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I am using Thomas Judson's open-sourced Abstract Algebra textbook, available at http://abstract.ups.edu/aata, and I'm having trouble with this problem:

Let $R$ represent a rotation element in $D_n$, the $n$th dihedral group. Prove that the order of $R^k \in D_n$ is $n/\gcd(n,k)$.

Here's what I've done/know so far:

  • Let $n,k \in \mathbb{N}$. Treat $k$ as if it were under (mod $n$); i.e. if $R^\kappa \in D_n$ such that $\kappa > n$, then rewrite $\kappa$ as $\kappa \equiv k (\text{mod} \; n)$. Otherwise, we already have that $R^k \in D_n$ such that $1 \leq k \leq n$.
  • Because $R \in D_n$, $R^n=1$ means $R$ is a counter-clockwise rotation of $2\pi/n$ radians, so $R^k$ would be a counter-clockwise rotation of $2\pi k/n$ radians.
  • For any arbitrary element $a$ in a group $G$, the order of $a$ is the smallest positive integer $n$ such that $a^n=1$. ($e$ is typically the identity element for this author; but, he tends to use $e=1$ for permutation groups).
  • By definition of greatest common divisor, there exist $a,b \in \mathbb{Z}$ such that $ak+bn = d = \gcd(n,k)$.

I don't know how to connect these things to the order of $R^k$.

Note: I don't know if any of these concepts are relevant, but references to these aspects of group theory are not (yet) developed for this author's course structure: cosets, Lagrange's theorem, Fermat's and Euler's theorems, isomorphisms, direct products, factor groups, normal subgroups, alternating groups, homomorphisms, automorphisms, Burnside's counting theorem, the class equation, and Sylow theorems. I'll get to those ideas in later chapters.

user372364
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1 Answers1

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Since $R$ generates a cyclic subgroup $C_n$ of $D_n$, the element $R^k$ has order $n/gcd(n,k)$, because this holds in general for cyclic groups of order $n$, see here for a proof.

Community
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Dietrich Burde
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  • Ah. So, $R$ is a generator of $D_n$ was the key piece I wasn't connecting in. I was thinking that I needed to get $k$ to look like $n$ somehow. Thank you. And the link for the proof is helpful, too. – user372364 Oct 13 '16 at 07:47
  • Well, $R$ is a generator of $C_n$, not $D_n$, because dihedral groups are not cyclic. For $D_n$ we have two generators $R$ and $S$ with $R^n=S^2=1$ and $RS=SR^{-1}$. But for the order of $R^k$ this does not matter. – Dietrich Burde Oct 13 '16 at 07:54
  • So, for example, if I consider $D_12$, $R=2\pi/n=\pi/6$, so $R^3=\pi/2$ and has order $12/\gcd(12,3)=12/3=4$. That makes sense to me, as $(R^3)^4=1$ because $\pi/2 \cdot 4=2\pi$ radians, which is 1 revolution. But this only grabs the rotation cycle symmetries, not the reflection cycle symmetries. – user372364 Oct 13 '16 at 08:09
  • *that was supposed to be $D_{12}$. – user372364 Oct 13 '16 at 08:11
  • I don't understand why your last sentence matters, but in any case $ord(R^3)=4$ is correct, right? – Dietrich Burde Oct 13 '16 at 08:11
  • 'Dihedral groups are not cyclic'. I'm chewing on that one for a minute (or two...). – user372364 Oct 13 '16 at 08:19
  • Don't chew! The presentation shows that we cannot have only $1$ generator (which is what cyclic means), but rather $2$ generators: see here. – Dietrich Burde Oct 13 '16 at 08:21
  • And yes, the elements generated off of $R^3={\pi/2, \pi, 3\pi/2, 1}$, which has order 4 and is a cyclic subgroup of $C_{12}$, which is part of the dihedral group $D_{12}$ – user372364 Oct 13 '16 at 09:41