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I am reading generators and relation part from Dummit and Foote's abstract algebra.Here group of symmetries of a regular n-gon is expressed as $$D_n=\langle r,s:r^n=s^2=e,rs=sr^{-1}\rangle\tag{A}$$where $r$ is the clockwise rotation of $\frac {2π}{n}$ radians

$s$ is the reflection about the line through center and the position of $1$ and $e$ is the idendity.

Now my question is that if we consider the r.h.s as a view of a abstract group(rather than the dihedral group $D_n$ then $r^n=e$ may not imply that $o(r)=n$.That is if $o(r)=k$ such that $k\mid n$ then the group(considering abstract view) $$H=\langle r,s:r^n=s^2=e,rs=sr^{-1}\rangle $$ is a subgroup of order $2k$ of some group $G$ containing $r,s,e$; $e$ being the identity.

But by $(A)$ , $\mid H\mid=2n$. Please clarify this confusion.

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It is implicitly assumed in a presentation that whenever something like $x^n=e$ is written that $n$ is the first positive integer to do it. Otherwise it is much too ambiguous as problems like yours arise.

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    $\begingroup$ Consider the group $K=\langle r,s:r^n=s^2=e,rs=sr^{2}\rangle$. This group is trivial group.So $o(r)\ne n$ and $o(s)\ne 2$ $\endgroup$ – Supriyo Halder Aug 16 '17 at 15:57
  • $\begingroup$ @user438576 What do you mean by "this group is trivial group"? That group doesn't seem like its only element is $e $. $\endgroup$ – Andrew Tawfeek Aug 16 '17 at 18:29
  • $\begingroup$ Sorry I have given a wrong counterexample.Consider this example $K=\langle r,s:r^5=e,rs=sr^2\rangle$.In this group $r=e$ condition is a hidden condition which can be established .So $r^5=e$ does not imply that $o(r)=5$.....The another example is $G=\langle r,s:r^4=s^3=e,rs=s^2r^2\rangle$ .This group has only element $e$ $\endgroup$ – Supriyo Halder Aug 17 '17 at 4:36

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