Let $f \in 8^8$ where the permutation is given in two line form:

Asked Apr 15 '15 at 22:30

Active Apr 16 '15 at 11:55

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I'm having trouble understanding how one would answer question's like this.

$$\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 3&4&5&8&1&2&7&6\end{pmatrix}$$

Find $f^{80}(1)$ and $f^{80}(2)$

Could someone please explain how to go about solving problems similar to these?

I believe with $f^p(x)$ is the number in the domain that you start at and that $p$ denotes the length of the cycle, or how many times you move through the cycle to arrive at an answer in the codomain.

Edit: I think I understand now can someone please confirm?

f =(1,3,5)(2,4,8,6)(7)

f^80(1) = 5

f^80(2) = 2

f = (1,3,5)(2,4,8,9,6)(7)

f^100(1) = 3

f^100(2) = 2

edited Apr 16 '15 at 11:55

asked Apr 15 '15 at 22:30

Socrox

Hint: write in cycle notation first. – vadim123 Apr 15 '15 at 22:43
f = (1,3,5),(2,4,8,6),(7) – Socrox Apr 15 '15 at 22:49
Okay, now calculate the powers of $f$; you will notice patterns very quickly. – vadim123 Apr 15 '15 at 22:50
For the example is this saying we would start at 1, and where would we be if we moved 80 places forward? It would stay in the (1,3,5) correct? – Socrox Apr 15 '15 at 22:50
I believe every 10th position we are at the value 3, so does f^80 = 3? – Socrox Apr 15 '15 at 22:57
Yes, your answers are correct. – Litho Apr 16 '15 at 12:32

Let $f \in 8^8$ where the permutation is given in two line form:

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