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let $\tau = (1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$ $\sigma = (1,4,12,11,8)(3,9,5,10)$

I have to calculate $\tau \circ \sigma^5 \circ \tau$. Question: Is there a more efficient way to calculate that than doing each composition?

I thought that I could do some calculation optimazation according to that post. But $\tau$ has the order $20$ so it does not help..

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jublikon
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    The order of $\tau$ is $2$ and $\tau(123)\tau^{-1}=(\tau(1)\tau(2)\tau(3))$. – bfhaha Mar 20 '17 at 12:18
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    note that $\sigma^5 = (3,9,5,10)$. – Abstraction Mar 20 '17 at 12:18
  • @bfhaha can you explain it a bit more detailed, please? Wolfram alpha returns $(1,12)(2,11)(3,4,5,9,8,10)(6,7)$. I see there must be a link but I cannot explain it... – jublikon Mar 20 '17 at 12:27
  • $\tau(1)\overset{\tau^{-1}}\to 1\overset{\sigma}\to \sigma(1) \overset{\tau}\to \tau(\sigma(1))$, so, say, if $1$ is mapped by $\sigma$ to $2$, then $\tau(1)$ is mapped by $\tau \sigma \tau^{-1}$ to $\tau(2)$. – zhoraster Mar 20 '17 at 15:35

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You can verify the identity $\tau\sigma\tau^{-1}=\tau(i_1,i_2, ...,i_n)\tau^{-1}=(\tau(i_1),\tau(i_2),...,\tau(i_n))$ directly or figure it out by the following commutative diagram.

\begin{align*} i & \stackrel{\tau}{\rightarrow} \tau(i)\\ \sigma\downarrow & ~~~~~~~\downarrow\tau\sigma\tau^{-1} \\ j & \stackrel{\tau}{\rightarrow} \tau(j)\\ \end{align*}

As Abstraction says, $\sigma^5=(3, 9, 5, 10)$. Note that $\tau^2=1$ and $\tau^{-1}=\tau$. So $$\tau\sigma^5\tau =\tau(3,9,5,11)\tau =\tau(3,9,5,11)\tau^{-1} =(\tau(3),\tau(9),\tau(5),\tau(11)).$$

bfhaha
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