Assume that the prime $\displaystyle p > 2$ has an odd primitive root $\displaystyle g$. Prove that $\displaystyle 2p$ must have a primitive root.
for any prime, $\displaystyle 2p$ always has a primitive root because if p = 2, then $\displaystyle U(\mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ and if p > 2, then $\displaystyle U(\mathbb{Z}/2p\mathbb{Z}) \cong U(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/(p-1)\mathbb{Z}.$
[$\displaystyle U(\mathbb{Z}/n\mathbb{Z})$ here is the (multiplicative) group of units modulo $\displaystyle n.$]