Solve $\sigma^2=(1345)(2687)$ in $S_8$

Question

I would like to solve the equation $\sigma^2=(1345)(2687)$ in $S_8$.

If we reason the same way that I did here we get that $\sigma$ is a cycle of length $8$. Here come the problems. How are we sure that we write all the solutions in this case?

We have the following formula:

$(i_1 i_2 i_3 i_4 i_5 i_6 i_7 i_8)^2=(i_1 i_3 i_5 i_7)(i_2 i_4 i_6 i_8)$.

Ok, so this tells us that one sigma is $(1 2 3 4 6 4 8 5 7)$.

But how do we get all of them? I think that we need to fix one of the 4-cycles and then write all the cyclic permutations of the other one. But I can't show that this actually produces all the solutions (I mean a formal argument, not just enumerating everything and seeing that this works). Furthermore, how would I do this if my $\sigma$ were the product of more cycles?

EDIT: I think that one possible argument for why fixing one cycle works is that if we do the procedure with the cyclic permutations we make sure, for instance, that $1$ goes to everything it can go (and the same applies to all the elements, but I picked $1$ for making myself clear). If we started permuting the other cycle too, we would obviously get the same solutions, because we have already sent each of its elements to all the possible images. Is this argument all right? How would this work for more cycles?

Answer 1

You know that $\sigma$ has no odd cycles, since then $\sigma^2$ would have one, too. You can also observe, that $\sigma$ has no fixed points nor 2-cycles, since then $\sigma^2$ will have a fixed point. It's left to rule out case of two 4-cycles, but then $\sigma^2$ would be the product of four transpositions.

So $\sigma$ is indeed an 8-cycle.

Now, let $c = \sigma(1)$. Then our permutation is

$$ 1 \to c \to 3 \to \sigma^2(c) \to 4 \to \sigma^4(c) \to 5 \to \sigma^6(c) \to 1 $$

and it's both clear that $c \in \{2, 6, 7, 8\}$ and that all such $c$ will give rise to four different solutions for $\sigma$.

Answer 2

As we can cyclically rotate any cycle, we may assume wlog that $i_1=1$. Then it follows that $i_3=3$, $i_5=4$, $i_7=5$. Consequently, $i_2$ must be one of $2,6,7,8$. For each of the four possible cases, we find exactly one solution, namely $i_4$ must be the next term in the second 4-cycle and so on.