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.


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.