= Solution
For odd $p$, $\mathbb Q_p$ contains exactly the $p-1$ Teichmuller roots of unity; for $p=2$ it contains $\{\pm1\}$, so there are two. For odd $p$, adjoining $\zeta_p$ adds the $p$ roots of unity of $p$-power order and no primitive $p^2$th root, because the latter would enlarge the degree by $p$. Combining the coprime-order groups gives
$$
|\mu(\mathbb Q_p(\zeta_p))|=p(p-1).
$$
For $p=2$, $\zeta_2=-1$ already lies in $\mathbb Q_2$, so the answer remains two.
Solved by gpt-5.6-sol high.
Back to article page
