Show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$)

Question

I'm trying to show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$), where $p$ is an odd prime.

Since $a^p \equiv 1$ (mod $p^n$) $\iff aa^{p-1} \equiv 1$ (mod $p^{n-1}p$), I was hoping to use Fermat's theorem, which states that $a^{p-1} \equiv 1$ (mod $p$). I therefore tried substituting $a^{p-1}$ with $pk + 1$ in the equation but I didn't manage to solve it like that.

I've been trying to solve it algebraically for some time but is no where close so I could use some help and ideas.

Particular case of https://math.stackexchange.com/questions/2933720/prove-that-a-n-equiv-bn-pmodpk-if-and-only-if-na-b-equiv-0-pmodpk for $b = 1$ and $n = p$. — darij grinberg, Jan 08 '20 at 17:31
Yes, you need that $p$ is an odd prime. You have $3^2\equiv 1\pmod{2^3}$ but $3\not\equiv 1\pmod{2^2}.$ — Thomas Andrews, Jan 08 '20 at 17:36
If $a^p≡1 \mod (p^n)$ then $\phi (p^n)=p$ ⇒ $p^{n-1}(p-1)=p$ ⇒ $p^{n-2}(p-1)=1$ ⇒ $\phi(p^{n-1}=1$ ⇒ $a^{\phi(p^{n-1})}=a^1≡1 \mod (p^{n-1})$. — sirous, Jan 08 '20 at 18:06
You are right, I forgot the part about p > 2. Will edit! @darijgrinberg — virreand, Jan 08 '20 at 18:36

score 3 · Accepted Answer · 2020-01-08T18:50:00.477

3

Let $a=1+p^iX$, where $X$ is coprime to $p$.

Then $a^p-1=\begin{pmatrix}p\\1\\\end{pmatrix}p^iX+ ... +p^{pi}X^p$, where all the coefficients on the RHS except possibly the last are divisible by $p$.

We know that $p^{n} $ divides the LHS so $i>0$. Then, on the RHS, the first term has the lowest power of $p$ and so its power, $i+1$, must be at least $n$.

edited Jan 08 '20 at 18:50

answered Jan 08 '20 at 17:49

The first term has power $i+1$. All the other terms have power $ki+1,k>1$ except the last which has power $pi$. Where's the problem? – Jan 08 '20 at 18:30
+1 Ok. So the argument is (1) the last term must be divisible by $p$ (because all the others are), so $i>0$. Hence all terms except the first are divisible by $p^{2i}$, except the first which is divisible by $p^{i+1}$. So if $i<n-1$ the sum will not be divisible by $p^n$, whereas the lhs is divisible by $p^n$. – almagest Jan 08 '20 at 18:47
Yes, I've now written it to make this clearer. – Jan 08 '20 at 18:50

lhf · Answer 2 · 2020-01-09T16:04:35.277

Actually,

$a^p \equiv 1 \bmod p^n \iff a \equiv 1 \bmod p^{n-1}$

The key relevant fact is this:

If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of order $d$: $H=\{ x \in G : x^d=1 \}$.

The group in question is $G=U(p^n)$, the group of units mod $p^n$. Then $G$ has order $\phi(p^n)=p^{n-1}(p-1)$, a multiple of $p$. Let $H$ be the subgroup of order $p$.

If $a \equiv 1 \bmod p^{n-1}$ then $a=1+bp^{n-1}$ and so $a^p=(1+bp^{n-1})^p=1+cp^n \equiv 1 \bmod p^n$. In other words, $a=1+bp^{n-1} \in H$ for $b=0,\dots,p-1$. This gives us $p$ elements in $H$. Therefore, they are all the elements of $H$ since $H$ has order $p$.

See also https://math.stackexchange.com/questions/410389/subgroups-of-a-cyclic-group-and-their-order — lhf, Jan 09 '20 at 22:25

Show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$)

2 Answers2