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

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 math.stackexchange.com/questions/2933720/… for $b = 1$ and $n = p$. Jan 8, 2020 at 17:31
• This is false . Take $p=2 , a=3$ and $n = 3$ Jan 8, 2020 at 17:31
• Ah yes, you should require that $p>2$. Jan 8, 2020 at 17:32
• Yes, you need that $p$ is an odd prime. You have $3^2\equiv 1\pmod{2^3}$ but $3\not\equiv 1\pmod{2^2}.$ Jan 8, 2020 at 17:36
• You appear to have accepted an incorrect answer.
– user502266
Jan 8, 2020 at 20:11

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 $$n$$.

• 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?
– user502266
Jan 8, 2020 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 $p^n$. Jan 8, 2020 at 18:47
• Yes, I've now written it to make this clearer.
– user502266
Jan 8, 2020 at 18:50

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 $$p$$.

• – lhf
Jan 9, 2020 at 22:25