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In my notes, it says that $\sigma = (1 5)(2 4 7 5)(1 4 6)(2 3) = (1 3 2 6)(4 7 5)$

However the answer I get is $\sigma = (1 5)(2 4 7 5)(1 4 6)(2 3) = (1 4 6)(2 3)(7 5)$. What I did was I started with the smallest element, 1, and worked through the cycles from right to left and saw what each element was mapped to. So $1 \rightarrow 4$ in the third cycle, and $4 \rightarrow 6$. Then $2 \rightarrow 3$ in the last cycle, and all we had left was $4 \rightarrow 7 \rightarrow 5$ in the second cycle.

I know my approach must be incorrect, but it's all I can find when I look through my notes and search online. Help would be appreciated :)

mathstack
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  • It seems like the first one has been evaluated left to right. It seems to send $1 \to 3,3 \to 2, 2 \to 6, 6 \to 1$, and $ 4 \to 7, 7 \to 5, 5 \to 4$. Even your computation seems incorrect. – Sarvesh Ravichandran Iyer Apr 25 '17 at 03:56

1 Answers1

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I actually disagree with both you AND the book. The book, it seems, has worked left-to-right instead of right-to-left; you have made an error in your computations.

To read this right-to-left, remember to work cycle-by-cycle:

$1$ is fixed by $(23)$. It is mapped to $4$ by $(146)$. $4$ is mapped to $7$ by $(2475)$. And $7$ is fixed by $(15)$; so, in all, $1\mapsto7$. Continuing in this way, you can write $(15)(2475)(146)(23)=(17)(23465)$.

Nick Peterson
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  • I wouldn't disagree with the book for its left-leaning practices :P But to rephrase the OP's specific error: $(146)$ doesn't send $1 \mapsto 4 \mapsto 6$ all in one go; once that cycle gets a turn sending $1 \mapsto 4$, it's done, and you move on to the next one. – pjs36 Apr 25 '17 at 03:59
  • I understand the mistake the pjs36 is talking about, however because of this I'm unsure why Nick Peterson says the 5 is present in (175)? – mathstack Apr 25 '17 at 04:11
  • Typo. :-) Should be $(17)(23465)$. I'll correct that. – Nick Peterson Apr 25 '17 at 15:09