Square-zero unit subgroup ($1+I$)

If $I^2=0$, then $1+I=\{1+a:a\in I\}$ is an abelian subgroup of the unit group. Multiplication satisfies $(1+a)(1+b)=1+a+b$, and $(1+a{)}^{-1}=1-a$, so $a\mapsto 1+a$ identifies the additive group of $I$ with $1+I$.