Primitive root mod $2p^k$.

Question

We say that $g$ is a primitive root mod $n$ if $\langle g\rangle=U_n$.I want to prove the following theorem:

Theorem

If $g$ is a primitive root modulo $p^k$,then either $g$ or $g+p^k$ is a primitive root modulo $2p^k$.Moreover,we can guarantee that the odd one out of $g$ and $g+p^k$ will be the one which is primitive root.

Can someone provide me hints on how to proceed.I think I can do it myself if some hint is provided.

Hint: only the odd one can be a primitive root, then count the number of relative primes modulo $2p^k$. — abiessu, Dec 14 '21 at 05:55
@KishalaySarkar FYI, a couple of closely related questions on this site are Primitive roots of odd primes and Proving that $r+p^{\alpha}$ is a primitive root of $2p^{\alpha}$. — John Omielan, Dec 14 '21 at 06:24

markvs · Accepted Answer · 2021-12-14T06:22:15.763

Note that the Euler function $\phi(2p^k)=(p-1)p^{k-1}=\phi(p^k)$. Suppose that $g$ is odd. Then it is coprime with $2p^k$. Let $s$ be the order of $g \pmod{2p^k}$. Then $s$ divides $(p-1)p^{k-1}$, and $g^s\equiv 1\pmod{p^k}$. Then $s$ is divisible by $\phi(p^k)=(p-1)p^{k-1}$. Hence $s=(p-1)p^{k-1}$ and $g$ is a primitive root $\pmod{2p^k}$. Now suppose $g$ is even. Then $g+p^k$ is odd and $\equiv g\pmod{p^k}$, so $g+p^k$ is an odd primitive root for $p^k$ and we can proceed as before: $g+p^k$ is a primitive root for $2p^k$.

Primitive root mod $2p^k$.

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