Isomorphism of group of units to ring of $p$-adic integers

Question

For every $n>0$, let $U_n=1+p^n\mathbb{Z}_p$ be the group of units in $\mathbb{Z}_p$ which are $\equiv1\pmod{p^n}$. I know the proof of following theorem:

Let $x\in U_n$ with $n>0$ if $p\neq2$ and $n>1$ if $p=2$. There is a unique morphism of groups $f_x:\mathbb{Z}_p\to U_n$ such that if $a=(a_r)_{r>0}$ (with $a_r\in\mathbb{Z}/p^r\mathbb{Z}$) and $x=(x_r)_{r>0}$ (with $x_r\in U_n/U_{n+r}$), then $f_x(a)=(x_r^{a_r})_{r>0}$. If $x\notin U_{n+1}$, then $f_x$ is an isomorphism.

From this how can I prove the following corollary:

The group $U_1$ is isomorphic to $\mathbb{Z}_p$ for $p\neq2$ and the group $U_2$ is isomorphic to $\mathbb{Z}_2$ for $p=2$.

Also, I am not able to comprehend the point of showing isomorphism between a group and a ring.

Edit: Can following fact (derived from main theorem) be helpful:

$U_n$ is a free $\mathbb{Z}_p$-module of rank 1.

Is $\mathbb{Z}_p$ is the $p$-adic integers? the additive group as @arctic tern ask? the multiplicative group? — boaz, Jan 03 '17 at 07:19
@boaz $\mathbb{Z}_p$ is the ring of $p$-adic integers. In my lecture notes there is no mention to consider $\mathbb{Z}_p$ as additive group, but I think it should be so. $U_n$ is a multiplicative group. — rationalbeing, Jan 03 '17 at 07:24

score 3 · Answer 1 · answered Jan 03 '17 at 08:00

The point is that the additive structure of $\mathbb{Z}_p$ (i.e. the additive group of $\mathbb{Z}_p$ under addition) is relatively easy to understand.

By showing this isomorphism, it makes the multiplicative structure of $\mathbb{Z}_p$ relatively easy to understand as well.

Also, you can define $p$-adic versions of $\log$ and $\exp$ functions that compute these maps; these maps are just as useful in doing $p$-adic calculations as their real analogs are.

It may help to compare this to the fact that the multiplcative group of positive real numbers is isomorphic to the additive group of all real numbers.

score 2 · Accepted Answer · answered Jan 03 '17 at 07:42

The key sentence:

If $x\not\in U_{n+1}$, then $f_x$ is an isomorphism.

Suppose that $p$ is odd. Then $x = 1 + p\in U_1$, so by the proposition you stated, there is a homomorphism of groups $$ f_x : \Bbb Z_p\to U_1. $$ Moreover, because $1 + p\not\in U_2$, $f_x$ is an isomorphism (again by the proposition), so $\Bbb Z_p\cong U_1$ (as groups). If $p = 2$, the proof is exactly the same, taking $x = 1 + 2^2\in U_2\setminus U_3$.

To address the edit: $U_n$ is a free $\Bbb Z_p$-module of rank one, this means exactly that $U_n\cong\Bbb Z_p$ as groups (plus some extra data about the $\Bbb Z_p$-action on the group).

Isomorphism of group of units to ring of $p$-adic integers

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