8

I have this permutation $A$:

$$ \left(\begin{array}{rrrrrrrrrr} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 10 & 8 & 5 & 2 & 3 & 1 & 6 & 4 & 7 & 9 \end{array}\right) $$

I want to calculate $A^9$. Is it ok to calculate it in this way?

$$A*A*A*A*A*A*A*A*A$$

where $A*A$ is defined as $A$ composed with $A$.

Thanks much in advance!!!

Thomas Andrews
  • 177,126
XXXX
  • 83
  • @ThomasAndrews, thanks, looks much better now. – Andreas Caranti Mar 23 '13 at 19:57
  • Just discovered the hard way what happens when two people edit the same question simultaneously - the earlier bird gets toasted! GMTA. Hope my (later) changes are okay. Great reference on MathJax here, or see the link "MathJaX help" on the right when you edit. – Bob Stein Mar 23 '13 at 20:48
  • 1
    @BobStein-VisiBone I changed yours slightly - the standard two-line notation for permutations does not have a solid line between and has parentheses around it. Left in your right-alignment, which was better than what I had done. – Thomas Andrews Mar 23 '13 at 21:00
  • @ThomasAndrews remarkable how similar our changes were. – Bob Stein Mar 23 '13 at 22:04

2 Answers2

8

Write it as the product of disjoint cycles $$ A = (1, 10, 9, 7, 6) (2, 8, 4) (3, 5), $$ and then it's much easier, as a $k$-cycle has period $k$, so that $$ A^9 = (1, 10, 9, 7, 6)^{-1} (3, 5) = (1, 6, 7 ,9 ,10) (3, 5) = \begin{pmatrix} 1 &2& 3& 4 &5 &6 &7 &8 &9 &10\\ 6 &2& 5& 4& 3 &7& 9& 8& 10& 1\end{pmatrix}. $$

PS Apologies, I write my permutations left-to-right.

Andreas Caranti
  • 68,873
  • @XXXX, you're welcome, please note that I have just fixed a mistake in my calculation. – Andreas Caranti Mar 23 '13 at 20:06
  • putting 1 at first position after 5-cycle inversion really confused me at first. This is valid because we can write cycle elements in any offset. Hope this comment will help someone. – Grigory Feb 04 '18 at 13:56
3

That's ok, but it's not the fastest way. The fastest way (without using tools such as Lagrange's theorem) is to calculate by repeated squaring: $A^9 = (((A^2)^2)^2)A$.

Edit: Okay, not sure if my way is faster than the solution posted by Andreas (factoring as a product of cycles), I guess both are useful to know.

Yoni Rozenshein
  • 6,817