Primitive root mod $2p^k$.
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 $2p^k$.