Problem

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

Use the generators and relations above to 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 \(2\).

Solution

Let \(x\) be an arbitrary element of \(D_{2n}\) which is not a power of \(r\). We can write \(x\) as \(x = sr^i\) for some integer \(i \ge 0\).

Since \(x\) is not the identity, \(|x| > 1\).

If \(i = 0\), then \(x = s\) and thus \(|x| = |s| = 2\).

If \(i > 0\), then

$$ (sr^i)^2 = sr^isr^i = sr^{i-1}(rs)r^i = sr^{i-1}(sr^{-1})r^i = sr^{i-1}sr^{i-1} = (sr^{i-1})^2 $$

By repeatedly applying this, we can see that $$ x^2 = (sr^i)^2 = (sr^{i-1})^2 = … = s^2 = 1 $$

Therefore, \(|x| = 2\).

$$\tag*{$\blacksquare$}$$

Since \(D_{2n} = \langle r, s\rangle\), by definition of generator, any element \(x \in D_{2n}\) can be expressed as a finite product of \(r\) and \(s\).

Note

$$ s(sr) = (ss)r = s^2r = r $$

That is, \(r\) can be expressed as a product of \(s\) and \(sr\). Therefore, any element \(x \in D_{2n}\) can be expressed as a finite product of \(s\) and \(sr\).

Hence, by definition of generator,

$$ D_{2n} = \langle s, sr \rangle $$

$$\tag*{$\blacksquare$}$$