Assume $g^{p^n}\pmod{p}=1$, with $p$ prime.
- Is $g$ a primitive root mod $p$? If not why?
- How about the converse? If not why?
This is not a homework question.
My attempt:
Update - Corrections to my attempts 1 & 2
- Since $g^{p^n}\pmod{p}=1$, then order of $g | p^n$.
- Assume $g$ a primitive root mod $p$, then there is no exponent $a$ s.t $g^a \pmod{p} =1$. By definition this is an element of order $p−1$.
Assuming $p$ is prime, then $g= g^p\pmod{p}$ for all $g\in\mathbb{Z}.$ This seems simple enough, but there is an important extension. Given $n>0$, define $r(g)$ such that $\;r(g)= g^{p^{n-1}}\pmod{p^n}.$ Then if $g\ne 0\pmod{p},\; r(g)\;$ is a $(p\!-\!1)$th root of unity. This is because the order of the multiplicative group $\!\pmod{p^n}$ is $p^{n-1}(p-1)$ and so $g^{p^{n-1}(p-1)}=1\pmod{p^n}$ implies $r(g)^{p-1}= 1\pmod{p^n}.$ Therefore, $\;g^{p^n}=r(g)\pmod{p^n},\;$ but $\;r(g)\;$ has $\;p-1\;$ values including $1$.
Now if $g_1$ is a primitive $(p\!-\!1)$th root of unity, then ${g_1}^k$ for all $0\le k<p\!-\!1\;$ are all the $(p\!-\!1)$th roots of unity.
Notice that if $g\ne 0\pmod{p},$ then $\;r(g) = 1\pmod{p}$ by Fermat's little theorem, but for $k\in\mathbb{Z}$ $(pk)^{p^n}=0\pmod{p}$.
Notice that in a finite field with $p^n$ elements that $g^{p^n}=g$ for all $g,\;$ but the multiplicative group order here is $p^n-1,$ implying that if $g\ne 0,$ then $\;g^{p^n-1}=1.$