Prove that in a Banach algebra, the connected component of the set of invertible elements containing the identity is a normal subgroup

Question

Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.

I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) \subseteq G_0(A)$. $B(e;1) \subseteq G_0(A)$ follows from $$(e-x)^{-1} = \sum_{n=0}^{\infty} x^n $$ But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.

Answer 1

This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.