Let $p$ be an odd prime. Suppose that $a$ is an odd integer and also $a$ is a primitive root mod $p$. Show that $a$ is also a primitive root mod $2p$. Let $p$ be an odd prime. Suppose that $a$ is an odd integer and also
$a$ is a primitive root modulo $p$. Show that a is also a primitive root modulo
$2p$.
Any hints will be appreciated. Thanks very much.

The natural projection $\pi:U(2p) \to U(p)$ given by $x \bmod 2p \mapsto x \bmod p$ is a homomorphism of groups.
The kernel of $\pi$ is trivial because $x \equiv 1 \bmod p$ implies $x \equiv 1 \bmod 2p$ since we must have $x \equiv 1 \bmod 2$ if $x \in U(2p)$. By the Chinese remainder theorem, we then have $x \equiv 1 \bmod 2p$.
Thus, $\pi$ is injective and so it is surjective, because $U(2p)$ and $U(p)$ have the same size.
Therefore, $\pi$ is an isomorphism and must map generators to generators.
Concretely, take $a$ as in the question. Then $a \in U(2p)$ and $\pi(a)$ has order $p-1$ in $U(p)$. Therefore, $a$ also has order $p-1$ in $U(2p)$, which is the order of $U(2p)$. Thefore, $a$ is a primitive root mod $2p$.
