Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425) $
Or, in two row format, get: $$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
Then, $$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
Due to disjoint cycles, they can commute as no affect on map produced.
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} \begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
My answer is:
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix} = e.\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix} = \begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix} $$
But, the answer is different: $(245)$?
Here
$$\begin{align} \sigma^2&=\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\cdot\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\\ &=\begin{pmatrix}1&2&3&4&5\\1&4&3&5&2\end{pmatrix}\\ &=(245). \end{align}$$