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My assignment asks me to prove $\left(\mathbb{Z}/p^{d} \mathbb{Z}\right)^{\times}$ is cyclic for prime $p>2$ and for any positive integer $d$.

They propose proving this by induction.

The base case:

I set $d=1$, then Fermat's Little Theorem states that:

$$\exists \hspace{3pt} x \in \left(\mathbb{Z}/p^{} \mathbb{Z}\right)^{\times} \hspace{7pt} s.t \hspace{6pt} x^{p-1} \equiv 1 \pmod p $$

Therefore, the statement is true for $d=1$. I am confused with how to move forward from here. I know that I need to show: assuming the statement is true for some $d$, this then implies that it is true for $d+1$.

I tried to prove directly: $\exists x$ such that

$$x^{p-1} \equiv 1 \pmod {p^d} $$

$$\Rightarrow x^{p-1} - kp^{d} = 1 \hspace{17pt} k \in \mathbb{Z}$$

Multiplying by $p$, I get: $$\Rightarrow px^{p-1} - kp^{d+1} = p \hspace{17pt} k \in \mathbb{Z}$$

However, it seems that this strategy brings me nowhere. Does anyone know of any other approaches to this that are hopefully simpler?

I thought about using Chinese Remainder Theorem with the $p$ and $p^{d}$ case, which would imply the $p^{d+1}$ case, but this only works when $p$ and $p^d$ are coprime, right? (which is clearly never the case).

Any strategies, approaches, insight or advice would be greatly appreciated. Thanks!

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    $\begingroup$ Any $\mathbf Z/n\mathbf Z$ is cyclic, and you don't need to call lil' Fermat to see this. $\endgroup$ – Bernard Feb 4 at 22:14
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    $\begingroup$ The first step is to say for $p$ odd prime : $(1+p)^{p^k} = \sum_{j=0}^{p^k} {p^k \choose j} p^{p^k-j} \equiv 1+p^{k+1} \bmod p^{k+2}$, from which you find the subgroup of $\mathbb{Z}/p^d \mathbb{Z}^\times$ generated by $1+p$. Then look at the surjective morphism $\mathbb{Z}/p^d \mathbb{Z}^\times \to \mathbb{Z}/p \mathbb{Z}^\times$ (what is its kernel ?) to see $\mathbb{Z}/p^d \mathbb{Z}^\times $ has an element of order $p-1$. $\endgroup$ – reuns Feb 4 at 22:29
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    $\begingroup$ Your proof for $d=1$ isn't correct: in $(\mathbb{Z/2Z})^2$, every $x$ satisfies $4x = 0$ in particular there exists one that does; but it's not cyclic for that matter $\endgroup$ – Max Feb 4 at 22:41
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    $\begingroup$ $p$ is odd though $\endgroup$ – Mike Feb 4 at 22:43
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    $\begingroup$ The Chinese remainder theorem is about relatively prime moduli. Therefore it is not realistic to think it will help you combine an argument mod $p$ and $p^d$ in a direct way to say something about modulus $p^{d+1}$. (Of course $p-1$ and $p^{d-1}$ are relatively prime, so maybe CRT is useful when working with those moduli.) $\endgroup$ – KCd Feb 4 at 23:21
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A sketch:

As mentioned in @reuns' comment, you first prove that

$$(1+p)^{p^k}\equiv 1+p^{k+1}\mod p^{k+2}$$ by induction on $k$ (you'll need the multinomial formula for that). this proves that in $(\mathbf Z/p^d\mathbf Z)^\times$, the class of $1+p$ has order $p^{d-1}$.

On the other hand, in the cyclic group $(\mathbf Z/p\mathbf Z)^\times$, there exists an integer $n\bmod p$ with order $p-1$ , hence the order of $n\bmod p^d\:$ is a multiple of $p-1$, so that some power $n^r\bmod p^d$ has order exactly $p-1$. As $p-1$ and $p^{d-1}$ are coprime, $n^r(1+p)\bmod p^d$ has order $(p-1)p^{d-1}=\varphi(p^d)$.

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  • $\begingroup$ Your notation is sloppy. Don't say "$n$ has order $p-1 \bmod p$" since it looks like you are writing about the order reduced modulo $p$. Instead say "$n \bmod p$ has order $p-1$." Fix the same kind of awkward writing when discussing orders of integers modulo $p^d$. Also, I feel the reasoning you give to get a unit mod $p^d$ with order $p-1$ needs more work: when $n \bmod p$ has order $p-1$, why should some power $n^r \bmod p^d$ have order $p-1$? Of course it might be okay to leave this for the OP to think about, but there is definitely something more needed there for a full argument. $\endgroup$ – KCd Feb 4 at 23:16
  • $\begingroup$ I recall it's only a sketch, not a detailed solution. Roughly speaking, theorder $p-1$ is the generator of the subgroup of periods, so the order $\bmod p^d$ is a multiple of $p-1$, namely it is $r(p-1)$ for some $r$, in which case $n^r\bmod p^d$ has order $p-1$. $\endgroup$ – Bernard Feb 4 at 23:22
  • $\begingroup$ I agree. The edit looks good. $\endgroup$ – KCd Feb 4 at 23:35
  • $\begingroup$ Wait, why does the fact that the ord$(n^r)$ $=p-1$ coprime to ord$(1+p) =p^{d-1}$ imply that the order of $n^r(1+p)$ is the product? Couldn't there be some $\ell$ of the form $\ell = (p-1)p^r; r<d-1$ such that $(n^r(1+p))^{\ell} \equiv 1$ $\mod p^d$> $\endgroup$ – Mike Feb 5 at 2:25
  • $\begingroup$ This results from the Chinese remainder theorem and the structure theorem for finite abelian groups* $1+p$ and $n^p$ are not in the same component of $\mathbf Z/p^d\mathbf Z$, hence the order of he product is the l.c.m. of the h, i.e. their product. $\endgroup$ – Bernard Feb 5 at 10:34

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