Let $\mathbb{Z}_p$ the ring of $p$-adic numbers. It's known that the multiplicative unit group $\mathbb{Z}_p ^\times$ can be set theoretically described as $\bigcup _{1 \le a \le p-1} a+ p\mathbb{Z}_p$.

I want to know how to see that we have also the group isomorphism $\mathbb{Z}_p ^\times \cong C_{p-1} \times (1+ p \mathbb{Z}_p)$.

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    $\begingroup$ The "reduction modulo $p$" map defines a short exact sequence, $$0\to 1+p\mathbb{Z}_p \to \mathbb{Z}_p^\times \to C_{p-1} \to 0.$$ It remains to split this exact sequence. The splitting copy of $C_{p-1}$ is the group of $(p-1)^{\text{st}}$ roots of unity in $\mathbb{Z}_p^\times$. You can use Hensel's Lemma to prove that this subgroup surjects isomorphically to $C_{p-1}$ via the "reduction modulo $p$" homomorphism. $\endgroup$ – Jason Starr Jan 31 '18 at 17:45
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    $\begingroup$ You can do this explicitly. Following @JasonStarr's hint, it suffices, for each $n \in (\mathbb Z/p)^\times$, to construct a $(p-1)$th root of unity whose mod-$p$ reduction is $n$. You can do this algorithmically. The "1s digit" of your number is $n$, and you check $n^{p-1} = 1 \mod p$ by Fermat's little theorem. Now you need to compute the "ps digit": you want $(ap + n)^{p-1} = 1 \mod p^2$; this simplifies to some linear equation in $a \in \mathbb Z/p$, which you can solve. Rinse and repeat. The algorithm is analogous to "long square roots" xlinux.nist.gov/dads/HTML/squareRoot.html. $\endgroup$ – Theo Johnson-Freyd Jan 31 '18 at 17:53
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    $\begingroup$ en.wikipedia.org/wiki/Teichm%C3%BCller_character $\endgroup$ – Vladimir Dotsenko Jan 31 '18 at 17:59
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    $\begingroup$ The procedure in Theo's comment also works to prove Hensel's lemma for any polynomial to which it applies. $\endgroup$ – Asvin Jan 31 '18 at 20:04

This is probably more suitable to MathStackExchange, although there may well be researchers in allied areas that aren't aware of this fact. So in that spirit, here's the standard proof. There is an exact sequence $$ 1 \to 1+p\mathbb Z_p \to \mathbb Z_p^\times \to \mathbb F_p^\times \to 1 .$$ So one needs split this exact sequence, since it is well known that $\mathbb F_p^\times$ is cyclic. The idea is that Hensel's lemma tells us that every root $\bar a$ of $X^p-X$ in $\mathbb F_p$ lifts to a unique root $a\in\mathbb Z_p$ satisfying $a\equiv\bar a\pmod p$. Discarding $0$, we see that every element of $\mathbb F_p^\times$ (all of which are roots of $X^{p-1}-1$ by Fermat's little theorem, or more intrinsically, by Lagrange's theorem) lifts to a unique root of $X^{p-1}-1$ in $\mathbb Z_p$. These lifts are $(p-1)$st roots of unity in $\mathbb Z_p^\times$, and this lifting map $\mathbb F_p^\times\to\mu_{p-1}\subset\mathbb Z_p^\times$ is clearly a homomorphism due to the uniqueness of the lift. The lifting map $\mathbb F_p^\times\to\mu_{p-1}\subset\mathbb Z_p^\times$ is called the Teichmuller character, as has been noted by others in the comments.

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    $\begingroup$ Maybe it is also worth pointing out that, for $p$ odd, the standard Taylor series for logarithm and exponential give isomorphisms between the multiplicative group $1+p \mathbb{Z}_p$ and the additive group $p \mathbb{Z}_p$. $\endgroup$ – David E Speyer Jan 31 '18 at 18:32

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