p-adic integers, isomorphism, proof explanation I am studying the following proof and need some help at some points.
While going through the proof, I numerate my questions, to refer to them at the end.
For the notation note, that $U_n=1+p^n\mathbb{Z}_p$ and $U=\mathbb{Z}_p^\times$ (the group of p-adic units).
Theorem:
If $p\neq 2$, $U_1$ is isomorphic to $\mathbb{Z}_p$.
If $p=2, U_1=\{\pm 1\}\times U_2$ and $U_2$ is isomorphic to $\mathbb{Z}_2$.
Proof:
Consider first the case $p\neq 2$.
Choose an element $\alpha\in U_1- U_2$, for example $\alpha=1+p$. (1)
By the lemma:

Let $x\in U_n- U_{n+1}$ with $n\geq 1$ if $p\neq 2$ and $n\geq 2$ if $p=2$. Then $x^p\in U_{n+1}- U_{n+2}$

we have $\alpha^{p^i}\in U_{i+1}- U_{i+2}$.
Let $\alpha_n$ be the image of $\alpha$ in $U_1/ U_n$.
We have $\alpha_n^{p^{n-2}}\neq 1$ and $\alpha_n^{p^{n-1}}=1$. (2)
But $U_1/ U_n$ is of order $p-1$, hence it is a cyclic group generated by $\alpha_n$.
Now denote by $\theta_{n,\alpha}:\mathbb{Z}/p^{n-1}\mathbb{Z}\to U_1/U_n$ the isomorphism $z\mapsto \alpha_n^z$.
The diagram:
$\require{AMScd}$
\begin{CD}
\mathbb{Z}/p^n\mathbb{Z} @>{\theta_{n+1,\alpha}}>> U_1/U_{n+1}\\
@VVV @VVV\\
\mathbb{Z}/p^{n-1}\mathbb{Z} @>{\theta_{n,\alpha}}>> U_1/U_n
\end{CD}
is commutative.
From this one sees that the $\theta_{n,\alpha}$ define an isomorphism (3)
$\theta:\mathbb{Z}_p\to U_1$ hence the proposition for $p\neq 2$.
Suppose now that $p=2$. 
Choose $\alpha\in U_2- U_3$, that is $\alpha\equiv 5\mod 8$. (4)
Define as above isomorphism $\theta_\alpha:\mathbb{Z}_2\to U_2$. 
On the other hand, the homomorphism $U_1\to U_1\setminus U_2\cong\mathbb{Z}/2\mathbb{Z}$ induces an isomorphism of $\{\pm 1\}$ onto $\mathbb{Z}/2\mathbb{Z}$ (5)
From this we get $U_1=\{\pm 1\}\times U_2$
Questions:
(1) 
It appears that the choice of $\alpha\in U_1-U_2$ is free, since the given lemma we get always a generator of the cyclic group for every $x\in U_1-U_2$ as long as $x\neq 1$.
(2)
The image of $\alpha$ in $U_1/U_n$ is given by $\alpha_n\mod (1+p^n\mathbb{Z}_p)$. 
What can we say about $\alpha_n$? Holds $\alpha_n\equiv 1\mod p^n$?
How can we see, that $\alpha_n^{p^{n-2}}\neq 1$ and $\alpha_n^{p^{n-1}}=1$?
Why is this enough to see, that $U_1/U_n$ is cyclic.
I guess, that $\alpha_n^{p^{n-1}}=1$ and $\alpha_n^{p^{n-2}}=\alpha_n\neq 1$ ensues from Fermat's little theorem. Am I right?
(3)
How exactly do we get the isomorphism $\theta:\mathbb{Z}_p\to U_1$
It is $\mathbb{Z}_p=\displaystyle{\lim_{\leftarrow}}~ \mathbb{Z}/p^{n-1}\mathbb{Z}$ and $U_1=\displaystyle{\lim_{\leftarrow}}~ U_1/U_n$
I see that we can extend the diagram and reduce $p^{n}$ any further until $n=0$.
But how do we get the projective limit? We get finite isomorphisms, but is that enough for the projective limit?
(4)
Why is for $\alpha\in U_2-U_3$ that $\alpha\equiv 5\mod 8$. 
$\alpha\in 1+4\mathbb{Z}_2$ that means $\alpha\equiv 1\mod 4$. Since $\alpha\notin (1+8\mathbb{Z}_2)$ it is $\alpha\not\equiv 1\mod 8$ which means, that $\alpha\equiv 5\mod 8$. 
(5)
We define an isomorphism $\theta_\alpha: \mathbb{Z}_2\to U_1$ as above. That is clear.
How do we see from the homomorphism $U_1\to U_1/U_2\cong \mathbb{Z}/2\mathbb{Z}$
That we get an isomorphism from $\{\pm 1\}\to \mathbb{Z}/2\mathbb{Z}$
How do we see that $U_1/U_2\cong \mathbb{Z}/2\mathbb{Z}$?
I want to give a function $f: U_1\to\mathbb{Z}/2\mathbb{Z}$ which is surjective and has kernel $U_2$.
Can this be done by $\alpha\mapsto \alpha\mod 4$? I am struggeling with the elements of the form $1+2\mathbb{Z}_2=U_1$. We can not say, that $U_1\cong 1+\mathbb{Z}/2\mathbb{Z}$, can we?
Thanks in advance for your help and comments.
 A: I see that you managed to confuse yourself quite a bit over it, so let's start with simple things. Forget about $p=2$ altogether (it just there to complicate every step by a special case; you can return to it later once you figured the rest out).
1) Yes, you can take anything in $U_1\setminus U_2$ though the condition $x\in U_1\setminus U_2$ is much stronger than $x\ne 1$. It means $x=1+yp+\dots$ for some "$p$-adic digit" $y\ne 0$.
2) $\alpha_n$ is the full remainder of $\alpha$ modulo $p^n$, so it is just the first $n$ digits of $\alpha$ (compare with the standard decimal representation; if you go modulo $10^n$, you just operate on the last $n$ digits ignoring all transfers to higher positions; $p$-adic integers are the same, just infinite and written low to high powers instead of high to low in the decimal notation. Obviously $\alpha_n\ne 1\mod p^n$. Note that $U_1/U_n$ is the multiplicative group (compare to the multiplicative group of last $n$ decimal digits of numbers ending in $1$).
$\alpha_n^{p^{n-2}}$ is the equivalence class of $\alpha^{p^{n-2}}\in U_{n-1}\setminus U_n$, so it is definitely not $1$ in $U_1/U_n$ (the $n$-th digit is still not $0$) but the next raising to the power $p$ will kill it.
$U_1/U_n$ has exactly $p^{n-1}$ elements (the first digit is $1$ and you have $p$ choices for each of the remaining $n-1$ digits. So, you have a finite group of order $p^{n-1}$ and an element $\alpha_n$ in it that has exactly the same order (recall that the order of any group element is a divisor of the group order and we have just excluded any divisor of $p^{n-1}$ strictly less than $p^{n-1}$. Thus, all powers of $\alpha^n$ from $0$ to $p^{n-1}-1$ are different and thus cover the whole group once.
(3) We can use this isomorphism $(Z_p/p^{n-1}Z_p,+)\ni z\mapsto \alpha_n^z\in (U_1/U_n,\times)$ to define the $n$ digits of $\theta(z)$ from the $n-1$ digits of $z$ and vice versa. The diagram shows that this process is consistent in the sense that getting more digits with larger $n$ doesn't change the digits we obtained with smaller $n$. So, in order to obtain the $n$-th digit we can just use any $\theta_m$ with $m>n$ with the same result.
This is indeed called something I don't remember what in algebra but the story is merely that you try to define a function that maps infinite sequences to infinite sequences from functions that tell you how to get the first $n$ elements of the output sequence from the first $n$ elements in the input sequence in a consistent way.
I'll stop here for now. Hope it helps a bit. :-)    
