Iwahori versus Bruhat decompositions I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $\mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K = 
\left( 
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & p & p & \mathcal{O}
\end{array}
\right)
$$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition
We have
$$
K = 
\left( 
\begin{array}{cccc}
1 &  & &  \\
p & 1 & &  \\
p &  & 1 &  \\
p & p & p & 1
\end{array}
\right)
\left( 
\begin{array}{cccc}
\mathcal{O}^\times & &  &  \\
 & \mathcal{O} & \mathcal{O} &  \\
 & \mathcal{O} & \mathcal{O} &  \\
 &  &  & \mathcal{O}^\times
\end{array}
\right)
\left( 
\begin{array}{cccc}
1 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & 1 &  & \mathcal{O} \\
 &  & 1 & \mathcal{O} \\
 &  &  & 1 
\end{array}
\right)
$$
so that in particular by decomposing the left subgroup we should obtain
$$
\left( 
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O}
\end{array}
\right)
=
\bigsqcup_{a, b, c \in \mathcal{O}/p} 
\left( 
\begin{array}{cccc}
1 &  &  &  \\
a & 1  & &  \\
b & & 1 & \\
c & b & -a & 1
\end{array}
\right)
K
$$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,

$$[K_0:K] =N(p)^3$$

Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q = 
\left( 
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & & & \mathcal{O}
\end{array}
\right)
$$
the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q 
\sqcup Qx 
\left( 
\begin{array}{cccc}
1 & k &  & \\
 & 1 &  & \\
 & & 1 & k \\
 & & & 1
\end{array}
\right)
\sqcup 
Qxy 
\left( 
\begin{array}{cccc}
1 & & k  &  \\
 & 1 & k & k \\
 & & 1 &  \\
 & & & 1
\end{array}
\right)
\sqcup 
Qxyx
\left( 
\begin{array}{cccc}
1 & k & k & k\\
 & 1 &  & k\\
 & & 1 & k \\
 & & & 1
\end{array}
\right)
$$
where the transformations $x$ and $y$ are defined by
$$x= 
\left( 
\begin{array}{cccc}
 & 1&  & \\
1 &  &  & \\
 & &  & 1 \\
 & &1 & 
\end{array}
\right)
$$
$$y =
\left( 
\begin{array}{cccc}
1 &  &  &\\
 &  & 1 & \\
 & -1 & &  \\
 & & & 1
\end{array}
\right)
$$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say

$$[K_0:K] =(1+N(p))(1+N(p)^2)$$

Here is the question following from this discussion:

Both results are different, what is
  happening?

 A: Your Iwahori-decomposition computation is a bit too free with quotient computations.  It would work fine for vector spaces, which is, in some sense, why you get the correct leading term in your count; but there are additional subtleties on the group level that it does not take into account—among other things, that some of the entries denoted by $\mathcal O$ can be $0$, but some cannot!
The same computation would suggest that the quotient of $\operatorname{GL}_2(k)$ by its Borel subgroup of upper-triangular matrices should have as coset representatives the lower uni-triangular matrices $u_-(c) = \begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}$, whereas actually there is another coset, the one containing $w_0 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.  Actually a better way to think about it is that most coset representatives are of the form $u_+(c)w_0$, where $u_+(c)$ is the transpose of $u_-(c)$, and only the trivial coset is not of this form; it just happens that $u_+(c)w_0$ and $u_-(c^{-1})$ lie in the same coset when $c \ne 0$.
Your count via the Bruhat decomposition shows what’s going on here:  you get $(1 + N(p))(1 + N(p)^2) = 1 + N(p) + N(p)^2 + N(p)^3$, which suggests, correctly, that we have stratified $\mathbb P^3k$ by affine spaces of the obvious dimensions, corresponding to the Weyl-group elements $1$, $x$, $x y$, and $x y x$ (which are minimal-length representatives in the quotient by an appropriate parabolic subgroup, which is doubtless what led you to choose them!).
