# Thread: Congruence modulo and equivalence classes

1. ## Congruence modulo and equivalence classes

Hi,

problem: Suppose that $M$ is a subspace of a vector space $V$. Prove that a subset of $V$ is an equivalence class modulo $M$ if and only if it is a coset of $M$

attempt:
Let $N$ be a subset of $V$.
1st direction: If $N$ is a coset of $M$, then it is an equivalence class modulo $M$.
$N=x+M$.
Now, I don't really know how to go from N beeing a coset of M to it beeing an equivalence class modulo M. Any help is greatly appreciated!

Thanks.

2. Hi,

The only thing you can really do is show that

$N = x+M = [x] = \{y\in V : y-x \in M\}.$

This isn't too hard, you just need to keep the definitions in mind:

$y \in [x] \Leftrightarrow y - x \in M \Leftrightarrow y \in x + M = N.$

Just make sure you keep appealing to the definitions, and make sure you use the closure of $M$ under addition and taking inverses (owing to the fact that it is a subspace) to convince yourself that $x\sim y \Leftrightarrow x-y\in M$ actually does define an equivalence relation.