Let \(\alpha\) be an element of the topological group \(G\). Show that the maps \(f_\alpha, g_\alpha: G \to G\) defined by
\[ f_\alpha(x) = \alpha \cdot x \quad \text{and} \quad g_\alpha(x) = x \cdot \alpha \]
are homeomorphisms of \(G\). Conclude that \(G\) is a homogeneous space. (This means that for every pair \(x, y\) of points of \(G\), there exists a homeomorphism of \(G\) onto itself that carries \(x\) to \(y\).)
Solution
Let \(m\colon G\times G\to G\) be the continuous multiplication.
1. \(f_\alpha\) and \(g_\alpha\) are continuous for every \(\alpha\in G\).
The maps \(x\mapsto(\alpha,x)\) and \(x\mapsto(x,\alpha)\) from \(G\) to \(G\times G\) are continuous, since their coordinates are constant or the identity. Composing with \(m\) gives \(f_\alpha\) and \(g_\alpha\).
2. \(f_\alpha\) and \(g_\alpha\) are homeomorphisms.
\(f_{\alpha^{-1}}\circ f_\alpha=f_\alpha\circ f_{\alpha^{-1}}=\operatorname{id}_G\) and \(g_{\alpha^{-1}}\circ g_\alpha=g_\alpha\circ g_{\alpha^{-1}}=\operatorname{id}_G\) by associativity, and the inverses are continuous by step 1.
3. Q.E.D.
Given \(x,y\in G\), put \(\alpha=yx^{-1}\). By step 2, \(f_\alpha\) is a homeomorphism of \(G\) onto itself, and \(f_\alpha(x)=yx^{-1}x=y\). So \(G\) is homogeneous.