Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm starting to study topological groups, and I noticed that Every single theorem in topological groups I have to use the following statement:

Let $G$ be a topological group and U an open subset of G, if $g\in G$, then $gU$ is an open subset of G.

I can't prove it, please anyone can help me please.


share|cite|improve this question
Both the multiplication and the inversion in the group are continuous. This means that the map that multiplies by a fixed element from the group is in fact a homeomorphism, so restricting it to any open subset gives a homeomorphism onto the image. – Tobias Kildetoft Oct 12 '12 at 9:34
up vote 4 down vote accepted

In fact the map $\varphi_g:G\to G:x\mapsto gx$ is a homeomorphism. It’s clearly a bijection, since $\varphi_g^{-1}=\varphi_{g^{-1}}$. To see that it’s continuous, let $U\subseteq G$ be open. The group operation is continuous, so $V=\{\langle x,y\rangle\in G\times G:xy\in U\}$ is open in $G\times G$. Let $\pi:G\times G\to G:\langle x,y\rangle\mapsto y$, and let $G_g=\{g\}\times G$; $\pi\upharpoonright G_g:G_g\to G$ is a homeomorphism, and $V\cap G_g$ is open in $G_g$, so $\pi[V\cap G_g]$ is open in $G$. But

$$\pi[V\cap G_g]=\{x\in G:\langle g,x\rangle\in V\}=\{x\in G:gx\in U\}=\varphi_g^{-1}[U]\;,$$

so $\varphi_g^{-1}[U]$ is open in $G$, and $\varphi_g$ is continuous. Since $g$ was arbitrary, it follows immediately that $\varphi_g^{-1}=\varphi_{g^{-1}}$ is also continuous and hence that $\varphi_g$ is a homeomorphism. In particular, then, $gU=\varphi_g[U]$ is open for every $g\in G$ and open $U\subseteq G$.

share|cite|improve this answer

Multiplication is continuous, so the map $g^{-1}:G\to G, u\mapsto g^{-1}u$ is continuos. Thus the preimage of an open set $U\subset G$ is open, and the preimage equals $gU$.

Hope I got it right.

share|cite|improve this answer
It’s right, but you probably should point out that there’s quite a bit of work hidden in the jump from multiplication is continuous to $u\mapsto g^{-1}u$ is continuous. Multiplication, after all, is a function from $G\times G$ to $G$, while you’re looking at a function from $G$ to $G$. – Brian M. Scott Oct 12 '12 at 9:45
@BrianM.Scott Thanks, I am sorry I forgot about that. – lka Oct 12 '12 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.