TITLE: The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$
QUESTION [5 upvotes]: I want know some information about the group 
\begin{equation*}
\frac{(1+p\mathbb Z_p)}{(1+p^{n}\mathbb Z_p)}
\end{equation*}
(the Quotient group).
What is the order of this group? I guess $p^{n-1}$ But how can I prove this?
How does an element of this group look like?
$\mathbb Z_p$ is the ring of the p-adic integers.

REPLY [2 votes]: The group $U_n:=1+p^n\mathbb{Z}_p$ is the kernel of $U \to (\mathbb Z/p^n \mathbb Z)^*$, the map being the (sujective) projection on the "n-th coordinate". Hence you have the exact sequences :
$$1\to U_1 \to U\to (\mathbb{Z}/ p \mathbb{Z})^*\to 1, \quad (1) $$
$$ 1\to U_n \to U\to (\mathbb{Z}/ p^n \mathbb{Z})^* \to 1, \quad (2)  $$ 
The first sequence gives : 
$$ 1 \to U_1/U_n \to U/U_n \to (\mathbb{Z}/ p \mathbb{Z})^*\to 1,\quad (3)$$
since $U_n\subset U_1$ (isomorphism theorem). 
On the other hand, we know that (using the sequence (2)) that $$\vert U/U_n \vert = \vert (\mathbb{Z}/ p^n \mathbb{Z})^*\vert = \varphi(p^n)=(p-1)p^{n-1}$$ ( $\varphi$ is the Euler ..).
Finaly,the last sequence (3) gives $\vert U/U_n \vert =\vert U_1/U_n \vert\times \vert(\mathbb{Z}/ p \mathbb{Z})^* \vert$. Hence $\vert U_1/U_n \vert=p^{n-1}$. 
PS: Serre's book "A course in Arithmetic" is a very good reference for p-adic numbers.