A bit stuck on this question: Let $G=D_{12}$=$({r^{i}s^j}:0\leq i \leq 5, 0 \leq j \leq 1)$ and $sr^{i}=r^{6-i}s$ Find a non-cyclic subgroup of $D_{12}$ of order 4. List the elements of the subgroup explicitly.
so far I understand that $D_{12}=({1,r,r^2,r^3,r^4,r^5,s,sr,sr^2,sr^3,sr^4,sr^5})$ and that I have to find four elements that satisfy the group axioms, what I don't understand is how to find them.
Any advice/hints would be appreciated (fairly new to group theory).
Best Answer
Geometrically, we can interpret $D_{12}$ as the symmetry group of a regular hexagon. Then we see the symmetry group of a rectangle within this hexagon, which has the following elements:
Here is a picture to demonstrate this.
These correspond to elements $\{1,s,sr^3,r^3\}$ in your notation. I leave it to you to verify that these elements form the sole non-cyclic order-4 group, the Klein four-group.