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: $(245)$?