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    primitive root

    Assume that the prime p > 2 has an odd primitive root g. Prove that 2p must have a primitive root.
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    Quote Originally Posted by dori1123 View Post
    Assume that the prime p > 2 has an odd primitive root g. Prove that 2p must have a primitive root.
    for any prime, 2p always has a primitive root because if p = 2, then U(\mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z} and if p > 2, then U(\mathbb{Z}/2p\mathbb{Z}) \cong U(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/(p-1)\mathbb{Z}.

    [ U(\mathbb{Z}/n\mathbb{Z}) here is the (multiplicative) group of units modulo n.]
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