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Is there any method to solving such an equation: $$f_1\circ f = f_2$$ Where $f_1, f, f_2 \in S_7$ and:
$f_1 = (1234)(5)(6)(7)$
$f_2 = (172536)(4)$

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  • $\begingroup$ the permutations form a group so $f=f_1^{-1} \circ f_2$ $\endgroup$ Apr 26, 2017 at 21:08
  • $\begingroup$ Thanks! That may help @DavidHolden $\endgroup$
    – Hendrra
    Apr 26, 2017 at 21:12
  • $\begingroup$ do you know how to find the inverse of a permutation? $\endgroup$ Apr 26, 2017 at 21:34
  • $\begingroup$ I posted my solution. I would really appreciate if you could check it @DavidHolden $\endgroup$
    – Hendrra
    Apr 26, 2017 at 21:39

3 Answers 3

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you are nearly there. note that the repeated $1$ in your $(14321)$ is incorrect notationwise. also unless there is a reason not to do so we usually omit cycles consisting of a single element. so $$ f_1 = (1234) $$ and $$ f_1^{-1} = (4321) = (1432) $$ therefore (following the usual convention of evaluating from right to left) $$ f = f_1^{-1} \circ f_2 \\ =(1432)\circ (172536) \\ = (17)(25)(364) $$

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  • $\begingroup$ Alright, I will "repair" the mistakes :) Thanks for your help! $\endgroup$
    – Hendrra
    Apr 26, 2017 at 21:51
  • $\begingroup$ most welcome. the "arithmetic" of permutations needs a little practice, as it is an unfamiliar mode of computation, but this notation is very useful. $\endgroup$ Apr 26, 2017 at 21:53
  • $\begingroup$ That's true. I've been learning that field of Mathematics not for a long time so some things are still a bit confusing for me. $\endgroup$
    – Hendrra
    Apr 26, 2017 at 22:00
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So to my mind that would be the solution:
1) first we are to find $f_1^{-1}$:
$$f_1^{-1}\circ f = id \rightarrow f_1^{-1} = (13421)(5)(6)(7)$$ 2) now we are to solve:
$$f_1^{-1} \circ f_2 = (17)(254)(36)$$ And that is our permutation $f$ we were looking for.

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You can also solve it by writing explicitly the table of transformations.

Note that since $f_1$ order is $4$ then $(f_1)^{-1}=(f_1)^3$.

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 1 & 2& 3 & 4 & 5 & 6 & 7 \\ \hline f_2(x) & 7 & 5 & 6 & 4 & 3 & 1 & 2\\ \hline f_1\circ f_2(x) & 7 & 5 & 6 & 1 & 4 & 2 & 3\\ \hline f_1\circ f_1\circ f_2(x) & 7 & 5 & 6 & 2 & 1 & 3 & 4\\ \hline f_1\circ f_1\circ f_1\circ f_2(x) & 7 & 5 & 6 & 3 & 2 & 4 & 1\\ \hline \end{array}$

Now $(f_1)^3\circ f_2=(f_1)^3\circ f_1 \circ f=Id\circ f=f$

So that last line is $f$ : $1\to 7\to 1$ and $2\to 5\to 2$ and $3\to 6\to 4\to 3$

Which in proper notation is $f=(17)(25)(364)$

I always find it difficult to figure out where the elements are going without a table.

Note that if you want to check your result, wolframalpha considers that $\sigma\tau$ is $\tau(\sigma(\cdot))$ so you have to inverse the order of cycles.

https://www.wolframalpha.com/input/?i=permutation+(172536)(1234)%5E(-1)

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