Define the relation $\rho$ on $\Bbb R$ by the rule: $\forall x, y \in \Bbb R, x \rho y$ if and only if $\exists n \in \Bbb Z$ such that $y = x + n\pi$. Let $G$ denote the set of all equivalence classes under $\rho$. Let $+ : G \times G \to G$ be defined as $[x] + [y] = [x + y]$ and let $H = \{ [0], [ \pi/2 ] \}$. Prove that $H$ is a subgroup of $G$.
This was one of the assignments at school and the answer to the question was already given. In order to prove it is closed under the operation, you add two elements of the set and see if it produces another element of the set. The answer provided said that [ π/2 ] + [ π/2 ] = [0]. This is where I get confused. Can anyone please explain why the answer is [0]? Maybe this is super easy to some but I have just started studying the group theory and have a bit of trouble understanding.
-Edited to add one more question:
H ={ [0], [π/4], [π/2], [3π/4]} is a subgroup of G.
I am still a bit confused about it being closed under the operation..especially these two:
[3π/4] + [3π/4]=????
[3π/4] + [π/2]=????
In short, $y \sim x + \pi n$, which mean that two numbers are the same if they are an integer multiple of $\pi$ away from each other. So $\pi/2 + \pi/2 = \pi$, which is an integer multiple of $\pi$ away from $0$. So $\pi \sim 0$.