Let \(A \subset B\) be an extension of commutative rings containing \(\mathbb{Q}\). Suppose that this extension is subintegral in the sense of R. G. Swan [see J. Algebra 67, 210-229 (1980; Zbl 0473.13001)]. In their main result the authors construct a natural group homomorphism \(\xi_{B/A}:B/A \to{\mathcal I}(A,B)\), \({\mathcal I}(A,B)\) denotes the group of invertible \(A\)-modules of \(B\), which is shown to be an isomorphism provided \(A\) is an excellent \(\mathbb{Q}\)-algebra of finite Krull dimension. In the case that \(A\) is a reduced \(G\)-algebra containing \(\mathbb{Q}\) and \(B\) is the seminormalization of \(A\) (here a graded commutative ring \(A=\bigoplus_{n \geq 0}A_ n\) with \(A_ 0\) a field and finitely generated as an \(A_ 0\)-algebra is called \(G\)-algebra) then \(A\) is excellent of finite Krull dimension, \(A \subseteq B\) is subintegral and \({\mathcal I}(A,B)=\text{Pic} A\). So the main result implies \(B/A \simeq \text{Pic} A\), which yields a result of B. H. Dayton [see J. Pure Appl. Algebra 59, No. 3, 237-253 (1989; Zbl 0697.13004)]. An interesting result towards the proof of the main theorem is an elementwise characterization of the subintegralness of \(A \subseteq B\). Because of the use of exponential and logarithmic series the assumption that \(A\) contains \(\mathbb{Q}\) is necessary. In fact, it is shown by an example that the main result does not hold without this assumption. In a concluding remark the authors announce a proof of the main result where they can drop that \(A\) is excellent of finite Krull dimension.