Equation with a permutation composition Is there any method to solving such an equation:
$$f_1\circ f = f_2$$
Where $f_1, f, f_2 \in S_7$ and:
$f_1 = (1234)(5)(6)(7)$
$f_2 = (172536)(4)$
 A: you are nearly there. note that the repeated $1$ in your $(14321)$ is incorrect notationwise. also unless there is a reason not to do so we usually omit cycles consisting of a single element. so
$$
f_1 = (1234)
$$
and
$$
f_1^{-1} = (4321) = (1432)
$$
therefore (following the usual convention of evaluating from right to left)
$$
f = f_1^{-1} \circ f_2 \\
=(1432)\circ (172536) \\
= (17)(25)(364)
$$
A: So to my mind that would be the solution:
1) first we are to find $f_1^{-1}$:
$$f_1^{-1}\circ f = id \rightarrow f_1^{-1} = (13421)(5)(6)(7)$$
2) now we are to solve:
$$f_1^{-1} \circ f_2 = (17)(254)(36)$$
And that is our permutation $f$ we were looking for.
A: You can also solve it by writing explicitly the table of transformations.
Note that since $f_1$ order is $4$ then $(f_1)^{-1}=(f_1)^3$.
$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & 1 & 2& 3 & 4 & 5 & 6 & 7 \\
\hline
f_2(x) & 7 & 5 & 6 & 4 & 3 & 1 & 2\\
\hline
f_1\circ f_2(x) & 7 & 5 & 6 & 1 & 4 & 2 & 3\\
\hline
f_1\circ f_1\circ f_2(x) & 7 & 5 & 6 & 2 & 1 & 3 & 4\\
\hline
f_1\circ f_1\circ f_1\circ f_2(x) & 7 & 5 & 6 & 3 & 2 & 4 & 1\\
\hline
\end{array}$
Now $(f_1)^3\circ f_2=(f_1)^3\circ f_1 \circ f=Id\circ f=f$
So that last line is $f$ : $1\to 7\to 1$ and $2\to 5\to 2$ and $3\to 6\to 4\to 3$

Which in proper notation is $f=(17)(25)(364)$

I always find it difficult to figure out where the elements are going without a table.
Note that if you want to check your result, wolframalpha considers that $\sigma\tau$ is $\tau(\sigma(\cdot))$ so you have to inverse the order of cycles.
https://www.wolframalpha.com/input/?i=permutation+(172536)(1234)%5E(-1)
