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For convenience, consider the symmetries of a square. When finding/specifying the reflection generator of a dihedral group does it matter which relfection I choose? Dummit/Foote say to choose the reflection about the line through vertex 1 (the first labeled vertex) but is that always the fixed line (y=x in case of the square), or does the line my reflection generator obeys rotate with the square? I.e would my generator change to reflection about the line y=-x after one rotation since 1 moved clockwise, or would I keep the line y=x for my reflection?

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No, it does not matter what reflection you choose in order to generate the dihedral group. This is because a dihedral group $Dih_n$ is isomorphic to the semidirect product of the cyclic group $C_n$ and the group $\mathbb{Z}_2$, which becomes the subgroup generated by a particular reflection, combined with the fact that all subgroups of $Dih_n$ generated by reflections are conjugate to each other, and so the semidirect product construction works for any such subgroup.

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you have a rotation, say $(1234)$ and a reflection (pick one, say $(12)(34),(14)(23),(24)$ or $(13)$ (with vertices of the square labeled counterclockwise). if you pick one through the sides, not fixing a vertex you get $(1234)(12)(34)=(13)$ and pick up a reflection through the corners anyway, and vice versa $(1234)(13)=(14)(23)$.

You can check that that with $r=(1234)$ and $s$ any reflection, you get the relations $srs=r^{-1}$ eg $(12)(34)(1234)(12)(34)=(1432)$

so to answer your question, it doesnt matter what reflection you choose.

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