= Solution
One useful form of <Hensel lemma> is this: if $R$ is a complete <discrete valuation ring>, $f\in R[X]$, and
$$
f(a_1)\equiv0\pmod\pi,
\qquad
f'(a_1)\not\equiv0\pmod\pi,
$$
then there is a unique $a\in R$ such that $f(a)=0$ and $a\equiv a_1\pmod\pi$. Indeed, after constructing $a_r$ with $f(a_r)\equiv0\pmod{\pi^r}$, choose the unique $t$ modulo $\pi$ for which
$$
f(a_r)+\pi^rtf'(a_r)\equiv0\pmod{\pi^{r+1}}
$$
and put $a_{r+1}=a_r+\pi^rt$. The resulting <Cauchy sequence> converges by completeness, and the same first-order congruence proves uniqueness.
Apply this to $X^{p-1}-1$. Every nonzero class in $\mathbb F_p$ is a simple root, so it has a unique <Teichmuller representative> in $\mathbb Z_p$. These give all roots of unity of order prime to $p$. For odd $p$, the group $1+p\mathbb Z_p$ has no nontrivial torsion: if $u\ne1$, then the binomial theorem gives $v_p(u^p-1)=v_p(u-1)+1$, which is incompatible with finite $p$-power order. Hence
$$
\mu(\mathbb Q_p)\cong C_{p-1}qquad(p\text{ odd}).
$$
For $p=2$, the subgroup $1+4\mathbb Z_2$ is torsion-free by the same argument, while $-1$ supplies the extra torsion element. Thus $\mu(\mathbb Q_2)=\{\pm1\}\cong C_2$. This describes the <roots of unity in a p-adic field> for $K=\mathbb Q_p$.
Solved by gpt-5.6-sol high.
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