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How would one go about proving/disproving whether $g$, a multiplicative generator mod $p$, is also a multiplicative generator mod $p^2$?

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I assume $p$ is an odd prime.

It suffices to check if $g^{p-1}$ is $1$ mod $p^2$ or not. If it is, then it is not a generator. (Clear as its order is $p-1$ and not the needed $(p-1)p$.)

If it is something else, then it is. This is because its order mod $p^2$ must be a multiple of the order mod $p$, which is $p-1$ and a divisor of $p(p-1)$. So if it is greater than $(p-1)$, it can only be $p(p-1)$.

Both cases can happen. One can show though that at least on of $g$ and $g+p$ will be a generator mod $p^2$.

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  • $\begingroup$ it looks close to the lifting in Hensel lemma ? $\endgroup$ – reuns Nov 1 '16 at 22:10
  • $\begingroup$ Yes, it's closely related to this. $\endgroup$ – quid Nov 1 '16 at 22:24

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