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Isn't this impossible...? Because this permutation goes from 3 --> 2 ---> 5 ---> 3 according to the first cycle, but goes from 2 --> 5 ---> 4 ---> 2 according to the second cycle. So 5 can't go to 3 and 4.

Chris
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  • What is table notation? Do you do the multiplication left to right? – Em. Apr 22 '16 at 05:02
  • It's where you write the permutations like I did with the arrow, except instead of 2--->5 you write 2 on top and 5 directly below that – Chris Apr 22 '16 at 05:03
  • Read it from right to left in terms of what each permutation does to a particular element. $(3,2,5)(2,5,4) = \begin{pmatrix} 1&2&3&4&5\ 1&3&2&5&4\end{pmatrix}$. This is because look at what happens to $2$. After the first cycle it gets changed to five, and while five it gets changed to a three by the second permutation, so altogether $2$ gets mapped to $3$. Similarly for the other entries. – JMoravitz Apr 22 '16 at 05:09
  • Chris, you really have to tell which convention you assume for the product of permutations, as both are used (left-to-right and right-to-left). JMoravitz and probablyme answers are both correct. – Jean-Claude Arbaut Apr 22 '16 at 05:21
  • I believe it's left-to-right. Sorry, I'm still not very familiar with permutations. – Chris Apr 22 '16 at 05:30

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Assuming you do multiplication left to right, then $$(3\quad2\quad5)(2\quad5\quad4) = \begin{pmatrix}1&2&3&4&5\\ 1&5&2&4&3 \end{pmatrix}\begin{pmatrix}1&2&3&4&5\\ 1&5&3&2&4 \end{pmatrix} = \begin{pmatrix}1&2&3&4&5\\ 1&4&5&2&3\end{pmatrix}$$

Em.
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