Decomposition of Cycle of Odd Length

this is the first time I've posted on here. I've been having issues with this particular homework problem for my group theory class. I'm not too sure how to begin the problem and, as it is a homework problem, a hint or a nudge in the right direction as opposed to a full solution would be greatly appreciated. The problem, as taken exactly from my assignment, is as follows:

Problem: Show that every cycle $\beta$ of odd length $k$ can be written as $\gamma^{2}$ where $\gamma$ is a cycle.

The problem also gives us a hint which is to consider $\langle \beta \rangle$. I'm also including two lemmas and a corollary which we proved in lecture. I'm doing this because I have a feeling I may need to use some or all of them, but I could be wrong.

Lemma 1
a) A cycle $\gamma = (a_{1}\, a_{2} \, a_{3} \, \dots \, a_{k})$ of length $k$ has order $k$.
b) $\gamma^{m}$ decomposes into $d=\mathrm{gcd}(m,k)$ cycles, each of length $\frac{k}{d}$

Lemma 2
Let $\langle g \rangle$ have order $m$, and $k$ be any integer. Then $\langle g^{k} \rangle$ has order $\frac{m}{d}$ where $d= \mathrm{gcd}(k,m)$

Corollary
If a permutation $\pi$ has cycle decomposition $\gamma_{1} \, \gamma_{2} \, \dots \, \gamma_{s}$, then the order of $\pi$ is $\mathrm{lcm}(|\gamma_{1}|, |\gamma_{2}|, \dots , |\gamma_{s}|)$

What I've Done

So far I know that the order of $\langle \beta \rangle$ is the same as the order of $\beta$ which is $k$. Since $k$ is odd we can write it as $k=2n+1$ for some $n \in \mathbb{Z}$. I also know that $\beta^{m}$ will decompose into $d=\mathrm{gcd}(m, 2n+1)$ cycles each of length $\frac{2n+1}{d}$. I considered letting $m=1$ but then all I get is that $\beta$ can be decomposed into $1$ cycle of length $2n+1$, but this doesn't really get me anywhere. And that is about as far as I've gotten. I'm really just not sure where to start. Thanks in advance for any help.

• Consider $\gamma \mapsto \gamma^2$ from $\langle \beta \rangle$ to itself. The map is bijective, and so $\exists \gamma \in \langle \beta \rangle$ such that $\gamma^2 = \beta$. Not sure why $\gamma$ should be a cycle though! – Prahlad Vaidyanathan Oct 17 '13 at 5:12

Take $r=n+1$. Then $(\beta^r)^2=\beta$. Note that $gcd(n+1,2n+1)=1$. By Lemma 1, part (b), you get that $\beta^r$ is a cycle. So your $\gamma$ is $\beta^r$.
• I don't understand what you prove here because of course $\beta^r$ is cycle... because $ord(\beta)=r$. – CS1 Nov 26 '13 at 9:12
• $ord(\beta)=2n+1$, not $r$. – D. N. Nov 26 '13 at 9:51
• Where is the definition of $\beta$? – CS1 Nov 26 '13 at 9:52