Trying to understand how the statement "$2$ is a primitive root for $3^k,k\geq 1$" is proved with the use of the lifting lemma

Question

If you want to jump straight into the end of the preamble or the start of the proof steps I have written, go to End of preamble/start of proof.

There are a few posts on this site addressing the statement in the title: $2$ is a primitive root for $3^k,k\geq 1$, but so far I have not really been able to understand the proofs to the detail. Some examples of such posts include e.g. 2 is a primitive root mod $3^h$ for any positive integer $h$ or Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$..

To my eye the use of the so called lifting lemma seems the most straightforward way to understand the statement and in this question we are focusing on understanding it. One of the lemmas, that can be found e.g. here: Lifting The Exponent, is as follows

Theorem 1 (First Form of LTE). Let $x,y$ be (not necessarily positive) integers, let $n$ be a positive integer, and let $p$ be an odd prime such that $p\mid x - y$ and none of $x$ and $y$ are divisible by $p$. We have $$v_p(x^n - y^n) = v_p(x - y) + v_p(n)$$

where $v_p(x) = m$ is the grestest integer such that $p^m\mid x$, i.e. $p^m\mid x$ but $p^{m+1}\nmid x$.

One given argument with the use of the LTE is given in the post Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$. as a response https://math.stackexchange.com/a/332768/877219. Note: The author of the response handles the case where $k\geq 2$, but it could also work for $k = 1$ -- this will be made more clear when I explain what I know/don't know. The author gives the following response:

Let $g$ be a primitive root $\pmod{p^2}$. Then $p|(g^{p-1}-1)$ by Fermat's little theorem and $p^2 \nmid (g^{p-1}-1)$ since $g$ is a primitive root. Thus by Lifting the Exponent Lemma, $p^{n-1}\|((g^{p-1})^{p^{n-2}}-1)$ and $p^n\|((g^{p-1})^{p^{n-1}}-1)$, so $g$ is also a primitive root $\pmod{p^n}, n>1$.

Edit: More details: Let $d$ be the order of $g \pmod{p^n}, n>1$. Since $g$ is a primitive root $\pmod{p^2}$, we have $p(p-1) \mid d$. By above, $d|p^{n-1}(p-1), d\nmid p^{n-2}(p-1)$, so $d=p^{n-1}(p-1)$, and thus $g$ is a primitive root $\pmod{p^n}, n>1$.

End of preamble/start of proof: I have failed to understand in the given response why $\mathrm{ord}_{p^{n}}(g) = p^{n-1}(p-1)$. I will write below what I know to make this more transparent: Let us work with the case where $k\geq 1$ instead of the $k\geq 2$.

Let $g$ be a primitive root modulo $p$. I will be using the notation $(.)$ as a shorthand for $\mathrm{mod .}$. Then by Fermat's little theorem we know that

$$g^{p-1}\equiv 1 (p)$$

Hence with the $v_p$ notation we have that $$v_p(g^{p-1} - 1) = 1.$$ By the prior LTE lemma we have that for any $n\geq 1$,

$$v_p((g^{p-1})^n - 1^n) = v_p(g^{n(p-1)} - 1) = v_p(g^{p-1} - 1) + v_p(n) = 1 + v_p(n)$$

so by taking $n(l) = p^l, l\geq 0$ (since $v_p(1) = 0$) we have that

$$v_p(g^{n(l)(p-1)} - 1) = 1 + l$$

That is

$$g^{p^l(p-1)}\equiv 1 (p^{1 + l})$$

So if we could now conclude that the order of $g$ in $\mathbb{Z}/p^{l+1}\mathbb{Z}$ is $p^l(p-1)$, that would show that indeed $g$ is a primitive root modulo $p^l$. But how do we actually do it? The congruence equality

$$g^{p^l(p-1)}\equiv 1 (p^{1 + l})$$

implies that

$$\mathrm{ord}_{p^{l+1}}(g)\mid p^l(p-1)$$

And I do not know how to continue from here.

Answer 1

The equality $g^{p^{l}(p-1)}\equiv 1\mod p^{1+l}$ is obvious, since $|(\mathbb{Z}/p^{1+l}\mathbb{Z})^{\times}| = p^{l+1} - p^{l}$, isn't it?

We need the equality $\nu_{p}(g^{n(l)(p-1)}-1) = 1+l$. For $n(l) = p^{l}$ it yields $g^{p^{l}(p-1)} = p^{1+l}s + 1,\;\gcd(p,s) = 1$. Suppose that $d\mid p^{l-1}(p-1)$. Then setting $n(l) = p^{l-1}$ we obtain $g^{p^{l-1}(p-1)} = p^{l}s' + 1,\;\gcd(p,s') = 1$, but $g^{d}-1\equiv 0\mod p^{1+l}$, thus $g^{d}-1\mid g^{p^{l-1}(p-1)}-1\implies g^{p^{l-1}(p-1)}= p^{1+l}s'' + 1$. But then $p^{1+l}s'' = p^{l}s'\implies p\mid s'$, which contradicts $\gcd(p,s') = 1$.

Now we have $p(p-1)\mid d$ and $d\mid p^{l}(p-1)\wedge d\nmid p^{l-1}(p-1)$. This means that $\nu_{p}(d) = l$ and $p-1\mid d$, thus $d = p^{l}(p-1)$.