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My instructor introduced to us about concept of Dihedral group yesterday. I am though having couple of doubts.

As i first saw example using triangle and then square.It is clear to me tht if:

r denotes rotation clockwise thorugh origin of $\frac{2\pi}{n}$ radians. Then rotating n times i get $r^n=1$. Also if S be reflection and so $s^2=e$

But i am not able to see why $s \neq r^i$ for $i=0,1,2,..n-1$ and why $sr^i \neq sr^j$ for all $0 \leq i$ , $j \leq n-1$ with i not equal to j.

Also instructor told us about representation of dihedral group in which there is $rs=sr^{-1}$. She told that using this and two other (which i mentioned in beginning) we can find reduce any other permutation to one's we have in reprentation. But i do not understand from where we get this $rs=sr^{-1}$

Thanks

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    $\begingroup$ If you view the symmetries as linear transformations (the origin must be at the center of the $n$-gon for that to be the case), then the rotations all have determinant $+1$, but the reflections have determinant $=-1$. $\endgroup$ – Jyrki Lahtonen Aug 7 '16 at 11:58
  • $\begingroup$ A symmetry has a fixed line, a rotation has only a fixed point. $\endgroup$ – Bernard Aug 7 '16 at 12:55
  • $\begingroup$ @Bernard Rotation has 1 fixed point means? $\endgroup$ – Gathdi Aug 8 '16 at 3:51
  • $\begingroup$ The centre of rotation doesn't move, as far as I know :- J $\endgroup$ – Bernard Aug 8 '16 at 9:14
  • $\begingroup$ @Bernard Can you provide some reference for this stuff? $\endgroup$ – Gathdi Aug 8 '16 at 10:35
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Regarding your first question, $s \ne r^i$ follows from the fixed points each leaves. Note that $s$ will leave a whole rect fixed, whereas a rotation around a point moves every other point in the plane.

For $sr^i \ne sr^j$ we must note first that $s$ is an injective function, meaning no two different points get mapped to the same one, so $sr^i=sr^j \implies r^i=r^j$, which clearly implies $i=j$.

Finally, $sr = r^{-1}s$ is more a visual one, or at least is how I understand it. Imagine a rect and a point to the left of that rect. Rotate it left around a point in the rect and mirror it ($sr$). Now imagine that same first point, mirror it first and now apply the same rotation but to the right ($r^{-1}s$). You then should get the exact same point you obtained with the previous transformation. Play around with some other configurations to get an intuition of it.

Hope it helps.

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