commutativity of rotations and reflections

Question

The question is as follows:

[Concerning the square embedded in the plane,] prove that the $90^{\circ}$ clockwise rotation $\sigma$ and the reflection through the north/south axis $\rho$ do not commute.

I'm generally just confused on how one would go about proving something like this. I have represented the square by numbering it's vertices as $\{1,2,3,4\}$ and I can show that the composition of transformations do not equal each other, but I have a feeling this is insufficient. I would really appreciate some guidance on the form a proof like this might take. Thanks a bunch!

Answer 1

Your approach is fine. If, as you wrote, the vertices are $1$, $2$, $3$, and $4$ (labelled clockwise), and your reflection $r$ is with respect to the line defined by $1$ and $3$, and your rotation is called $\rho$, then

• $\rho\bigl(r(1)\bigr)=\rho(1)=1$;
• $r\bigl(\rho(1)\bigr)=r(2)=3$.

Since $\rho\bigl(r(1)\bigr)\neq r\bigl(\rho(1)\bigr)$, then $\rho\circ r\neq r\circ\rho$. So, $\rho$ and $r$ do not commute.