This is one of the most popular unofficial solution guides. It’s well-typeset in LaTeX and covers many exercises from Chapter 4. You can view the PDF directly on Greg Kikola's Personal Site
Solution: $(\Rightarrow)$ Suppose $aH = bH$. Then $a = ae \in aH = bH$, implying $a = bh$ for some $h \in H$. Thus, $ab^-1 = h \in H$. abstract algebra dummit and foote solutions chapter 4
If a proof feels too abstract, test the logic against S3cap S sub 3 D8cap D sub 8 . If it doesn't work there, your logic is flawed. Where to Find Detailed Solutions This is one of the most popular unofficial solution guides
