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Let $\sigma = (15793)(2468) ∈S_9$

Find all $τ ∈ S_9$ such that $τ^3=σ$. And prove there are no more such $τ ∈S_9$.

I found $τ=(17359)(8642)$. Is this the only solution ? Or are there more solutions possible ?

In general, if $x^3=y \ne e$. And you want to solve for $x$, is it possible to have more then one solution ? I can't think of an example that ${x_1}^3={x_2}^3 \ne e$, and $x_1 \ne x_3$ in group theory.

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(123)(78) and (456)(78) both cube to (78) –  user58512 Mar 24 '13 at 22:26

2 Answers 2

up vote 6 down vote accepted

Observation 1) The numbers $1,5,7,9,3$ and the numbers $2,4,6,8$ must be in the same cycle of $\tau$, indeed $\tau^3(1)=5$, $\tau^3(5)=7$ etc.

Observation 2) $\tau$ can not consist of only one cylce, since then $\sigma$ would consist of three $3$-cycles.

We can conclude that $\tau$ consists of exactly two cycles and is then uniquely determined. $$\tau=(17359)(2864)$$

Edit: To clarify a bit more why $\tau$ is uniquely determinded, we can try to construct the 5-cycle. It will be of the form $(1??5?)$ since $\tau^3(1)=5$. So it will be of the form $(17?5?)$ as $\tau^3(5)=7$, etc.

Edit 2 (concerning the comment): Well, the reasoning above works similarly for the general case. All numbers in a cycle of $\sigma$ must be in the same cycle of $\tau$. Also taking the cube can do two different things to an $n$-cycle of $\tau$. Either you get another $n$-cycle back, containing the same numbers (this happens, when $n$ and $3$ are coprime), or if $3$ divides $n$ you end up with three cycles of length $\frac{n}{3}$.

So given any $\sigma$ you will either find one cyle of length $n$ for every cycle of length $n$ in $\tau$ (n>1), or any three cycles of the same length can be combined to one cycle of triple that length. Note that this also works for '$1$-cycles', i.e. fixed points of $\sigma$.

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Thanks for your answer. I'm trying to think of how I can see in general if there are possible more solutions to such equations as given here. Could you help me with that ? –  Kasper Mar 24 '13 at 22:37
    
@Kasper i added something at the end of my answer which clarifies this extra question you are asking.... –  N. S. Mar 24 '13 at 22:49
    
Observation 2) is wrong. The cube of any 9-cycle is three 3-cycles. –  TonyK Mar 24 '13 at 22:53
    
Thanks @TonyK, I changed this. Hope it's correct now. –  Simon Markett Mar 25 '13 at 0:32
    
I really like your edits ! Thank you very much much ! –  Kasper Mar 25 '13 at 14:50

Note that the order of $\sigma$ is 20. It follows that the order of $\tau$ is either 20 or 60.

Write $\tau$ as a product of distinct cycles. then the order of $\tau$ is the least common multiple of the length of the cycles, and it cannot be $60$ (hint: $3+4+5 >9$.) Thus $\tau$ must have order $20$.

Then

$$\tau=\tau^{21}=\sigma^7$$

It follows that $\tau=\sigma^7$ is the only solution.

P.S. Note that in $S_{12}$ there are permutations of order $60$. If you solve the same problem in $S_{12}$ there are at least three solutions: $$\{ (17359)(2864), (17359)(2864)(10 \, 11 \, 12), (17359)(2864) (10 \, 12 \, 11) \} \,.$$

[It is easy to show these are the only ones.]

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