 Solution

The set is the ideal quotient $K=(I:J)$. It contains zero. If $r,s\in K$, then $(r-s)J\subseteq I$, since $I$ is an subgroup. If $a\in R$ and $r\in K$, then $(ar)J=a(rJ)\subseteq I$ because $I$ is an ideal. Thus $K$ is an ideal.

Write $J=(a)$ and $K=(b)$. Because $I\subseteq J$, each $x\in I$ has the form $ar$. Then $rJ=r(aR)=xR\subseteq I$, so $r\in K$, and $x\in aK$. Conversely $aK\subseteq I$ by the definition of $K$. Thereforeso $I$ is a principal ideal. This argument uses only the commutative ring convention, not cancellation or an integral domain hypothesis.