Prove that products of permutations $ab$ and $ba$ have the same cycle length For example:
Let $a,b \in S_3$.
$a = (1,2,3), b = (1,2)(3)$,
then $ab = (1,3)(2) $ and $ba = (2,3)(1) $.
These are two different permutations, but their cycle type is the same.
How could we prove this for a general case?
I have some idea of what conjugate cycles are and that they might be required to prove this, but if there is a proof without them, that would be even better.
 A: I'm sorry but I'm gonna give you a proof using conjugation since it is just the easiest. Like you I would indeed be interested in seeing a different proof.
Lemma Let $(x_1, x_2, \ldots, x_n)$ be a cycle (so each $x_i$ is an integer between 1 and $N$ for some $N \geq n$), and let $a$ be a permutation in $S_N$. Then the permutation $a(x_1, x_2, \ldots, x_n)a^{-1}$ can also be written as a cycle of length $n$.
proof The easiest way to show that it can be written in this form is by doing it. For this we need a notational convention: I denote by $a(x_i) \in [1, \ldots, N]$ the number to which $x_i$ is sent by permutation $a$. (So I view $a$ as a function from $[1, \ldots, N]$ to itself.)
Now the permutation $a(x_1, \ldots, x_n)a^{-1}$ equals the cycle $(a(x_1), \ldots, a(x_n))$.
To see that just check where both permuations send a given number $z$. If the two answers are the same for every $z$ we know the permuations are equal.
We know that $z$ equals $a(y)$ for some (unique) $y$. We consider two cases: $y$ is one of the $x_i$ or it is not.
In the first case let's see what $a(x_1, \ldots, x_n)a^{-1}$ does do to $z$. First we let $a^{-1}$ act that sends it to $y = z_i$. Then we have the cycle that sends $z_i$ to $z_{i+1}$ (or to $z_1$ if $i = n$). Finally we have $a$ that sends $x_{i+1}$ to $a(x_{i+1})$. So the answer is $a(x_{i+1})$.
On the other hand what does the permuation $(a(x_1), \ldots, a(x_n))$ do with $z$? Well since $z = a(x_i)$ by assumption, this cycle sends it so $a(x_{i+1})$. So the answers are equal as expected.
In other case, when $y$ is not one of the $x_i$ what do both permuations do? $a(x_1, \ldots, x_n)a^{-1}$ acts by first letting $a^{-1}$ send $z$ to $y$. Then $y$ gets sent to itself by the cycle and finally $a$ sends $y$ to $a(y) = z$, so that $z$ is sent to itself. It is clear that the permutation $(a(x_1), \ldots, a(x_n))$ also sends $z$ to itself when $z$ is not one of the $a(z_i)$ so also in this case the two answer coincide. End proof.
Corollary Let $p$ be any permutation, then $p$ and $apa^{-1}$ have the same cycle type.
proof Let $p = c_1 c_2 \cdots c_n$ with each $c_i$ a cycle. We know from the lemma that $ac_ia^{-1}$ is a cycle of the same length as $c_i$ for each $i$. But now we can just write
$$a^{-1}pa = a^{-1}c_1(aa^{-1})c_2(aa^{-1})c_3(aa^{-1}) \cdots (aa^{-1})c_na = (a^{-1}c_1a)(a^{-1}c_2a)\cdots (a^{-1}c_na)$$ which is a product of cycles of the same length as the $c_i$ appearing in the cycle decompostion of $p$. End proof.
I love this trick.
Now how does this relate to your question?
$$ba = a^{-1}(ab)a$$
So we can apply the corollary with $p = ab$.
