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I learned that a subgroup of $D_n = \langle r,s \mid r^n=s^2=(rs)^2=1 \rangle$, the dihedral group of order $2n$, is either cyclic or dihedral itself, and that a subgroup of the latter kind is of the form $\langle r^d, sr^i\rangle =: G(d, i)$, where $d\mid n$ and $0\le i < d$.

I would like to interpret this fact geometrically. I feel that $G(d,i)$ is the set of congruent transformation that preserves the regular $n/d$-gon that one gets by picking every $d$-th vertex of the original $n$-gon that the elements of $D_n$ preserves. However, in this picture I do not know how to interpret the generator $sr^i$ of order two of $G(d,i)$; since it is of order 2, I think it is a rotation about some axis, but I am not sure whether it preserves the $n/d$-gon in question.

I would be grateful if you could give some geometric intuition about this matter.

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$sr^i$ is not a rotation, but a reflection (since it has determinant $-1$). If you write down the matrix for it, then the axis should be apparent –  Prahlad Vaidyanathan Sep 5 '13 at 2:13
    
@PrahladVaidyanathan I mean rotation in the 3-space along some line (but the word reflection is more appropriate as you said). –  Pteromys Sep 5 '13 at 2:18

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The element $sr^i$ can be thought of as reflection about a different axis. $s$ is of course interpreted as a reflection about a given axis, and if you look at $sr$ you'll see that it is the same as reflecting about an axis rotated half the angle by which $r$ rotates (direction depends on the order of operations you take $sr$ to mean).

So for your subgroup $G(d,i)$, you'll think of $sr^i$ as reflection through a different axis. Call this new axis of reflection $l'$. Then take any $n/d$-gon which is symmetric about $l'$ and sharing the same center as the original $n$-gon. Then you can think of $G(d,i)$ as the dihedral group for that $n/d$-gon. The thing to note is that the isometries preserving this $n/d$-gon also preserve the original $n$-gon.

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