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I know the question has been asked and answered many times, but I am trying to shore up my understanding of this concept. Given the questions here and here, does this mean that I could rearrange the bottom row of the array form and each rearrangement can be written in the form of 2-cycles? Also, each rearrangement is guaranteed to commute? Essentially, $\sigma\times(12)(34)(56)=(12)(34)(56)\times \sigma$ for all $\sigma$?

Examples:

$\begin{equation} \sigma=\left(\begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6\\ a_1 & a_2 & a_3 & a_4 & a_5& a_6 \end{array}\right)=(a_1a_2)(a_3a_4)(a_5a_6)\end{equation}$

$\begin{equation} \sigma_1=\left(\begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6\\ 1&2&3&5&4&6 \end{array}\right)=(12)(35)(46)\end{equation}$

$\begin{equation} \sigma_2=\left(\begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6\\ 1&2&3&6&4&5 \end{array}\right)=(12)(36)(45)\end{equation}$

frierfly
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    For $\sigma$ to commute with (12)(34)(56) you need $\sigma \times (12)(34)(56)=(12)(34)(56) \times \sigma.$ Note how the $\sigma$ occurs on the left on one side, on the right on the other. So your equation with $\sigma$ only there once is not what is to be looked at. – coffeemath Mar 22 '15 at 22:57
  • Oops, fixing post now. – frierfly Mar 23 '15 at 02:49
  • The cycle versions to the right of your three examples are not right. For example $\sigma_1=(45)$ since it only switches 4,5 and fixes all the others. – coffeemath Mar 23 '15 at 10:41

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