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Let $\sigma$ be a cycle with length $n$ where $\sigma \in S_m$.

How do i prove that $|\langle \sigma \rangle |$ is $n$?

Martin Sleziak
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John. p
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  • Induction...? And then you may want to subdivide in odd and even lenghts... – DonAntonio Apr 20 '14 at 23:31
  • @DonAntonio Yes, I have tried to apply induction, but i can't find where to fix and do induction. – John. p Apr 20 '14 at 23:36
  • @DonAntonio Note that $(123)$ is a cycle of length 3 but $(123)^2$ is again a cycle of length 3 – John. p Apr 20 '14 at 23:38
  • Might be worth adding a clarifying note that "order" here means its order in the permutation group, i.e. the smallest $n$ such that $\sigma^n$ is the identity permutation. (Someone had proposed an edit to the question with wording like "order of the cycle generator", which I find even more confusing.) – ShreevatsaR Apr 20 '14 at 23:38
  • I know that, @John.p...and that's why I proposed to check odd and even ones separatedly. – DonAntonio Apr 20 '14 at 23:43
  • @Donantonio I don't get it.. Would you please give me some more hints? For any odd cycle $\sigma$, $\sigma^2$ is an even cycle. I don't know in which way these to be separated.. – John. p Apr 20 '14 at 23:46

2 Answers2

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The order of a permutation $\sigma$ is the smallest positive integer $n$ such that $\sigma^n$ is the identity permutation. (This is sometimes a definition, sometimes a consequence of a different definition; I will assume it is known).

Now suppose we have an $n$-cycle $\sigma=(a_0\;a_1\;a_2\;a_3\ldots a_{n-1})$. Its order cannot be less than $n$, because if $0<k<n$ then $\sigma^k(a_0)=a_k$ and it's implicit in the concept of a cycle that $a_0\ne a_k$.

On the other hand, $\sigma^n$ is certainly the identity permutation, because it acts on each element of the cycle by moving it around the entire cycle one time. Formally one can prove (by induction) that $\sigma^k(a_j) = a_{(j+k)\bmod n}$, and since $n\equiv 0\pmod n$ we have $(j+n)\bmod n=j$ whenever $0\le j < n$. And elements that are not in the cycle at all are left alone by $\sigma$ and therefore also by $\sigma^n$.

Since $\sigma^n=e$ and $\sigma^k\ne e$ when $0<k<n$, the order of $\sigma$ is $n$.

(There's no need to divide into odd and even cases as Don Antonio suggests).

hmakholm left over Monica
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  • ''because it acts on each element of the cycle by moving it around the entire cycle one time.'' Aren't we assuming that $σ^k$ is one cycle for any k. Which is not the case ! What I'm missing? I just can't see how this observation is independent of whether the power and the number of elements are even or odd. – I0_0I Nov 02 '19 at 14:31
  • Although I did the induction proof and it doesn't require any assumption as you said, but I just feel it's just too easy to be true, where did the complexity of the problem go! Depending on the power you rais the permutation to you could have 1 or 2 or more disjoint cycles how is all that didn't appear in the induction proof! – I0_0I Nov 02 '19 at 15:16
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    @yousefmagableh: Each step in the iteration just applies the original cycle. In this part of the argument we don't worry about precisely what the grand-scale nature of $\sigma^k$ is, just that it brings everything back to its original position when $k=n$. – hmakholm left over Monica Nov 02 '19 at 18:17
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For example: even length, so that we know that transpositions have order $\;2\;$ and any cycle of length less than $\;2n\;$ has the appropiate order, so that:

$$\sigma=(i_1\,i_2\,\ldots\,i_{2n})\implies \sigma^2=(i_1\,i_3\,\ldots\,i_{2n-1})(i_2\,i_4\,\ldots\,i_{2n})$$

But by the Inductive Hypothesis (IH), we have that

$$\text{ord}(i_1\,i_3\,\ldots\,i_{2n-1})=n=\text{ord}(i_2\,i_4\,\ldots\,i_{2n})$$

and since both $\;n$-cycles are disjoint (and thus they commute), we get

$$\sigma^{2n}=(\sigma^2)^n=(i_1\,i_3\,\ldots\,i_{2n-1})^n(i_2\,i_4\,\ldots\,i_{2n})^n=1$$

But...why can't the order be less than $ \;2n\;$ above? Because that'd mean one of the two $\;n$-cycles above has order less than $\;n\;$ ...

Try now a similar argument for the odd length case (I think it is simpler as you always get, upon exponentiation, again a cycle of the same length...!)

DonAntonio
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  • "I think it is simpler as you always get, upon exponentiation, again a cycle of the same length" - you don't. Try a cycle of length 9, taken to the third power. – user2357112 Apr 21 '14 at 11:26
  • Your argument that the order can't be less than $2n$ only shows that it can't be an even number less than $2n$. If the order were to be odd, then you wouldn't be able to convert $\sigma^{order}$ to a power of $\sigma^2$. Similarly, I don't think the argument generalizes to the odd-length case, especially the prime-length case, since no power of $\sigma$ will allow you to apply the inductive hypothesis. – user2357112 Apr 21 '14 at 11:33
  • I don't get the argument for 'why can't the order be less than $2n$ above? ' please someone explain me. – ramanujan Jun 30 '18 at 19:28