Algebra Solutions: Exhibit an alternate presentation for Dih(2n)

Show that $\langle a,b \ |\ a^2 = b^2 = (ab)^n = 1 \rangle$ gives a presentation for $D_{2n}$ in terms of the generators $a = s$ and $b = sr$ computed in a previous exercise.

Given that $a = s$ and $b = sr$ and that $a^2 = b^2 = (ab)^n = 1$ , we have $s^2 = a^2 = 1$ , $r^n = (ssr)^n = (ab)^n = 1$ , and $rsr = s(sr)(sr) = sb^2 = a = s$ , so that $rs = sr^{-1}$ .
Given that $a = s$ and $b = sr$ and that $r^n = s^2 = 1$ and $rs = sr^{-1}$ , we have $a^2 = s^2 = 1$ , $(ab)^n = (ssr)^n = r^n = 1$ , and $b^2 = srsr = ssr^{-1}r = 1$ . $\blacksquare$

Algebra Solutions

Exhibit an alternate presentation for Dih(2n)

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