Finding a permutation given a composite of permutations Given the permutations $x = (1, 2)(3, 4)$ and $y = (5, 6)(1, 3)$ find a permutation $a$ such that $a^{−1}xa = y$.
I have an answer to this but I would like to know the process involved in getting the answer.
My thoughts thus far are that $a^{-1}(12)(34)a =(56)(13)$ which means that $a^{-1}((12)(34)a) =(56)(13)$. So $(a^{-1}(1(a)), a^{-1}(2(a))) (a^{-1}(3(a)), a^{-1}(4(a))) = (5,6)(1,3)$.
Any help would be appreciated.

Let $\sigma=[i_1,\ldots, i_n]$ be a cycle.  Use the fact that $\tau\sigma\tau^{-1}=[\tau(i_1),\ldots,\tau(i_n)]$.  So, what permutation takes $1$ to $5$, $2$ to $6$, $3$ to $1$ and $4$ to $3$.  The inverse of this permutation is the answer.
To find this, we can expess it as a matrix:
$$a^{-1}=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 6 & 1 & 3 & c & d
\end{pmatrix}$$
Here the image of a number in the first row, is the corresponding number in the second row.
We see that there are two possibilities, $c=2$, $d=4$ and $c=4$,$d=2$.  
Let's take the case, $c=2$ and $d=4$.
So,
$$a^{-1}=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 6 & 1 & 3 & 2 & 4
\end{pmatrix}$$
and, 
$$a=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 5 & 4 & 6 & 1 & 2
\end{pmatrix}$$
Writing these as cycles gives:
$$a=(134625)\quad\text{and}\quad a^{-1}=(152643).$$
