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With $\alpha = (12345)$ in the cycle notation, I should interpret it as:

$1\mapsto 2 \mapsto 3\mapsto 4\mapsto 5\mapsto 1$

I need to find out $\alpha^2$ and write it in cyclic notation. As I am not quite apt at it, I used the two row notation to solve it:

$\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{pmatrix}$

So $\alpha ^2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 1 & 2 \end{pmatrix}$

The answer provided is $\alpha^2 = (13524)$. Similarly, the answers to $\alpha^3$ etc were not matching. What am I doing wrong?

  • 1
    you're answer is the same as the one provided. Also \alpha^3 = \left(\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 5 & 1 & 2 & 3\end{smallmatrix}\right) = (14253)2011-09-11

1 Answers 1

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You're not doing anything wrong; your answer agrees with the book's answer:

$\alpha^2 = (13524)$

means that $\alpha^2$ acts by

$1\mapsto 3 \mapsto 5\mapsto 2\mapsto 4\mapsto 1$

which is precisely what the two-column form

$\alpha ^2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 1 & 2 \end{pmatrix}$

says (or rather, it says $1\mapsto 3$ and $2\mapsto 4$ and $3\mapsto 5$ and $4\mapsto 1$ and $5\mapsto 2$, which we simplify by writing $1\mapsto 3 \mapsto 5\mapsto 2\mapsto 4\mapsto 1$).

  • 5
    Don't worry, happens to all of us :)2011-09-11