I was reading the second answer given in 2 is a primitive root mod $3^h$ for any positive integer $h$
However, I don't see the connection between this lemma and $2$ being a primitive root. Why are they equivalent? How can I prove? Any help is appreciated
If $x$ is a primitive root $\!\bmod p^n$ and there's some $k$ such that $x^k\equiv1+p^n\pmod{p^{n+1}}$ then $x$ is a primitive root $\bmod p^{n+1}$.
Hint: apply the following, noting that the hypothesis excludes the first possibility $\,k = j$.
Lemma $ $ If $\,a\,$ has order $\,j,\,$ mod $\,p^n$ then $\,a\,$ has order $\,k = \color{#c00}j\,$ or $\,\color{#0a0}{pj},\,$ mod $p^{n+1}$
Proof $\,\ \ \ \ \ a^k\equiv 1\pmod{p^{n+1}}\,\Rightarrow\, a^k\equiv 1\pmod{p^{n}}\Rightarrow \color{#c00}j\mid k$
By $\mu$LTE $\:\!\ \ a^j\equiv 1\pmod{p^{n}}\Rightarrow a^{pj}\equiv 1\pmod{p^{n+1}}\Rightarrow k\mid \color{#0a0}{pj}$