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show that if $g$ and $g'$ are primitive roots modulo an odd prime $p$, then $gg'$ is not primitive root of $p$. I know that $g^\phi\equiv 1 \pmod {p}$ .Similarly for $g'$ , but what can I do further.

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    $\begingroup$ here $\phi$ replaces with $\phi(p)$ $\endgroup$ – Abhishek Kumar Oct 5 '15 at 7:35
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Let $g^{\phi(p)/2}= a$, by Fermat's Little Theorem, $p|(a^2-1)=(a-1)(a+1)$

As ord$_pg=\phi(p),p\nmid(a-1)\implies g^{\phi(p)/2}\equiv-1\pmod p$

$$\implies (gg')^{\phi(p-1)/2}\equiv?$$

$\implies$ord$_p(gg)'\le\dfrac{\phi(p)}2<\phi(p)$

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As $g$ is a primitive root, we can write $g'=g^i$ for some $i$ with $(i,p-1)=1$. In particular, $i$ must be odd (as $(2j,p-1)≥2$). But then $i+1$ is even so $gg'=g^{i+1}$ is not a primitive root.

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