How to solve conjugation equations in group theory

Question

Given the permutations $(12)(34)$ and $(56)(13)$

find $a$ such that

$$a^{-1}xa = y$$

I just realized that I don't know how to solve this exercise. My book don't even give examples of how to solve 'cycles' equations. It's confusing to me. Multiplying by $a$ and $a^{-1}$ doesn't help at all.

I've already solved an exercise in cycles euqations that kinda looks like this, which does not require solving. In fact, it delas with the order of the left and rigth sides.

In the exercise below, I know that the rigth side must have order $6$, and so does the left side. But I can't show it's impossible to make the left side having order $6$ too.

Show that there isn't $a$ such that $a^{-1}(123)a = (13)(578)$

Answer 1

You can choose a permutation $a$ (others will work, too) such that:

$a(1) = 5, a(2) = 6, a(3) = 1$ and $a(4) = 3$.

It doesn't matter what we choose for $a(5),a(6)$, as long as we don't pick from the set $\{1,3,5,6\}$, since those values are already "taken".

$a(5) = 2$, and $a(6) = 4$ will do.

Thus $a = (1\ 5\ 2\ 6\ 4\ 3)$ is one possibility.

Answer 2

The key fact to note is that two elements of the symmetric group are conjugate if and only if they have the same number of $k$-cycles in their cycle decomposition for all $k$. In particular, every conjugate of a $3$-cycle is a $3$-cycle, which solves your second exercise.

You have shown in a previous question that $$a(i_1i_2\cdots i_k)a^{-1}=(a(i_1)a(i_2)\cdots a(i_k))$$ For any $a$, $$a(12)(34)a^{-1}=(a(12)a^{-1})(a(34)a^{-1})=(a(1)a(2))(a(3)a(4))$$ So you just need to find a permutation $a$ such that $a(1)=5$, $a(2)=6$, $a(3)=1$, and $a(4)=3$. I promise this is not a hard thing to do, try it.