0

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!

Henno Brandsma
  • 242,131
maramath
  • 108
  • 1
    Do you want the union to be over $g \in K$ rather than $g \in G$? If so, then yes because $gK = K$ if $g \in K$. – Michael Albanese Jun 21 '20 at 16:20
  • You don't say what $K^C$ means. At first I thought maybe "closure". Then I decided you meant complement. In a group $G$, if $K$ is a subgroup and $g$ is not in $K$ then $gK$ is disjoint from $K$ (unequal cosets are disjoint), so certainly the complement of $K$ is the union you write. But this has nothing to do with topological groups. It's a property of all groups, so I suspect you might be missing a hypothesis to make the theorem interesting. – KCd Jun 21 '20 at 16:23

1 Answers1

2

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.

Matt Samuel
  • 58,164