Subrings of finite index and units If $S$ is a subring of $R$ of finite index as an abelian group, does it follow that the subgroup of units $S^\star$ has finite index in $R^\star$ as a multiplicative group?
 A: In the case $R$ is commutative the answer is yes.
Let $R \supset S$ be rings such that the quotient $(R,+)/(S,+)$ is finite. 
Let $I = \{s \in S: sr \in S \mbox{ for all } r \in  R\}.$ We claim $I$ is an ideal of $R.$ Clearly, $I$ is an $S$-module. We must show $I$ is stable under multiplication by elements of $R$. Let $r \in R.$ Then if $s' \in I,$ then $rs' = s'r \in S$ such that $rs'r' =s(rr') \in S$ for any $r' \in R.$ Hence, $rI \subset I.$ It follows $I$ is an ideal of $R.$ 
Now consider the ring $R/I.$ We claim $|R/I|$ is a finite. Observe that the $S$ action on $R$ descends to an action on $(R,+)/(S,+)$ and the kernel of this action is $I.$ Hence, $S/I$ acts faithfully on $(R,+)/(S,+)$ via left multiplication. As the set of endomorphisms of $(R,+)/(S,+),$ as an abelian group, is finite, it follows $S/I$ is a finite ring. Then as $S/I$ has finite index in $R/I,$ we obtain the ring $R/I$ is a finite.
The group $R^{\times}$ acts on $R/I$ via left multiplication as $R$-module automorphisms. Consider the kernel of this action, $K.$ Because the group $\mathrm{Aut}_R(R/I)$ is finite, the index $[R^{\times}:K]$ is finite. We claim $K \subset S^{\times}.$ Let $r \in K.$ Because $r$ acts on $R/I$ as the identity, $r = 1 + s$ for some $s\in I \subset S.$ Hence, $r \in S.$ An identical argument shows $r^{-1} \in S.$ It follows $r \in S^{\times}$ and therefore $K \subset S^{\times}.$ 
We conclude
$$[R^{\times}:S^{\times}] \le [R^{\times}:K] < \infty.$$    
