Dummit and Foote's proof that $(\mathbb{Z}/p^{\alpha}\mathbb{Z})^*$ is cyclic of order $p^{\alpha - 1}(p-1)$ Here is the proof from Dummit and Foote. I follow the proof until the part at the end where it says that all Sylow subgroups are cyclic, but I do not see how that implies that $(\mathbb{Z}/p^{\alpha}\mathbb{Z})^*$ is cyclic.
This is $(3)$ of Corollary $20$ of section $9.5$ on page $314$


Help would be really appreciated.
 A: There are two key facts that explain why all Sylow subgroups being cyclic implies that the group itself is cyclic. I'll let $G = (\mathbb Z / p^\alpha \mathbb Z)^\times$.
1) $G$ is the internal product of its Sylow subgroups.
2) The product of cyclic subgroups of $G$ of relatively prime order is a cyclic subgroup.
In case you're unfamiliar, I'll define the product of two subgroups. Let $H, K \subseteq G$ be subgroups. Then $HK = \{hk : h \in H, k \in K\}$. This definition makes sense for all groups. Note that this is not, in general, a subgroup of $G$. If, say $K$ is a normal subgroup, then $HK = KH$ and it is a subgroup. Of course, in the abelian case, (which is where we are) all subgroups are normal.
The statement (1) above turns out to be true for any finite abelian group. In fact, this condition is called being nilpotent. Statement (2) is a completely general fact of groups, so long as one of the cyclic subgroups is normal.
To prove statement (1), we can appeal to a counting result. Indeed, for $H, K$ subgroups of $G$, $|HK| =|H||K|/|H \cap K|$. Hence, if $H \cap K = \{e\}$, then $|HK| = |H||K|$. Indeed, by induction, of we have $H_1, \dots, H_k$ subgroups of $G$ such that each $H_i \cap H_j = \{e\}$, then $|H_1 \cdots H_k| = |H_1| \cdots |H_k|$. Let $H_q$ be the distinct Sylow subgroups of $G$, ranging over $q \mid |G|$. These certainly intersect trivially, as they have coprime order. Hence, $|G| = \prod_{q \mid |G|} |H_q| = |\prod_{q \mid |G|} H_q|$. Thus, $\prod_{q \mid |G|} H_q$, i.e. the product of the Sylow subgroups of $G$, is a subgroup of $G$ with the same cardinality. Hence, $\prod_{q \mid |G|} H_q = G$.
For (2), let $H = \langle a \rangle$ and $K = \langle b \rangle$ be cyclic subgroups of $G$. Let $|H| = n$, $|K| = m$. Assume that $n, m$ are coprime. Then as $a$ and $b$ commute and have coprime order, the order of $ab$ is $nm$. Furthermore, $|HK| = nm$ by the counting result stated above. Hence, $\langle ab \rangle \subseteq HK$ is a subgroup of the same order, so $HK = \langle ab \rangle$, which is cyclic.
Putting these together, we see that by (1), $G = \prod_{q \mid |G|} H_q$, where $H_q$ denotes the unique Sylow $q$ subgroup of $G$. Proceed by induction on the number of prime divisors of $G$. The base case follows as each $H_q$ was proven to be cyclic in Dummit and Foote. Fix some $q \mid |G|$ and let $K_q = \prod_{Q \neq q} H_Q$. Of course, $H_q K_q = G$. By induction, $K_q$ is cyclic. Furthermore, let $|H_q| = q^k$. Then $K_q = |G|/q^k$, which is relatively prime to $q^k$ as $H_q$ is Sylow. Hence, by statement (2) above, $G = K_q H_q$ is cyclic.
