Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 4 / a / Solution

The two algebras have the same identity, so an inverse in $A$ is also an inverse in $B$. Therefore $G(A)\subseteq G_B(A)$.

For strictness, take $B=C_{\mathbb{C}}(\mathbb{T})$, where $\mathbb{T}$ is the unit circle, and let $A$ be the uniform closure of the polynomials in the coordinate function $z$. This is a closed unital subalgebra of the Banach algebra $B$. The coordinate $z$ has inverse $\overline{z}$ in $B$. But $\overline{z}$ cannot belong to $A$: for every polynomial $p$,and the same identity holds for every element of $A$ by uniform convergence. It fails for $\overline{z}$, since $z\overline{z}=1$. Thus $z\in G_B(A)\setminus G(A)$. This algebra is the boundary realization of the disk algebra.