Use the subgroup lattice structure of the symmetries of a square, $Sq$, and use it to find the subgroup lattice of all left cosets of the normal subgroup $\langle r^2\rangle$.
This is a problem from my intro algebra class.
In this case, we begin with the subgroup structure of $Sq$ and consider the normal subgroup generated by $r^2$, where $r$ is rotation by $\frac{\pi}{2}$ around the z axis and the square is centered at the origin and lies in the xy plane. (Don't think it matters, but a, b, c, and d are rotations by $\pi$ around the respective axes x, y, y=-x, and y=x).
My professor's claim was that since $\langle r^2\rangle$ is a normal subgroup, there exists a bijection between $Sq/\langle r^2\rangle$ and subgroups of $Sq$ containing $\langle r^2\rangle$, and thus we can basically preserve the structure of all groups containing $\langle r^2\rangle$, as in the diagrams below.
Can anyone explain the logic behind this? I feel as though we need something more than "there exists a bijection." I don't believe that this bijection would be an isomorphism, since I don't think you could assign a group structure to both $Sq/\langle r^2\rangle$ and $\langle r^2\rangle$.
Thanks.
The bijection is not between $Sq/\langle r^2\rangle$ and $\langle r^2\rangle$ but between the set of subgroups of $Sq/\langle r^2\rangle$ and the set of subgroups of $Sq$ "over" (containing) $\langle r^2\rangle$. This is an example of what is called a lattice morphism, as it preserves the partial ordering of inclusion.