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$(2,3)(4,6,5,1,2)=?$

The multiplication is from right to left. I don't know, where I make the mistake.

Denote $\tau=(2,3), \sigma=(4,6,5,1,2)$

$1\ \ 2\ \ 3\ \ 4\ \ 5\ \ 6$$\quad$ first apply $\sigma$

$2\ \ 4\ \ 3\ \ 6\ \ 1\ \ 5$$\quad$ then $\tau$

$2\ \ 3\ \ 4\ \ 6\ \ 1\ \ 5$

$\Rightarrow (2,3)(4,6,5,1,2)=(1,2,3,4,6,5)$

I think this form is correct (If I convert the cycle notation to one-line notation, compose the permutations and then convert it back to cycle-form), but I want to derive it directly from the cycle notation, why does it fail ?

$1)$ $4$ goes to $6$ and $\tau$ does nothing to $6$, $\Rightarrow (4,6,...)$

$2)$ $6$ goes to $5$ and $\tau$ again fixes $5$, $\Rightarrow (4,6,5,..)$

$\dots\Rightarrow (4,6,5,1,...)$

Now $1$ goes to $2$ and $\tau$ sends $2$ to $3$ $\Rightarrow (4,6,5,1,\color{red}{3}..)$

am I wrong ?

inequal
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    What you're doing wrong is in the first part you assume the permutation works on specific positions while in the second part you assume permutations work on specific elements. You can see this because in the first part, $\tau$ swaps $3$ and $4$, since they are the numbers in positions $2$ and $3$, while below you say "$1$ goes to $2$ and $\tau$ sends $2$ to $3$". – Arthur Jan 09 '15 at 00:48
  • @Arthur you mean the 3rd line of the first computation should be $3\ 4\ 2\ 6\ 1\ 5$ ? – inequal Jan 09 '15 at 00:52
  • Exactly which one of them is correct is up to your professor / the author of your book / your personal preference, but yes, that would make your two arguments agree. – Arthur Jan 09 '15 at 00:53
  • @Arthur. Thanks a lot, such a mistake could be fatal in future. – inequal Jan 09 '15 at 00:59

1 Answers1

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Yes, it's $(4,6,5,1,3,2)$. If you plot the big cycle and the small transposition, the effect is an implant on the initial cycle, giving a bigger one.

orangeskid
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