Normal subgroup of a compact topological group

Question

Is a normal subgroup of a compact topological group closed? What if the group is not compact?

Answer 1

Not necessarily, Take $(\Bbb Q,+)\subseteq (\Bbb R,+)$. Since the group is abelian, all subgroups are normal, but clearly $\Bbb Q$ is not closed in the usual topology as a subgroup of $\Bbb R$.

The same thing can be done for a compact counterexample: Take $\Bbb Q\big/\Bbb Z\subseteq\Bbb R\big/\Bbb Z$. The quotient map is a local homeomorphism, so the density is inherited from the previous example.

Answer 2

No, a standard counterexample is the skew torus: Pick $c \in \mathbb{R} - \mathbb{Q}$, and let $H \subset \mathbb{S}^1 \times \mathbb{S}^1$ be the image of the subgroup $\mathbb{R} \cdot (1, c) \subset \mathbb{R} \times \mathbb{R}$ under the homomorphism $\mathbb{R} \times \mathbb{R} \to \mathbb{S}^1 \times \mathbb{S}^1$ defined by $$(x, y) \mapsto (\exp(2\pi i x), \exp(2 \pi i y)).$$ Now, $\mathbb{S}^1 \times \mathbb{S}^1$ is abelian, and so $H$ is normal. Also, $H$ is proper but dense in $\mathbb{S}^1 \times \mathbb{S}^1$ and so is not closed.