Proof regarding the multiplicative group of integers modulo $p^m$

I have the following conjecture:

Let $p$ be a prime, and $g\in(\Bbb{Z}/p\Bbb{Z})^\times$. If $g$ generates $(\Bbb{Z}/p\Bbb{Z})^\times$, then $g$ also generates $(\Bbb{Z}/p^m\Bbb{Z})^\times$ for all $m\in\Bbb{N}$ with $m>1$.

Now I did attempt a proof, but I can't seem to figure out where to go at the end, or if there's a simpler proof than what I had in mind. Number theory and group theory aren't my strong suits. Here's my proof:

Proof: We will attempt prove this inductively. Assume $g$ generates $(\Bbb{Z}/p^m\Bbb{Z})^\times$. Thus $g^{\phi(p^m)}\equiv 1\ (\text{ mod }p^m)$, where $\phi(n)$ is Euler's totient function. We also know $g^{\phi(p^m)}$ is the smallest power of $g$ that satisfies this congruence. We need to show that the smallest power of $g$ that satisfies $g^k\equiv1\ (\text{ mod }p^{m+1})$ is $k=\phi(p^{m+1})=p^m(p-1)$.

Assume there exists such a $k$ that $g^k\equiv1\ (\text{ mod }p^{m+1})$, with $k\le\phi(p^{m+1})$. We want to show that the only possibility is $k=\phi(p^{m+1})$. By Lagrange's Theorem, $k\ |\ \phi(p^{m+1})$, since otherwise the order of the subgroup generated by $g$ would not divide the order of $(\Bbb{Z}/p^{m+1}\Bbb{Z})^\times$.

We can rewrite these equivalences as follows: $$1.\quad \frac{g^{\phi(p^{m+1})}-1}{p^{m+1}}\in\Bbb{N} \\ 2.\quad \frac{g^k-1}{p^{m+1}}\in\Bbb{N}$$

Now, Zsigmondy's Theorem states that if $(a, b)=1$ and $n\in\Bbb{N},\ n>1$, then there exists some prime $p$ that divides $a^n - b^n$ that doesn't divide any $a^k - b^k$ for $k < n$ (with some notable exceptions that we will be careful to avoid in this proof).

By setting $a = g,\ b=1,\ n=\phi(p^{m+1})$, in order to show that $k=\phi(p^{m+1})$ is the only solution, we must show that $g^{\phi(p^{m+1})}-1$ and $g^k - 1$ have all the same prime factors (with different multiplicities), because if this is true, we would be contradicting Zsigmondy's theorem if $k<\phi(p^{m+1})$. Thus $k=\phi(p^{m+1})$.

Currently, I have no clue how to prove that they have the same prime factors of different multiplicities. Write $g^{\phi(p^{m+1})}-1 = Cp^{m+1}$ and $g^k-1 = Dp^{m+1}$ with $C,\ D\in\Bbb{N}^2$. Then it's easy to see that $\frac CD\in\Bbb{N}$, since \begin{align} \frac CD = \frac{g^{\phi(p^{m+1})}-1}{g^k-1} & = \frac{(g^k)^{\phi(p^{m+1})/k}-1}{g^k-1} \\ & = \sum_{j=0}^{\phi(p^{m+1})/k-1}g^{jk} \end{align} and the last sum is a sum of integers (since $k\ |\ \phi(p^{m+1})$). Thus it suffices to show that $\omega\left(g^{\phi(p^{m+1})}-1\right) = \omega\left(g^k-1\right)$, where $\omega(n)$ counts the number of distinct prime factors of $n$.

This is where I'm stuck.

I can't help but think there's a much simpler proof. Also, it concerns me that I haven't used the fact that $g^{\phi(p^m)}\equiv 1(\text{ mod }p^m)$. I feel like if I could figure out how to use that fact, I could avoid the whole Zsigmondy's Theorem rigmarole (that, or I could use it to finish the proof in it's current state). Any ideas?

Also, if possible, can the proof avoid using Sylow Subgroups? I'm looking for a way to prove this without relying on Sylow's Theorems (I wasn't explicitly told that they would help solve this problem, but I have a hunch, since I saw a proof that $(\Bbb{Z}/p\Bbb{Z})^\times$ is cyclic which used Sylow's Theorems).

• Hopefully my memory isn't flawed, but I'm pretty sure you're conjecture isn't true. If $g$ generates $(\mathbb{Z}/p\mathbb{Z})^{\times}$, then one of $g$ or $g+p$ generates $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$. Try to prove that and see if you can get some sort of super-generator using that. Aug 21 '16 at 22:33

For instance, $14$ is a generator of $(\mathbb{Z}/29\mathbb{Z})^\times$, but $14$ is not a generator of $(\mathbb{Z}/29^2\mathbb{Z})^\times$.
But you can show that if $g$ is a generator for $(\mathbb{Z}/p\mathbb{Z})^\times$ then at least one of $g$ or $g + p$ is a generator of $(\mathbb{Z}/p^k\mathbb{Z})^\times$.
• @mixedmath, I think your statement might be missing the condition $p \neq 2$, as $(\mathbb Z/2^k \mathbb Z)^{\times}$ isn't even cyclic for $k \geq 3$. Aug 22 '16 at 3:38