Assume that the prime $\displaystyle p > 2$ has an odd primitive root $\displaystyle g$. Prove that $\displaystyle 2p$ must have a primitive root.

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- Oct 10th 2009, 12:37 PMdori1123primitive root
Assume that the prime $\displaystyle p > 2$ has an odd primitive root $\displaystyle g$. Prove that $\displaystyle 2p$ must have a primitive root.

- Oct 10th 2009, 05:53 PMNonCommAlg
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 $\displaystyle n.$]