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Let $a$ be the generator for the group of units $\bmod{p^k}$ for $p$ an odd prime and k a positive integer, i.e. $\langle a \rangle= (\mathbb{Z}/p^k\mathbb{Z})^{\times}$. Is it true that $a$ is also a generator for $(\mathbb{Z}/p\mathbb{Z})^{\times}$? I then want to be able to generalize for other powers of p. Thanks.

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    $\begingroup$ Dear Edison: You can take a look at this answer. $\endgroup$ – Pierre-Yves Gaillard Feb 3 '12 at 18:32
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    $\begingroup$ What other powers of $p$ than positive integer powers do you envisage to generalize to? Fractional powers? Negative powers? Imaginary powers? It doesn't seem to make much sense. $\endgroup$ – Marc van Leeuwen Feb 3 '12 at 18:57
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    $\begingroup$ More interesting is the fact that one can go the other way: if $a$ is a generator for the group of units mod $p^2$, then it is a generator for the group of units mod $p^k$ for all $k$. $\endgroup$ – Arturo Magidin Feb 3 '12 at 19:04
  • $\begingroup$ Plus, if $a$ is a primitive root modulo $p^e$, then either $a$ is odd and primitive mod $2p^e$, or $a$ is even and $a+p^e$ is primitive modulo $2p^e$. ($p$ odd.) $\endgroup$ – anon Feb 3 '12 at 19:16
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The answer to your question is trivially yes: if the powers of $a$ run through all classes modulo $p^k$ except those of the multiples of $p$, then by (further) reduction modulo $p$ they certainly also run through all classes modulo $p$ except that of $0$.

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