Find an alternate generating set for Dih(2n)

Recall that D_{2n} = \langle r,s \ |\ r^n = s^2 = 1, rs = sr^{-1} \rangle.

Show that every element of D_{2n} which is not a power of r has order 2. Deduce that D_{2n} is generated by the two elements s and sr, both of which have order 2.


Let x \in D_{2n} be a non power of r. By a previous exercise, we know that rx = xr^{-1}. Moreover, x = sr^i for some 0 \leq i < n. Thus x^2 = sr^isr^i = ssr^{-i}r^i = 1, hence |x| = 2.

Now note that s \in \langle s, sr \rangle and r \in \langle s, sr \rangle since r = s \cdot sr. Thus D_{2n} = \langle s, sr \rangle.

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