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Suppose we have disjoint cycles $\sigma=(1 3 2 4)(5 6)$ and $\tau=(1 4)(2 5 3 6)$. Can we find some $\alpha\in S_6$ such that $\alpha\sigma\alpha^{-1}=\tau$?

My solution: $\alpha\sigma\alpha^{-1}=(\alpha(1) \alpha(3) \alpha(2) \alpha(4))(\alpha(5) \alpha(6))=(2 5 3 6)(1 4)$, and I get$\alpha=(1 2 3 5)(4 6)$. But by checking with a calculator online, my answer is wrong. Can someone help me on identifying where I am wrong and write out the solution? Thank you.

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    In which direction do you compose permutations? – Wuestenfux Apr 16 '20 at 12:38
  • Hi, thanks for the quick reply. I don't understand what you mean. Can you write your comments to my solution? –  Apr 16 '20 at 12:41
  • "by checking with a calculator online" It is worth pointing out that as Wuestenfux alludes to that different authors use different conventions in regards to the order of applying permutations. Recall that $\sigma\circ \tau$ might be unequal to $\tau\circ\sigma$. It is possible that whatever book you are using multiplies permutations in the one direction while the calculator you used multiplies in the other direction. – JMoravitz Apr 16 '20 at 12:43
  • I see. From right to left. –  Apr 16 '20 at 12:45
  • Okay. But can you help me identify whether my method is correct or not? –  Apr 16 '20 at 12:47
  • The calculator is Wolframalpha. Just type in the command "permutation" and cycles of permutation you'd like to calculate. https://www.wolframalpha.com/ –  Apr 16 '20 at 12:52
  • @PseudodifferentialOperators and wolfram applies things left to right. $(1~2)(1~3)$ results in $(1~2~3)$, the opposite of what I interpret you as saying the convention you follow is. Note the fine print, "the product $\sigma \tau$ represents the permutation $\tau(\sigma(\cdot))$" – JMoravitz Apr 16 '20 at 12:55
  • Oh, I see. I checked by wolfram alpha again by correct order. It is correct. Thanks everyone! –  Apr 16 '20 at 12:57
  • From what we can tell, yes. If you were to just reverse the order in which you write the permutations (with a tad bit of laziness allowed in terms of disjoint cycles) in order to account for the differences in order in which permutations are applied between your convention and wolfram's, you can get wolfram to agree – JMoravitz Apr 16 '20 at 13:02
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    Does this answer your question? Composition of permutations left to right or right to left? – JMoravitz Apr 16 '20 at 13:05
  • Actually, I just figured it out by myself. Anyway, thanks for answering! –  Apr 16 '20 at 13:06
  • Also related: https://math.stackexchange.com/questions/1394423/conjugacy-in-s-n-with-composing-permutations-left-to-right-vs-right-to-left?rq=1 – JMoravitz Apr 16 '20 at 13:08

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