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Along the same vein of another question I've posted is another question that merges group theory with number theory. I havn't taken a formal course on group theory, and is likely why I'm stumped at these questions posed from elementary number theory. The question goes as

Let $p$ be an odd prime and $g$ a generator of the group $(\mathbb{Z}/p \mathbb{Z})^*$. I need to show that either $g$ or $g+p$ is a generator of $(\mathbb{Z}/p^2 \mathbb{Z})^*$

I'm not sure where to begin. Just need a push in the right direction.

Thanks!

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    Note that an element of $(\mathbb Z/p^2\mathbb Z)^*$ has order $p(p-1)$ (that is, it is a generator) if it has order >p. Note further that is order $\pmod {p^2}$ cannot be smaller than its order $\pmod p$.2012-11-05

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If $g$ is not primitive mod $p^2$, the fact that $\mathrm{ord}(g)$ divides $p(p-1)$ and is a multiple of $p-1$ implies that $\mathrm{ord}(g)=p-1$ exactly. The same holds for $g+p$.

So if neither $g$ nor $g+p$ is primitive mod $p^2$, we must have $g^{p-1} \equiv 1 \mod p^2$ and $(g+p)^{p-1} \equiv 1 \mod p^2$. (In particular, $g^{p-1} \equiv (g+p)^{p-1} \mod p^2$.) Yet $(g+p)^{p-1} \equiv g^{p-1} + \binom{p-1}{1}g^{p-2}p \equiv g^{p-1}-pg^{p-2}\mod p^2,$ and if this is congruent to $g^{p-1}$ mod $p^2$ it follows that $pg^{p-2} \equiv 0 \mod p^2$. This forces $g \equiv 0 \mod p$ as $\mathbb{Z}/p\mathbb{Z}$ is an integral domain. This is a contradiction, and our result follows.

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    More generally, let $\phi :G \to H$ be a group homomorphism. If $g^k =e_G$, then $\phi(g)^k = \phi(e_G)=e_H$, so $\mathrm{ord}(\phi(g))$ divides $\mathrm{ord}(g)$ (if both are finite). In our case, our map is projection $\pi: \mathbb{Z}/p^2 \mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$.2012-11-08