Why $G/N$ is discrete?

Question

I am reading the lecture notes. On page 15, Line -5, why $G/N$ is discrete? Thank you very much.

Answer 1

Since by assumption $N$ is open, singletons in $G/N$ are open (the preimage of the singleton under the projection is a coset of $N$, and cosets of open sets are open), hence every subset of $G/N$ is open (as a union of singletons, which are open), so $G/N$ is a discrete space.

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Alternative, more verbose explanation:

The quotient $G/N$ is endowed with the quotient topology, that means a subset $A \subset G/N$ is open if and only if $\rho^{-1}(A)$ is open in $G$, where $\rho \colon G \to G/N$ is the canonical homomorphism.

$\rho^{-1}(A)$ is a union of cosets of $N$, $$\rho^{-1}(A) = \bigcup_{x\in \rho^{-1}(A)} x\cdot N.$$

$N$ is open by assumption, and the translations are homeomorphisms in a topological group, hence $\rho^{-1}(A)$ is open in $G$ for every $A\subset G/N$, so every subset of $G/N$ is open.