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I have the follwing question:

Find the decomposition into cycles with disjoint supports, then determine the order and the signature.

I have the following cycle:

$$\sigma=\left(\begin{array}{rrr} 1&2&3&4&5&6&7&8&9\\3&4&9&7&6&5&1&2&8\end{array}\right)$$

I remark that $\sigma=\left(\begin{array}{rrr}1&3&9&8&2&4&7\end{array}\right)\cdot \left(\begin{array}{rrr}5&6\end{array}\right)$. Let us denote $c_1=\left(\begin{array}{rrr}1&3&9&8&2&4&7\end{array}\right)$ and $c_2=\left(\begin{array}{rrr}5&6\end{array}\right)$ We know that $$supp(c_1)=\{5,6\}, supp(c_2)=\{1,2,3,4,8,9\}$$Therefore their intersection is clearly empty. Futhermore $$ord(\sigma)=lcm{(l(c_1),l(c_2))}=lcm(7,2)=14$$ and $$sgn(\sigma)=(-1)^{9-1}=1$$

Does this work like this, the part where I am most unsure is the determination of the support.

Thank you for your help.

user123234
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    What formula are you using to compute the signature? I would have got $(-1)^{7-1}(-1)^{2-1} =-1$. – ancient mathematician Nov 20 '21 at 15:21
  • Also a comment: you say you have a "cycle" $\sigma$: you don't you have a permutation $\sigma$. – ancient mathematician Nov 20 '21 at 15:22
  • oh yes I computed it wrong sorry. – user123234 Nov 20 '21 at 15:25
  • Thank you but could I ask something more. Here we don't have "fixed points" what if we consider $$\sigma =\left(\begin{array}{rrr} 1&2&3&4&5&6&7&8&9\2&4&6&8&1&3&5&7&9\end{array}\right)$$ then we would have $$\sigma=(\begin{array}{rrr}1&2&4&8&7&5\end{array})(\begin{array}{rrr}3&6\end{array})$$ But then 9 would be a fixed point, so the support of the first cycle would be ${3,6,9}$ and the one of the second cycle would be ${1,2,4,5,7,8,9}$ and then the one of the last one would be ${1,2,3,4,5,6,7,8}$ right? so we can't use this decomposition? – user123234 Nov 20 '21 at 15:31
  • You are writing down the complements of the supports instead of the supports! In this case $\sigma=(1 2 4 8 7 5)(3 6)(9)$ of order $6$, supports/orbits ${1,2,4,7,8,5},{3,6},{9}$ and signature $(-1)^{6-1}(-1)^{2-1}(-1)^{1-1}=+1$. – ancient mathematician Nov 20 '21 at 15:57
  • I see that I missed the mistake before: sorry. Your supports in the original question are swapped. – ancient mathematician Nov 20 '21 at 15:58
  • So but then I also did it wrong in the previous exercise, i.e. I swaped the definition there – user123234 Nov 20 '21 at 15:59
  • Do I need to write (9) at the end of my decomposition? – user123234 Nov 20 '21 at 16:00
  • For the signature: note that it's a homomorphism (respects multiplication) and the signature of a cycle of odd length is $+1$ and of a cycle of even length (including transpositions) is $-1$. – Berci Nov 20 '21 at 16:00
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    About $(9)$: it is traditional to omit $1$-cycles, but it is safer (especially if counting things) to leave them in. On reflection ${9}$ is an orbit, but is not part of the support. – ancient mathematician Nov 21 '21 at 07:34

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