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If we have the following permutation: $\sigma = (2,1,4)(4,5,1,6) \in S_7$

Am I right in assuming it will permute in the following way?

$ \begin{array}{clcr} 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \\ 6 \ 1 \ 3 \ 5 \ 2 \ 4 \ 7 \end{array}$

And, is there a quick way of finding a $\tau \in S_7$ such that $\tau^{-1} \sigma \tau = (1,2)(3,4,5)$, i.e. not having to write out all the permutations and checking which one fits? I've tried solving it with the cancellation law but screwed up somewhere.

Thanks.

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  • $\begingroup$ How do you get $\sigma(5)=2$? We feed $5$ through $(4\;5\;1\;6)$ first, giving $1$, and then $(2\;1\;4)$ applied to $1$ gives $4$, not $2$. $\endgroup$ – hmakholm left over Monica Apr 15 '12 at 14:18
  • $\begingroup$ For your second question, the general way to solve that is to multiply out the permutations (i.e., what you're already doing if only you were doing it correctly) and then expand it as a product of disjoint cycles. Then to find $\tau$ just write the two disjoint-cycle expansions next to each other (matching the cycle lengths, which must be possible or they are not conjugates) and map the indices in the $\sigma$ expansion to the corresponding ones in $(1\; 2)(3\; 4\; 5)$. $\endgroup$ – hmakholm left over Monica Apr 15 '12 at 14:21
  • $\begingroup$ Okay, so if I have the following disjoint cycle for $\sigma = (1,6,2)(4,5)$ and I want to map it to $(3,4,5)(1,2)$, will the following cycle for $\tau$ do? $\begin{array}{clcr} 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \\ 3 \ 5 \ 3 \ 1 \ 2 \ 4 \ 7 \\ \end{array}$ ? I don't think it's correct that 3 should repeat? And I'm not sure how to accomodate the $\tau^{-1}$ in all this. $\endgroup$ – janvdl Apr 15 '12 at 14:41
  • $\begingroup$ Unrelated but important: "Permutation of a group" does not make sense whatsoever. Permutation of some objects make sense, note that "object s " occurs here in plural. Hope you understand that this abuse of terminology can cripple understanding, (yes, it is not the case here, but in general...?) Regards, $\endgroup$ – user21436 Apr 15 '12 at 15:10
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    $\begingroup$ You need to include the length-1 cycles such that all elements are accounted for, so match $(1\;6\;2)(4\;5)(3)$ up with $(3\;4\;5)(1\;2)(6)$. After you decide on a $\tau$, just invert it to get the mathing $\tau^{-1}$. $\endgroup$ – hmakholm left over Monica Apr 15 '12 at 15:12
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Your calculation is almost correct, but $5$ gets sent to $4$ and $6$ gets sent to $2$, so the final line should read $6 1 3 5 4 2 7$. Using this calculation, you can write $\sigma$ in disjoint cycle form: $\sigma = (1, 6, 2)(4, 5)$.

With this calculation in hand, can you find the $\tau$ you are looking for? Let me give you a start: $\tau (1, 2) \tau^{-1} = (\tau(1), \tau(2))$. (Think through why this is true.) You can similarly compute $\tau (3, 4, 5) \tau^{-1}$. Comparing with the expression for $\sigma$ will tell you how to choose $\tau$.

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  • $\begingroup$ Okay, so you have to start with the rightmost bracket first? So first you notice $(4,5,1,6)$ and 5 is mapped to $1$. Then you apply the second bracket which maps $1 \rightarrow 4$ and so $5 \rightarrow 1 \rightarrow 4$ ? $\endgroup$ – janvdl Apr 15 '12 at 14:33
  • $\begingroup$ Yep, that's it. Since you're really doing composition of functions, the convention is to work from right-to-left. $\endgroup$ – Michael Joyce Apr 15 '12 at 14:39

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