If $G$ is a finite abelian group, and $R$ a ring, then does every automorphism of $G$ induce the identity on the group $R[G]^\times/G$?

Question

Let $G$ be a finite abelian group, and let $R$ be a ring (commutative with 1). In particular I'm interested in the case where $R$ is $\mathbb{Z}/n\mathbb{Z}$ for some $n$. By functoriality, every automorphism $\alpha\in\text{Aut}(G)$ induces an automorphism of the group algebra $R[G]$, and hence an automorphism of the group of units $R[G]^\times$, which stabilizes the subgroup $G\subset R[G]^\times$.

I'd like to understand in general the units of $R[G]$ which don't come from $G$. In particular, must $\alpha$ induce the identity on $R[G]^\times/G$?

Answer 1

No.

Let $G=\mathbb{Z}_8=\langle w\rangle$ and set $v=2-w^4+(1-w^4)(w+w^{-1})$. This is a (normalized) unit in $\mathbb{Z} G$ by 10.8 in S. K. SEHGAL, Units in Integral Group Rings (with an appendix by A. Weiss), Pitman Monographs and Surveys in Pure and Applied Math.69, Longman Scientific & Technical, Harlow, 1993.

Consider $\phi\in\text{Aut}(G)$ given by $\phi(w)=w^3$. Expanding we have $$v = 2 - w^4 + w + w^{-1} - w^{-3}-w^3.$$ Therefore $$\phi(v) = 2-w^4+w^3+w^{-3}-w^{-1}-w.$$ It follows that there is no $x\in G$ such that $x\phi(v) = v$, and so $\phi$ does not give the identity on $(\mathbb{Z}G)^\times/G$.