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Permutation $ \pi$ has a signature $2^43^5$. Find number of permutation $\sigma$ such that $\sigma^4 = \pi$

Could you give me a clue ?

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  • $\begingroup$ I am not sure if your notation for signature is a general one... can you just define what do you mean when you say signature? $\endgroup$ – user87543 Sep 6 '14 at 19:43
  • $\begingroup$ There are four cycles lenght of 2, and five cycles length of 3, rest of cycles are lenght of 1 $\endgroup$ – xawey Sep 6 '14 at 19:58
  • $\begingroup$ The number will depend on the cardinality of the underlying set (which is 23 plus the number of cycles of length 1). $\endgroup$ – Dave Sep 6 '14 at 20:05
  • $\begingroup$ so Is there solution ? $\endgroup$ – xawey Sep 6 '14 at 20:16
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Hint: Consider the primes $2, 3$ separately.

What is the fourth power of a $3$-cycle?

Now note that for the $2$-cycle (transposition) bit, if you square a transposition, you will get the identity. When you square this fourth power, you get the identity in the $2$ bit of your element. Can you think of a kind of permutation which would have that property? Can you find a permutation of that kind whose fourth power is a product of four transpositions?

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  • $\begingroup$ fourh power of 3-cycle is the same 3-cycle. fourth power of 6-cycle yields twice 3-cycle $\endgroup$ – xawey Sep 6 '14 at 21:02
  • $\begingroup$ So can you see how to get the part of the fourth root $\sigma$ which gives you the $3^5$ part of $\pi$? $\endgroup$ – Mark Bennet Sep 6 '14 at 21:08
  • $\begingroup$ It may be: two 6-cycles + one 3-cycle or 5 *3-cycle+0*6-cycle or 3 * 3-cycle+1*6-cycle $\endgroup$ – xawey Sep 6 '14 at 21:12
  • $\begingroup$ @xawey Now you need to sort out the other part of the problem. $\endgroup$ – Mark Bennet Sep 6 '14 at 21:16
  • $\begingroup$ Ok, but when can I get 2-cycles ? It must be cycle of length 8 ? $\endgroup$ – xawey Sep 6 '14 at 21:31

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