${{\mathbb{F}}}_q^4$ has $\frac{q^4-1}{q-1}=q^3+q^2+q+1$ pairwise linearly independent vectors. Each one of the $q^3+q^2$ pairwise linearly independent vectors not in $X$ can be used to for a 3-subspace of ${{\mathbb{F}}}_q^4$. Each 3-subspace contain $\frac{q^3-1}{q-1}=q^2+q+1$ pairwise linearly independent vectors, i.e. $q^2$ additional vector to $X$. Each one of them will form the same 3-subspace when appended to $X$. Hence $X$ can be expanded in $\frac{q^3 + q^2}{q^2} = q+1$ distinct ways to a 3-subspace of ${{\mathbb{F}}}_q^4$.

\[lem:representDbyA\] Each 3-subspace of ${{\mathbb{F}}}_q^4$ (also of ${{\cal D}}$) contains a unique 2-subspace of the set ${{\cal A}}$.

If $X \in {{\cal A}}$ and $v \in {{\mathbb{F}}}_q^4$ is a vector such that $v \notin {{\cal A}}$, then $Y {\mbox{$\stackrel{\rm def}{=}$}}{{\left\langle { X \cup \{ v \} } \right\rangle}}$ is clearly a 3-subspace of ${{\mathbb{F}}}_q^4$. Since $Y$ is a 3-subspace and all the 2-subspaces of ${{\cal A}}$ are pairwise disjoint, it follows that $Y$ cannot contain two 2-subspaces of ${{\cal A}}$.

There are $q+1$ different 3-subspaces which contain $X$, $q^2+1$ different 2-subspaces in ${{\cal A}}$, and hence there are $(q^2+1)(q+1) = \frac{q^4-1}{q-1}$ 3-subspaces which contain 2-subspaces from ${{\cal A}}$. The total number of different 3-subspace of ${{\mathbb{F}}}_q^4$ is $\frac{q^4-1}{q-1}$. It implies that each 3-subspace of ${{\mathbb{F}}}_q^6$ contains a 