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Prove that $1, r, r^2,\ldots, r^{n-1}, t, tr, tr^2,\ldots, tr^{n-1}$ are $2n$ distinct symmetries of the regular n-gon.

It is clear that there are always $2n$ symmetries for any regular $n$-gon, with half of them being rotations and the other half being their reflections. Therefore for an $n$-gon with an even number of sides, every nontrivial symmetry in $D_n$, we shall call it, is either a reflection or rotation. Therefore $|D_n| = 2n$, as we have $2n$ elements, and each nontrivial element of $D_n$ is a reflection or rotation.

That was my thought, but I think I'm missing the mark. I haven't meddled with proofs for some time and am struggling with the seeming simplicity of this question.

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Prove that $1,r,r^2,\ldots,r^{n−1},t,tr,tr^2, \ldots,tr^{n−1}$, are $2n$ distinct symmetries of the regular n-gon.

I would focus more on proving that they are distinct.

If $r^i=r^j, j> i, \{i,j\} \subseteq \{ 0,n-1\}$, then $r^{j-i}=1$ but the order of $r$ is $n$.

We also check that we do not have $tr^i=r^j, \{i,j\} \subseteq \{ 0,n-1\}$, then we have $t=r^{j-i}$ but we know that $t$ is not a rotation.

The case to rule out $tr^i = tr^j$ should be trivial.

Also, I am not sure if you are required to prove that these are the only elements.

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You have showed that there are at least $2n$ symmetries, but there could be more. Note that these symmetries are distance preserving, in that if the distance between vertices $a$ and $b$ is $D$, then after applying the symmetry $s$ the vertices $s(a)$ and $s(b)$ will still have distance $D$.

Ok so now pick an arbitrary vertex $P$; the symmetry sends this to $Q$ ($n$ choices). But now the two vertices around it must be the two vertices adjacent to $P$, and we have two choices in where to put each. Fill in the details that after this, we no longer have any choice in our symmetry!

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