Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. 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: $(245)$?

We have
$$\begin{align}
\sigma^2&=(13)(254)(13)(254)\\
&=(13)(13)(254)(254)\\
&=(254)^2\\
&=(245)
\end{align}$$
because
$$\begin{align}
1&\xrightarrow{(254)}1\xrightarrow{(254)}1,\\
2&\xrightarrow{(254)}5\xrightarrow{(254)}4,\\
3&\xrightarrow{(254)}3\xrightarrow{(254)}3,\\
4&\xrightarrow{(254)}2\xrightarrow{(254)}5,\\
5&\xrightarrow{(254)}4\xrightarrow{(254)}2.\\
\end{align}$$
In row format,
$$\begin{align}
\sigma^2&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}^2\\
&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 4 & 3 & 5 & 2
\end{pmatrix},
\end{align}$$
which gives $(245)$ again.
