Let $\Bbb{Q}^Z$ be composition of all $\Bbb{Z}_p$-extension over $\Bbb{Q}$, which is defined as galois extension over $\Bbb{Q}$ whose galois group is isomorphic to $\Bbb{Z}_p$ as an abelian group.
I encountered an argument which says
A fact ' $\text{Gal}(\Bbb{Q}^Z/\Bbb{Q})$ is isomorphic to $\Bbb{Z}_p$ ' implies there is only one $\Bbb{Z}_p$-extension over $\Bbb{Q}$.
I cannot follow this argument. Why $\text{Gal}(\Bbb{Q}^Z/\Bbb{Q})$ is isomorphic to $\Bbb{Z}_p$ ' implies there is only one $\Bbb{Z}_p$-extension over $\Bbb{Q}$ ?
P.S My question is
Why $\text{Gal}(\Bbb{Q}^Z/\Bbb{Q})$ is isomorphic to $\Bbb{Z}_p$ ' implies there is only one $\Bbb{Z}_p$-extension over $\Bbb{Q}$ ?
Can anybody follow this logic(not asking the titled question's proof except for this logic) ?
Every finite abelian extension of $\mathbf Q$ lies in a cyclotomic extension of $\mathbf Q$, so if ${\rm Gal}(K/\mathbf Q) \cong \mathbf Z/p^n\mathbf Z$ then $K \subset \mathbf Q(\zeta_m)$ for some $m$. A number field inside a $\mathbf Z_p$-extension of $\mathbf Q$ is ramified only at $p$, and the subfields of $\mathbf Q(\zeta_m)$ that are ramified only at $p$ lie inside $\mathbf Q(\zeta_{p^r})$, where $p^r$ is the highest power of $p$ dividing $m$. Therefore $K$ is a subfield of a $p$-power cyclotomic extension of $\mathbf Q$.
For all $p$ the Galois group ${\rm Gal}(\mathbf Q(\zeta_{p^r})/\mathbf Q)$ is isomorphic to $(\mathbf Z/p^r\mathbf Z)^\times$ and has order $\varphi(p^r) = p^{r-1}(p-1)$. When $p \not= 2$ this group is cyclic, so $\mathbf Q(\zeta_{p^r})$ has a unique (cyclic) subextension of $\mathbf Q$ of each $p$-power degree that divides $p^{r-1}$. Since the fields $\mathbf Q(\zeta_{p^r})$ as $r$ grows form an increasing tower, each layer of a $\mathbf Z_p$-extension of $\mathbf Q$ has only one possibility. Therefore there is at most one $\mathbf Z_p$-extension of $\mathbf Q$. Since we know how to create at least one $\mathbf Z_p$-extension of $\mathbf Q$ (the cyclotomic $\mathbf Z_p$-extension), this extension exists and is unique.
For $p \not= 2$ the argument needs a little more work since $(\mathbf Z/2^r\mathbf Z)^\times$ is not cyclic when $r \geq 3$. Layers of a $\mathbf Z_2$-extension are ramified only at $2$, so these layers are inside the maximal real subfield $\mathbf Q(\zeta_{2^r})^+$, as this is the subfield fixed by complex conjugation: subfields unramified at the Archimedean place are contained in this subfield. The extension $\mathbf Q(\zeta_{2^r})^+/\mathbf Q$ is cyclic of degree $2^{r-1}$, so we can run through the same argument as above to see $\mathbf Q$ has a unique $\mathbf Z_2$-extension.