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I was trying to proof that the order of the dihedral group $D_{2n}$ is $2n$ given its presentation $$ D_{2n} = \langle r,s ~\vert~ r^n = s^2 = 1 ,~ rs = sr^{-1} \rangle $$ and I further assumed that the order of $r$ is precisely $n$ (someone please correct me if this is unnecessary assumption).
Anyway, I proved that there is at most $2n$ elements (by proving that all elements are of the form $s^ir^j$ for $0\le i<2$ and $0\le j<n$) and that no $2$ elements that have the same power of $(s)$ are equal (i.e. $s^ir^j \neq s^ir^k\hspace{0.2cm} \forall \hspace{0.1cm} j\neq k$), but regarding the third case (proving that $sr^i \neq r^j \hspace{0.2cm} \forall \hspace{0.1cm} 0\le i, j < n$), I am stuck at trying to prove that $s \neq r^k \hspace{0.2cm} \forall \hspace{0.1cm} 0\le k < n$.

My question is does the presentation of a group imply implicitly that any generator can't be an element of the cyclic group generated by another generator (i.e. $a_i \notin \langle a_j\rangle \hspace{0.2cm} \forall \hspace{0.1cm} a_i, a_j \in A $ and $ i \neq j$ where $A$ is the set of generators of the group). If not, can someone please tell me how to proceed from here.

Hesham Abdelgawad
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  • @DietrichBurde He didn't prove it, he just used it as a fact. The purpose of this question is to prove precisely this claim – Hesham Abdelgawad Jun 25 '23 at 19:29
  • This is proved here. The order is exactly $2n$, so all named elements must be different. – Dietrich Burde Jun 25 '23 at 19:30
  • I feel that the answer in here isn't complete you can clearly verify that the example of the group he gave is of order $2n$ but the map (I suppose he meant to define it as $\varphi : s^ir^j \rightarrow \sigma^i\rho^j$) may be not well defined because he didn't prove that all elements of this form are different – Hesham Abdelgawad Jun 25 '23 at 19:44
  • No, I mean Arturo's answer here. Once you have proved that this presentation yields a group isomorphic to the symmetry group $D_n$, defined by reflections and rotations, it is clear that $r^k$ and $s$ are different elements for a choice of $r$ and $s$. Usually a reflection is not a power of a rotation, but, say, for the square, $r^2$ is indeed a reflection. – Dietrich Burde Jun 26 '23 at 08:11

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