Classification of $ \mathbb{Z}/p$-actions on $ \mathbb{Z}/p^n$

Question

I have to prove the following lemma: let $p$ prime number, $n \in \mathbb{N}_{>1}$ and $a \in \mathbb{Z}$ s.t. $a^p \equiv 1$ modulo $p^n$, thus we have the following cases

1) if $p$ is odd then $a \equiv 1$ mod $p^{n-1}$;

2)if $p=2$ then $a \equiv \pm1$ mod $2^{n-1}$.

I'm given the hint to use Fermat little theorem to prove the equality $a=1+kp^d$ where $p$ doesn't divide $k$ and $d>0$: to prove this formula is very easy but I don't understand how to use it to get the final result in the firts case.

I dealt 2) in the following way: knowing $a=1+k2^d$ we have $ a^2=1+k2^{d+1}+k^2 2^{2d} $ and imposing the condition $a^2 \equiv 1$ mod $2^n$ I can get the result after various computations (here I omit them since they are trivial).

The problem is that I'm not sure this method works even for $p$ odd: I get by the usual formula $a^p=(1+kp^d)^p= 1+\binom{p}{1}kp^d+\dots + (kp^d)^p$ so I have to impose $\binom{p}{1}kp^d+ \binom{p}{2}k^2p^{2d}+\dots + (kp^d)^p\equiv 0$ mod $p^n$ but I don't get how this condition implies $d\geq n-1$.

Surely the point is that either my approach is not effective and I have to find another method or I'm passing over some basic number theory fact that would lead me to the conclusion.

Any help or advice is welcome.

Answer 1

I think you were almost there. We want to prove that if $p$ is odd then $$a^p\equiv 1\pmod{p^n}\Rightarrow a\equiv 1\pmod{p^{n-1}}.$$

Let's prove by induction that for all $1\leq k\leq n-1$ we have $a\equiv 1\pmod{p^k}$. $k=1$ is obvious by Fermat's little theorem. Now write $a=1+p^kb$ for some $1\leq k\leq n-2$ and the condition becomes $$a^p=(1+bp^k)^p=\sum_{j=0}^p {p\choose j}b^jp^{kj}\equiv 1\pmod{p^n}.$$ The first term in the sum is a $1$, the second is $bp^{k+1}$ and all the others are divisible by $p^{k+2}$ (because they contain a factor at least $p^{2k+1}$ thanks to the binomial coefficient and $2k+1\geq k+2$). Hence $1+bp^{k+1}\equiv 1\pmod{p^{k+2}}$ and so $p\mid b$. The thesis follows by induction on $k$.

Answer 2

HIT:

For $n\ge3,$ $$a^2\equiv1\pmod{2^n}\implies2^{n-2}\mid\dfrac{a-1}2\cdot\dfrac{a+1}2$$

But $\dfrac{a+1}2-\dfrac{a-1}2=1$ so $\dfrac{a+1}2,\dfrac{a-1}2$ have opposite parities

If $\dfrac{a+1}2$ is odd, $2^{n-2}\mid\dfrac{a-1}2\iff a\equiv1\pmod{2^{n-1}}$

For $p>2$ use discrete logarithm

$$a^p\equiv1\pmod{p^n}\implies p\cdot\text{ind}(a)\equiv0\pmod{\phi(p^n)}$$

Answer 3

$1)\ $ is in fact a problem from UNESCO Competition 1995.

This follows from Lifting the Exponent Lemma (LTE).

If $p$ is odd, $a\equiv b\not\equiv 0\pmod{p}$, then $\upsilon_p\left(a^n-b^n\right)=\upsilon_p(a-b)+\upsilon_p(n)$.

$a^p\equiv 1\pmod{p}$ and $a^{p-1}\equiv 1\pmod{p}$ by Fermat's Little Theorem.

Therefore $a^{\gcd(p,p-1)}\equiv a\equiv 1\not\equiv 0\pmod{p}$ (see here), so by LTE:

$\upsilon_p(a^p-1^p)=\upsilon_p(a-1)+1=n\implies \upsilon_p(a-1)=n-1$.

So in fact we've proved a stronger statement: $p^{n-1}\mid a-1,\, p^n\nmid a-1$.

$2)\ $ We'll prove ($n\ge 2$): $\,2^n\mid (a+1)(a-1)\implies 2^{n-1}\mid a+1$ or $a-1$.

$a+1,a-1$ are even, $\gcd(a+1,a-1)=\gcd(2,a-1)=2$, so

$\upsilon_2(a\pm 1)=1,\,\upsilon_2(a\mp 1)\ge n-1$, and we're done.