Suppose that $a \in (\mathbb{Z}/p\mathbb{Z})^{\times}$ with $a \not\equiv \pm 1$ (mod $p$). Prove that if $a$ is not a generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ then $-a$ is a generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}$.
Proving the additive inverse is a generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}$
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number-theory
prime-numbers
cyclic-groups
1 Answers
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This is false: if $p=13$ and $a=3$, then $a^3\equiv 1$ (mod $13$) and $(-a)^6\equiv 1$ (mod $13$). Hence neither $a$ nor $-a$ can generate $(\mathbb{Z}/13\mathbb{Z})^{\times}$.