Proof that rationals are normal in the reals: $(\mathbb{Q}, +) \trianglelefteq (\mathbb{R}, +)$

Question

While reading about Vitali sets I stubled onto the fact that the rational numbers are a normal subgroup of the real numbers (with respect to $+$).
I searched for a while but could not come up with an idea how to proof this and how to describe $\mathbb{R} / \mathbb{Q}$ with a corresponding homomorphism that sends $\mathbb{Q}$ to $0$.

Can someone help me out with an homomorphism or sources/links dealing with this question (on an undergraduate level).
Thanks!

Answer 1

ALL subgroups of an abelian group are normal.

Answer 2

$\mathbb R/\mathbb Z \cong S^1 = \{ z \in \mathbb C : |z|=1 \}$ via $t \mapsto e^{2\pi it}$.

Under this map, $\mathbb Q/\mathbb Z \cong \operatorname{tor} S^1$, the torsion subgroup of $S^1$.

So, $\mathbb R/\mathbb Q \cong (\mathbb R/\mathbb Z) / (\mathbb Q/\mathbb Z)$ is bound to be very complicated.