I'm trying to work through calculating the order of orthogonal groups in characteristic $\neq 2$. However there is one proof by induction used that i can't quite follow. Could someone help me understand where the formula for $z\_{m+1}$ comes from and how we know $U$ must contain $2q-1$ vectors with norm $0$ and $q-1$ vectors of every non-zero norm in the following extract:



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> Let $z\_m$ denote the number of non-zero isotropic vectors in an orthogonal space with dimension $2m$ or $2m+1$. We claim that:
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> $z\_m = q^{2m}-1$ for dimension $2m+1$
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> $z\_m = (q^m-1)(q^{m-1}+1)$ for plus type with dimension $2m$
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> $z\_m = (q^m-1)(q^{m-1}-1)$ for minus type with dimension $2m$
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> For our inductive step we look at a $n+2$ dimensional space $V$ to ensure all spaces remain of the same type. Split V into the direct sum of $U$ and $W$ where $U$ is a $2$-dimensional space of plus type and $W$ is an $n$-dimensional space with the same type as $V$. Any isotopic vector in $V$ can be written as $u+w$ for isotropic vectors $u\in U$ and $w\in W$. Either $u$ and $w$ both have norm $0$ (with one being non-zero) or $u$ has norm $\lambda \neq 0$ and $w$ has norm $-\lambda$. Since $U$ contains $2q-1$ vectors of norm $0$ (including the zero vector) and $q-1$ vectors of every non-zero norm we count:
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> $z\_{m+1}=(2q-1)(1+z\_m)+(q-1)(q^n-1-z\_m)-1$
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> The other three cases are similar.
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> 


Thanks



Calculating norms over a finite field (orthogonal groups).
It is the case that each isotropic vector in $V$ has the form
$u+w$ where $u\in U$ and $w\in W$ but $u$ and $w$ need not be isotropic.


To see where $2q-1$ and $q-1$ come from, the quadratic form on $U$ has
norm given by $(x,y)\mapsto xy$. Now count how many pairs of elements of
$\mathbb{F}\_q$ give zero, and any given nonzero element.


