# commutativity of rotations and reflections

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!

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.

• obg, but is there not a more rigorous way to do this other than just through examples? – Gwen Di Feb 23 at 20:57
• What is there about this approach which is not rigorous? Proving that $\rho\circ r\neq r\circ\rho$ means proving that $\rho\bigl(r(k)\bigr)\neq r\bigl(\rho(k)\bigr)$, for some $k\in\{1,2,3,4\}$. – José Carlos Santos Feb 23 at 21:00
• @gwen The way to disprove a statement is to give a special example called a counterexample. Perhaps you are confusing this with trying to prove statements with examples? – rschwieb Feb 23 at 23:05
• @JoséCarlosSantos: did you mean $\rho \circ r$ where you wrote $\rho \circ c$? – J. W. Tanner Feb 24 at 2:51
• @J.W.Tanner Yes! I've edited my answer. Thank you. – José Carlos Santos Feb 24 at 8:41