How to show that $|A/R|=|A|/n$.

Question

Suppose $A$ is a finite set and $R$ is an equivalence relation on $A$. Suppose also there is some positive integer $n$ such that for every $x\in A |[x_R]|=n$. Prove that $A/R$ is finite and $|A/R|=|A|/n$.

Proof:

Suppose $X\subseteq A$, since for all $x\in A$ we can always choose an $x\in A$ such that $X=[x_R]$ then $X\subseteq A/R$ and since $A$ is finite then $A/R$ is finite.

So i'm stuck on how to show $|A/R|=|A|/n$.

Answer 1

Your proof doesn't make any sense. There are more subsets of $A$ than equivalence classes in $R$.

HINT:

1. Find a surjection from $A$ onto $A/R$. Recall that the image of a finite set is finite.
2. Choose a system of representatives $x_1,\ldots,x_k$ from $A/R$, then $A=[x_1]_R\cup\ldots\cup[x_k]_R$. That is a disjoint union. Now use the assumption.

Answer 2

Hint $\ $ Recall that an equivalence relation on $A$ induces a partition of $A$ (into equivalence classes). By hypothesis, each class in the partition has the same size $n$. Hence $\,|A|\, =\, n\, |A/R|$.