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For odd $p$ there is a decompositioninto the valuation factor, the Teichmuller representative factor, and the group of principal units. If $x$ is a $((p-1)m)$th power for every $m$ coprime to $p$, its valuation is divisible by every such $(p-1)m$, and is therefore zero. Its residue in ${\mathbb{F}}_p^{\times }$ is a $(p-1)$st power, hence is one. Thus $x\in 1+p{\mathbb{Z}}_p$.

Conversely, exponentiation by any integer $n$ coprime to $p$ is an automorphism of $1+p{\mathbb{Z}}_p$. This follows either from the principal-unit logarithm, under which it becomes multiplication by $n$, or by applying the Hensel lemma to $Y^n-x$. Since $(p-1)m$ is coprime to $p$, every $x\in 1+p{\mathbb{Z}}_p$ has a $((p-1)m)$th root for every allowed $m$.

Let ${\mathfrak{m}}_K$ be the maximal ideal. For all sufficiently large $r$, the convergent p-adic logarithm and p-adic exponential give the principal-unit logarithm isomorphismMultiplication by a power of a uniformizer identifies the additive group ${\mathfrak{m}}_K^r$ with $({\mathcal{O}}_K,+)$. Since $1+{\mathfrak{m}}_K^r$ has finite index in ${\mathcal{O}}_K^{\times }$, it is the required subgroup.

For sufficiently large $r$, the series for the p-adic logarithm and p-adic exponential converge on $1+{\pi }^r{\mathcal{O}}_K$ and ${\pi }^r{\mathcal{O}}_K$ and are inverse homomorphisms. Hence the principal-unit logarithm giveswhere the last map is division by ${\pi }^r$.

Now take $K={\mathbb{Q}}_3({\zeta }_3)$ and $\pi ={\zeta }_3-1$. This is a uniformizer, the residue field is ${\mathbb{F}}_3$, and its Teichmuller units are ${\mu }_2=\{\pm 1\}$. ThusThe image of ${\zeta }_3$ generates $U_1/U_2$, and ${\mu }_3\cap U_2=1$, so $U_1={\mu }_3\times U_2$. The logarithm and exponential already converge inversely on $U_2$, giving $U_2\cong ({\pi }^2{\mathcal{O}}_K,+)\cong ({\mathcal{O}}_K,+)$. Since ${\mu }_2\times {\mu }_3\cong {\mu }_6$, this proves the unit group of Q3 zeta3 decomposition