Let ${\cal P}(\mathbb{F}\_q^n)$ be the set of all subspaces of the vector space $\mathbb{F}\_q^n$ (where $q$ is a prime power). 


Fix a $Z \in {\cal P}(\mathbb{F}\_q^n)$. Define a relation ~ on ${\cal P}(\mathbb{F}\_q^n)$ as follows: 


$A$ ~ $B$ iff $A+Z = B+Z$


It is easy to show that this is an equivalence relation. This means ${\cal P}(\mathbb{F}\_q^n)$ is partitioned into equivalence classes. 


My questions are:


1) What is the number of equivalence classes? If not an exact value, do we have a decent bound?


2) What can one say about the cardinality of an equivalence class? (Personally, I feel this question is hopeless!)


Thank you



Counting the number of equivalence classes of subspaces
1) It equals the number of subspaces of $\mathbb F\_q^n/Z$.


2) Fix a subspace $A$, denote its equivalence class by $[A]$. Consider the map $[A]\to\mathcal P(Z)$, $B\mapsto B\cap Z$. Its fiber over some $U\subseteq Z$ consists of all $B$ satisfying $B\cap Z=U$ and $B+Z=A+Z$. This set may be identified via $B\mapsto B/U$ with the set of all complements of $Z/U$ in $(A+Z)/U$. Summing over all possible $U$ gives the size of $[A]$.


