Let \(G\) be a topological group; let \(C\) be the component of \(G\) containing the identity element \(e\). Show that \(C\) is a normal subgroup of \(G\). [Hint: If \(x \in G\), then \(xC\) is the component of \(G\) containing \(x\).]
Solution
For each \(g\in G\), left translation \[ L_g:G\to G,\qquad L_g(x)=gx, \] is a homeomorphism. Therefore it sends components to components. Since \(L_g(C)=gC\) contains \(g\), it is exactly the component containing \(g\). Similarly, right translation gives that \(Cg\) is the component containing \(g\). Hence \[ gC=Cg \] for every \(g\in G\).
First, \(C\) is a subgroup. If \(x,y\in C\), then \(xC=C\), because both are components containing \(x\) and \(e\in C\) implies \(x\in xC\). Thus \(xy\in C\). Also inversion is a homeomorphism fixing \(e\), so it sends the component \(C\) of \(e\) to itself; hence \(x^{-1}\in C\).
Finally \(gC=Cg\) for every \(g\) implies \[ gCg^{-1}=C. \] Thus \(C\) is a normal subgroup of \(G\).