$$\sigma = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&5 &7 &4 &6 &9 &3 &2 &8 \end{bmatrix}$$
$$\sigma^{2} = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1& 6&3 &4 &9 &8 &7 &5 & 2 \end{bmatrix}$$
Given $$\sigma$$ I found $$\sigma^2.$$
I want to find $$\sigma^{-2}$$ but I'm being told that my permutation cycle for $\sigma^2$ is wrong. Very frustrating.
Please help... Thanks in advance..
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$\begingroup$Permutations are represented in two ways. One is a verbose description of the mapping: $$\sigma = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&5 &7 &4 &6 &9 &3 &2 &8 \end{bmatrix}$$ means that $1$ goes to $1$, $2$ goes to $5$, and so on.
We abbreviate this to cycle notation, which is more compact and also more revealing of important structure. This permutation abbreviates to $$(1)(25698)(37)(4)$$ which means the same thing as before, only more compactly. The $(25698)$ means that $2$ goes to $5$, $5$ goes to $6$, $6$ goes to $9$, $9$ goes to $8$, and $8$ goes back to $2$. Then we abbreviate further by dropping the $(1)$ and $(4)$, which can be inferred even if not written explicitly.
In the cycle notation, $\sigma^2$ is indeed written $(26859)$, as you should check. You are being asked to write $\sigma^{-2}$ in the compact cycle notation.
The inverse of a permutation $p$ is another permutation that un-does the effect of $p$. If $p$ takes $3$ to $7$, then $p^{-1}$ should take $7$ to $3$. Let's take a simple example:
$$p = \begin{bmatrix} 1 &2 &3 &4 &5 \\ 4&5 &2 &1 &3 \end{bmatrix}$$
The inverse of $p$, in mapping notation, is
$$p^{-1} = \begin{bmatrix} 4&5 &2 &1 &3 \\ 1 &2 &3 &4 &5 \\ \end{bmatrix}$$
because where $p$ took $2$ to $5$, $p^{-1}$ takes $5$ to $2$. As commented earlier, we have “flipped the notation upside down”. We usually arrange the columns in order:
$$p^{-1} = \begin{bmatrix} 1 &2 &3 &4 &5 \\ 4 & 3& 5 & 1 & 2 \\ \end{bmatrix}$$
In cycle notation, $p$ is written $(14)(253)$. Again, this says that $2$ goes to $5$. As you observed in the comments, you can find $p^{-1}$ by writing the cycle notation for $p$ backwards: $$p^{-1} = (352)(41)$$ which we would usually write as $$p^{-1} = (14)(235).$$ Again, notice that this says that $p^{-1}$ takes $5$ to $2$, as before.
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