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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?

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you're answer is the same as the one provided. Also $\alpha^3 = \left(\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 5 & 1 & 2 & 3\end{smallmatrix}\right) = (14253)$ –  Deven Ware Sep 11 '11 at 0:53

1 Answer 1

up vote 7 down vote accepted

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

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Ahh. Thanks. I feel silly now. –  kuch nahi Sep 11 '11 at 0:53
5  
Don't worry, happens to all of us :) –  Zev Chonoles Sep 11 '11 at 0:55

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