Primitive roots mod $p^k$

Question

I'm having trouble understanding why this is true. Say m is the order of $g$ mod $2p^k$, with $\gcd(g, 2p^k)=1$, i.e. $g^m \equiv 1$ mod $2p^k$. How do I know that $g^m \equiv 1$ mod $p^k$?

Answer 1

More generally, $a \equiv b \bmod mn \implies a \equiv b \bmod m$.