Recipe

Compare r∘s to s∘r in D_n

Use when: you're given the two generators r and s of D_n and must find s∘r's label, or check whether r and s commute

Goal: both composition orders written as single labeled elements, and a verdict on whether they're equal

Taught in Composing symmetries and non-abelian behavior

1. 1

   Recall the label rs is defined to mean apply s first, then r, so r\circ s = rs follows immediately, with no tracing required.

   Example

   In D_3: r\circ s = rs, straight from the labeling convention itself.

   Why: the r^k s labeling in section 2 was built specifically so "apply s then r" always reads off as the label rs.

2. 2

   Apply r first, then s, to every vertex of the polygon (or read the widget's tracked positions), and record where the whole set lands.

   Example

   In D_3, tracing the triangle's vertices through r then s lands on the element labeled r^2s.

   Why: unlike r\circ s, this order doesn't match any label's definition directly, so the actual motion has to be traced to see which single element it equals.

3. 3

   Since r^n=e, the power r^{n-1} always equals r^{-1}; rewrite Step 2's exponent in that form.

   Example

   In D_3, r^3=e, so r^2=r^{-1}, giving s\circ r = r^2s = r^{-1}s.

   Why: writing the exponent as -1 instead of n-1 exposes the same relation across every D_n, not a coincidence tied to one specific n.

4. 4

   Check whether Step 1's label and Step 3's label are the same element.

   Example

   In D_3, rs \ne r^{-1}s since r\ne r^2, so r and s do not commute, and D_3 is non-abelian.

   • r \ne r^{-1} (true for every n\ge3) — rs \ne r^{-1}s, so this pair doesn't commute and D_n is non-abelian.
   • r = r^{-1} (only n=1 or n=2) — rs = r^{-1}s, so this particular pair happens to commute.

   Why: two elements commute exactly when composing them in either order gives the identical result, so comparing the two labels directly answers the question.