Generators of $GL_2(\mathbb{Q}_p)$ A well known fact is that the  group $GL_2(\mathbb{Q}_p)$ is generated by the following matrices:
$1) \text{ } 
   w=
  \left( {\begin{array}{cc}
   0 & 1 \\
   1 & 0 \\
  \end{array} } \right)
$
$2) \text{ } \mathbb{Q}_p^{\star}
 \left( {\begin{array}{cc}
   1 & 0 \\
   0 & 1 \\
  \end{array} } \right)
$
$3) \text{ } 
 \left( {\begin{array}{cc}
   \mathbb{Z}_p^{\star} & 0 \\
   0 & 1 \\
  \end{array} } \right)
$
$4) \text{ } 
 \left( {\begin{array}{cc}
   p & 0 \\
   0 & 1 \\
  \end{array} } \right)
$
$5) \text{ } 
 \left( {\begin{array}{cc}
   1 & p\mathbb{Z}_p \\
   0 & 1 \\
  \end{array} } \right)
$
I need a reference for the above fact. In particular, for $b \in \mathbb{Q}_p$, I am trying to write lower unipotent elements $
 \left( {\begin{array}{cc}
   1 & 0 \\
   b & 1 \\
  \end{array} } \right)
$ in terms of the above matrices. Any help is welcome. Thanks.
 A: We first note that multiplication on the left by $(1)$ switches the rows of a matrix and multiplication on the right by $(1)$ switches the columns call these observations $(*)$. Using both parts of $(*)$ we see conjugation by $(1)$ transforms $\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}$ to $\begin{pmatrix} 1 & 0 \\ b & 1\end{pmatrix}$ so it is sufficient to make the upper-triangular matrix $\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}$.
First the case $b\in p\Bbb Z_p$ let $b= up^k$ for $u\in\Bbb Z_p^*$ and $k>0$. Then as $\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & nup \\ 0 & 1\end{pmatrix}$ we write $\begin{pmatrix} 1 & up^k \\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}^{p^{k-1}}$. So when the dust settles your expression is just
$$\begin{pmatrix} 1 & 0 \\ b & 1\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}1 & up \\ 0 & 1\end{pmatrix}^{p^{k-1}}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$$
Now when $k\le 0$, we just need to scale a bit. Note that multiplication by a multiple of the identity just scales the matrix, so if $b=up^{k}$, with $k\le 0$ we can write $\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}\begin{pmatrix}p^{k-1} & 0 \\ 0 & p^{k-1}\end{pmatrix}= \begin{pmatrix}p^{k-1} & b \\0 & p^{k-1}\end{pmatrix}$ using operation $(2)$. Then to scale this back down, first use $(*)$ to switch the top and bottom rows, then multiply by $\begin{pmatrix} p^{1-k} & 0 \\ 0 & 1\end{pmatrix}$ with $(4)$ and use $(*)$ again to switch the rows back and we have $\begin{pmatrix} p^{k-1} & b \\ 0 & 1\end{pmatrix}$. Now conjugate by $(1)$ to give $\begin{pmatrix} p^{k-1} & 0 \\ b & 1\end{pmatrix}$ and multiply by $\begin{pmatrix} p^{1-k} & 0 \\0 & 1\end{pmatrix}$ to arrive at the finished product.
