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I am trying to prove myself that $(1)(2)(3)(4) = (12)(12)(3)(4).$

So,

$\begin{pmatrix} 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$ $\begin{pmatrix} 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$ $\begin{pmatrix} 3 & 4 \\ 3 & 4 \\ \end{pmatrix} = (12)(12)(3)(4).$

Let $p = \begin{pmatrix} 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$. Then $p \circ p = p^2 = \begin{pmatrix} 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$ $\begin{pmatrix} 1 & 2 \\ 2 & 1 \\ \end{pmatrix}.$ So, right to left: $(p \circ p)(1) = p(p(1)) = p(2) = 1$ and $(p \circ p)(2) = p(p(2)) = p(1) = 2.$ Thus, $p \circ p = \begin{pmatrix} 1 & 2 \\ 1 & 2 \\ \end{pmatrix}.$

Let $q = \begin{pmatrix} 3 & 4 \\ 3 & 4 \\ \end{pmatrix}.$ Then $(p \circ p) \circ q = p^2 \circ q = \begin{pmatrix} 1 & 2 \\ 1 & 2 \\ \end{pmatrix}$ $\begin{pmatrix} 3 & 4 \\ 3 & 4 \\ \end{pmatrix} = (1)(2)(3)(4).$

Does that make sense to you? In particular, I am having a problem with $p^2 \circ q(1) = p^2(q(1))$ and $p^2 \circ q(3) = p^2(q(3))$ which are undefined? Generally, how do we compose $\begin{pmatrix} 1 & 2 \\ 1 & 2 \\ \end{pmatrix}$$\begin{pmatrix} 3 & 4 \\ 3 & 4 \\ \end{pmatrix}$?

user255751
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  • I do it in my head. For example, consider what $(12)(12)(3)(4)$ does to 1. First 1 goes to 2, then 2 goes to 1. So 1 goes to 1. But if you want to write it your war, then it should be $p = \pmatrix{1&2&3&4\2&1&3&4}$ and $q=\pmatrix{1&2&3&4\1&2&3&4}$. Note that all one-cycles are the identity. – Stephen Montgomery-Smith Jul 21 '15 at 03:11
  • @ Stephen Montgomery-Smith, $p(p(1)) = 1, p(p(2)) = 2, p(p(3)) = 3, p(p(4)) = 4$, so $p^2$ is identity. This works out great. Is $(12)(12)(3)(4)$ actually $[(12)(3)(4)][(12)(3)(4)][(1)(2)(3)(4)]$? – user255751 Jul 21 '15 at 03:25
  • It is. But it is much easier to write $(12)(3)(4)$ as $(12)$. – Stephen Montgomery-Smith Jul 21 '15 at 03:28

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