Square-zero unit subgroup
= Square-zero unit subgroup
{title2=$1+I$}
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 $a\mapsto1+a$ identifies the <additive group> of $I$ with $1+I$.