square of a permutation cycle $$\sigma = \begin{bmatrix}
1 &2  &3  &4  &5  &6  &7  &8  &9 \\ 
 1&5  &7  &4  &6  &9  &3  &2  &8 
\end{bmatrix}$$
$$\sigma^{2} = \begin{bmatrix}
1 &2  &3  &4  &5  &6  &7  &8  &9  \\ 
1&  6&3  &4  &9  &8 &7  &5  & 2 
\end{bmatrix}$$
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..

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.
