Subgroup of a topological group

Question

Let $G$ be a topological group and $K\subset G$ a subgroup of G.

Is it true that $K = \bigcup_{g\in K}gK$ ?

I'm asking this because in my notes I have that $K^{C} = \bigcup_{g\in G-K}gK$

Thank you in advance for your answer!

Answer 1

It's not true that $K=\bigcup_{g\in G}gK$. In fact, this is equal to $G$. Assuming $K^C$ is the complement of $K$, it is true that $K^C=\bigcup_{g\in G-K}gK$. This is a simple algebraic fact: the cosets $gK$ for $g\notin K$ are all disjoint from $K$, and every element that is not in $K$ is in such a coset. This is true for any group, not just topological groups.

To answer the edited question, it is indeed true that $K=\bigcup_{g\in K}gK$. In fact, $gK=K$ for all $g\in K$, so every term in the union is the same set.