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 $\mathbb{Z}\_p ^\times \cong C\_{p-1} \times (1+ p \mathbb{Z}\_p)$.



The Unit Group of $\mathbb{Z}_p$
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.


