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Let $f = (1 2 3 4 5 6) ◦ (7 8 9 10) ∈ S_{10}$. Does there exist a positive integer n such that, when $f^n$ is decomposed into disjoint cycles, one of the cycles has length 5? Justify your answer.

Any directions? Because I don't have the experience to play with powers of cycles...

Bernard
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    A few questions you may want to ask yourself: what are the orders of your two cycles? What about the powers of f? And the cycles composing it? (Use Lagrange’s theorem) – Noe Blassel Mar 25 '19 at 22:33
  • @NoeBlassel Hm, so the orders of the 2 cycles are equal to their respective lengths, so 6 and 5. But I don't know how to compute the order of a power of f, and unfortunately we will probably be taught Lagranges theorem sometime this week, so I don't know how to use it yet –  Mar 25 '19 at 23:17
  • 6 and 4. A corollary of Lagrange’s theorem is that the order of any power of a cycle divides the original order, and the order of any cycle in the decomposition of a permutation divides the order of the permutation. – Noe Blassel Mar 26 '19 at 10:35

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Just an example for the powers of a cycle: \begin{align} f&=(1\,2\,3\,4\,5\,6)& f^2&=(1\,3\,5)(2\,4\,\,6) \\ f^3&=(1\,4)(2\,5)(3\,6)&f^4&=(1\,5\,3)(2\,6\,4) \\ f^5&=(1\,6\,5\,4\,3\,2)=f^{-1}& f^6&=()=\operatorname{id} \end{align} Can see why, if $f$ is a cycle of length $\ell$, $f^k$ is decomposed as the product of $\gcd(\ell,k)$ disjoint cycles of length $\frac{\ell}{\gcd(\ell, k)}$?

Furthermore, powers of disjoint cycles are disjoint.

Bernard
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  • Last edit, I think I got this now: Since the cycles in which f is decomposed have orders 6 and 4, and $f^n = (c_1)^n (c_2)^n$ , and there are no integers m,n such that $6/m = 5$ or $4/n = 5$, then the answer to my question is no. –  Mar 25 '19 at 23:53
  • It's exactly so. – Bernard Mar 26 '19 at 09:37
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Let $G$ be a finite group, and $m,n$ positive integers such that gcd$(m,n)=1$. Let $x,y$ be two elements of order $m$ and $n$ respectively.

Then the two cyclic subgroups generated by $x$ and $y$ respectively will have only the identity element in common.

Now coming to symmetric groups, if a cycle of length $p$, a prime number, occurs in the disjoint cycle decomposition of a permutation $\sigma$, then some power of $\sigma$ will be of order $p$. [Verify this].

Now combining these two statements you can see that the answer is negative for your question.

P Vanchinathan
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