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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!

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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.

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  • $\begingroup$ obg, but is there not a more rigorous way to do this other than just through examples? $\endgroup$ – Gwen Di Feb 23 at 20:57
  • $\begingroup$ 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 $k\in\{1,2,3,4\}$. $\endgroup$ – José Carlos Santos Feb 23 at 21:00
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    $\begingroup$ @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? $\endgroup$ – rschwieb Feb 23 at 23:05
  • $\begingroup$ @JoséCarlosSantos: did you mean $\rho \circ r$ where you wrote $\rho \circ c$? $\endgroup$ – J. W. Tanner Feb 24 at 2:51
  • $\begingroup$ @J.W.Tanner Yes! I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Feb 24 at 8:41

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