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Show that the subgroup of $D_{2n}$ generated by the set $\{s,rs\}$ is $D_{2n}$ itself.

Here is my attempt:

$\langle s, rs \rangle = D_{2n}$ if and only if $\langle r, s \rangle = \langle s, rs \rangle$, so it is sufficient to show that the latter equality holds. The containment $\langle r, s \rangle \subseteq \langle s, rs \rangle$ holds: since $r, s \in \langle s, rs \rangle$ (cleary this is true for $s$, and also $r=(rs)s$), every element in $\langle r, s \rangle$ is also an element of $\langle s, rs \rangle$. The containment $\langle s, rs \rangle \subseteq \langle r, s \rangle$ is clear, because since $s,rs \in \langle r, s \rangle$ every element in $\langle s, rs \rangle$ is also an element in $\langle r, s \rangle$. Therefore $\langle r, s \rangle = \langle s, rs \rangle = D_{2n}$.

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  • $\begingroup$ Your solution is perfect! $\endgroup$ – LucaMac Sep 6 '18 at 19:50
  • $\begingroup$ There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . . $\endgroup$ – Shaun Dec 1 '18 at 17:45
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Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{\color{red}{-1}}.$$

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