to do a bit of work to set things up, but once you have the engine set up, it is indeed Hensel that does the hard lifting; it also points out exactly where the proof breaks down for $p=2$, as I recall).2011-03-08
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    Sometimes negative results are helpful: lots of rings similar to these rings do not have cyclic groups of units. For instance, replace Z by R=Z[√2], and take p=3 (which is still prime), then R/p^2R is a finite local ring whose group of units is 3×3×8 and so not cyclic.2011-03-09
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    @quanta: about the update, see exercise 19 in section 4.1 of LeVeque's *Fundamentals of Number Theory* for an discussion of the $U_m$ case (i.e., units mod $m$). For the general case, do you mean the product of the units in a finite commutative ring? If so, see Jack Schmidt's comment.2011-03-25
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    @lhf, it seems to depend on primitive roots? But mostly it's just an excise that says to prove the thing I was wondering if could be proved.2011-03-25

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