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The presentation of the Dihedral Group $$D_{2n} = \langle r,s\mid r^{n} = s^{2} = e , rs = sr^{-1} \rangle $$

Shows that the order of the $D_{2n}$ is 2n

as in $ r^{n} = s^{2} = e --(1)$ shows that it is at least 2n

While $rs = sr^{-1} --(2)$ shows that it is at most 2n

Why is this so? From the geometrical view of the dihedral group I understand that the order is precisely 2n

Is it the case that (1) relation shows that there is n degrees of freedom multiply by 2 ... hence there exist those 2n elements generated from the relation

While (2)... I am not sure

Thanks for the assistance!

We_Will_be_Happy
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  • Relation (2) tells you that you may write any element in $D_{2n}$ as $s^k r^m$ by bringing all $s$'s to the left. Geometrically, (2) tells you that mirroring a rotation $r$ gives you the inverse rotation $r^{-1}$. – MPos Jan 02 '24 at 07:33
  • I think you should read the approved answer to this question https://math.stackexchange.com/questions/4270475/prove-that-order-of-d-2n-is-exactly-2n – ancient mathematician Jan 02 '24 at 07:46
  • Hi @MPos thanks for reply , while I can see that mirroring a rotation is equal to inverse rotation of the mirror? $= s r^{-1} $ ... since (2) provides the upper bound... (is that the case)? I am guessing that (2) ensures that the order of each $ s^{k}r^{m} $ is 1 – We_Will_be_Happy Jan 02 '24 at 09:57
  • Hi @ancientmathematician thanks for the reply and link, somehow my searches did not lead to that well-posed question that said it seems the reply delves into the "at least" portion of the order of the Group.. and I looked up Von Dyck's theorem and note that every Group can be uniquely represented by its representation $\langle a \epsilon A \mid r \epsilon R $ up to isomorphism... and certainly we can show that a Group has order at least 2n by finding a surjective mapping to a Group order 2n (for example) . But what about the at most? – We_Will_be_Happy Jan 02 '24 at 10:17
  • The relations let you write every element as $r^i s^j$ with $i=0,1,\dots, n-1$ and $j=0,1$. – ancient mathematician Jan 02 '24 at 11:39

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