I am solving the following exercise:
- Show that ${\rm Aut}(\mathbb{Z}_{p^2})\cong \mathbb{Z}_{p^2 - p}$ for prime $p$. (Hint: Let $m$ be a primitive root modulo $p$, and show that either $m$ or $m + p$ is a primitive root modulo $p^2$ )
Now what confuses the hell out of me is this hint. It is not that I don't understand how we could solve the exercise using this machinery, but I don't understand why complicate things. Namely, the previous chapter contains the following result:
Proposition: Let $G=\langle x\rangle \cong \mathbb{Z}_n$ for $n\in \mathbb{N}$, and for each $0\leq m<n$ let $\sigma_m$ be the endomorphism of $G$ sending $x$ to $x^m$. Then ${\rm Aut}(G)$ consists precisely of those $\sigma_m$ for which $m\neq0$ and $\gcd(m,n)=1$.
which gives us the order of the ${\rm Aut}(G)$ for free as $\phi (p^2) = p^2 - p$, where $\phi$ is Euler's phi function. And combined with this result from the same chapter
Theorem: ${\rm Aut}(\mathbb{Z}_n)$ is cyclic iff $n=2$ or $4$, or $n=p^k$ or $2p^k$ for some odd prime $p$ and some $k\in \mathbb{N}$.
that tells us that it is a cyclic group, the result is direct.
This is a graduate book, so some understanding of number theory and basic group theory is assumed. Am I missing something, or are they unnecessarily complicating things?