Proof explanation of $P(p^e) \cong \mathbb{Z}/(p-1)\mathbb{Z} \times \mathbb{Z}/(p^{e-1})\mathbb{Z}$.

Question

The following is from Classical Theory of Algebraic Numbers by Paulo Ribenboim : $P(p^e)$ is the set of all nonzero residue classes a modulo m, where gcd(a, m) = 1.

My question underlined and in order of "reading text":

1- The text concludes $P(p^e)$ to be cyclic based on being 'component-wise' cyclic': Why this is true, i.e. Why if $A=B \times C$ with $B\cap C={\{1}\}$ and $B$ and $C$ are both cyclic so is their product $A$?

2- Why $B\cap C = {\{\overline{1}\}}$?

3- a. Why $\overline{a}^{p^{e-1}} \in B$ for every $\overline{a} \in P(p)$? And, b. Isn't it supposed to be $\overline{\overline{a}} \in P(p)$ or $\overline{a} \in P(p^e)$, because $\overline{a} \notin P(p)$? if so, still the question "a." without answer.

4- Why $\overline{\overline{a}}^{p^{e-1}} = \overline{\overline{a}}$? (If it was $\overline{\overline{a}}^{p} \equiv \overline{\overline{a}}$ yes true by Fermat's theorem.)

Simple clear explanation would be much appreciated.

Answer 1

1. Let $x$ be a generator of $B$ and let $y$ be a generator of $C$. Then $o(x, y) = \operatorname{lcm} (o(x), o(y)) = \frac{o(x) \cdot o(y)}{\gcd(o(x), o(y))}=o(x) \cdot o(y).$

2. We know from above that $|C|=p^{e-1}$, with $p$ prime, so every element of $C$ must have some power of $p$ as its order. On the other hand, every element of $B$ has order dividing $p-1$, so no element of $B$ can have a power of $p$ as its order, other than $1$.

3. If $x \in P(p)$, then $(x^{p^{e-1}})\in P(p)$. Also, $|P(p)| = p-1$, so $(x^{p^{e-1}})\in P(p) \Rightarrow (x^{p^{e-1}})^{p-1}=1 \Rightarrow (x^{p^{e-1}})\in B$ by the definition of $B$.

4. You can easily see this by induction. $x^{p^2}=(x^p)^p = x^p=x$, and $x^{p^{n+1}}=(x^{p^n})^p$. By inductive hypothesis $x^{p^n}=x$, so $x^{p^{n+1}}=x^p=x$.