Krdytkyu

So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations.
The question is:

Let $alpha_1$ and $alpha_2$ be cycles of the same length. Let $beta_1$ and $beta_2$ be cycles of the same length. Let $alpha_1$ and $beta_1$ be disjoint, and let $alpha_2$ and $beta_2$ be disjoint. [Prove that] There is a permutation, $piin S_n$, such that $alpha_1beta_1=pialpha_2beta_2pi^{-1}$.

That's the question.

But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.

Part 1 was:

Let $alpha=(a_1,...,a_s)$ be a cycle and let $pi$ be a permutation in $S_n$. Then $pialphapi^{-1}$ is the cycle $(pi(alpha_1),...,pi(alpha_s))$.

(a solution can be found here Proof for conjugate cycles)

Part 2 was:

Conclude from part 1: Any two cycles of the same length are conjugates of each other.

The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $pi$ is a permutation and by part 1, that's enough.

Parts 3 and 4 are straightforward.

Part 3:

If $alpha$ and $beta$ are disjoint cycles, then $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint cycles.

(this follows directly from part 1)

And Part 4:

Let $sigma$ be a product $alpha_1...alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $pisigmapi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.

(an easy generalization of part 3)

In the fifth part, I see how one can find a $pi$ such that $alpha_1=pialpha_2pi^{-1}$. I also see how one can find a $pi$ such that $beta_1=pibeta_2pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).

Any help appreciated.

Let

$alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
$beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.

Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.

Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.