What is this proposition trying to do? Proposition
Let $\alpha$ and $\beta$ be permutations in $S(n)$. The cycle decomposition of the element $\alpha\beta\alpha^{-1}$ is obtained from that of $\beta$ by replacing each integer $i$ in the cycle decomposition of $\beta$ with the integer $\alpha(i)$.
Proof
$\alpha\beta\alpha^{-1}(\alpha(i))=\alpha(\beta(i))$
Example
Conjugating the permutation $(1,2,3,4,5)$ by $(1,2)(3,4)$, the proposition shows that the result is $(2,1,4,3,5)=(1,4,3,5,2)$. Similarly, the permutation $(1,2)(3,4)$ is conjugate to $(1,3)(2,5)$ via $(2,3)(4,5)$.
Doubt
I don't understand anything because I don't know what "that of $\beta$ by replacing each integer $i$ in the cycle decomposition of $\beta$ with the integer $\alpha(i)$" means.
It may be because English is not my native language, although I understand it well. I have searched for an explanation in my language (Spanish), but I have not been successful.
I have also read other articles in English (some from this same forum) related to conjugate permutations (I think that is what the topic is about), but without success.
I understand the following:
$\alpha=(1,2,3,4,5)$
$\alpha^{-1}=(5,4,3,2,1)$
$\beta=(5,2,4,1,3)$
$\alpha\beta\alpha^{-1}=(\alpha(5),\alpha(2),\alpha(4),\alpha(1),\alpha(3))$
Is it correct? Could someone give me a more developed explanation so I can better see the proposition? Thanks.
 A: You're exactly right! Let's go through a few concrete examples together, and that will probably clarify things.
Let

*

*$\beta = (1,2,3)(4,5)(6)$

*$\alpha = (3,6)$
Then $\alpha \beta \alpha^{-1} = (1,2,6)(4,5)(3) = (\alpha 1, \alpha 2, \alpha 3)(\alpha 4, \alpha 5)(\alpha 6)$.
Notice $\beta$ and $\alpha \beta \alpha^{-1}$ look exactly the same, except we swapped $3$ and $6$. This is because $\alpha$ swaps $3$ and $6$.
Next, let

*

*$\beta = (1,2,3)(4,5)(6)$

*$\alpha = (2,4)(5,6)$
can you guess what the $\alpha \beta \alpha^{-1}$ will be? I'll put it under a fold:

 $\alpha \beta \alpha^{-1} = (1,4,3)(2,6)(5)$. Because we swap $2$ and $4$, then we swap $5$ and $6$, inside $\beta$.

Now for something a bit harder:

*

*$\beta = (1,2,3)(4,5)(6)$

*$\alpha = (2,4,6)$
Thankfully, the same idea works!
$$\alpha \beta \alpha^{-1} = (1,4,3)(6,5)(2) = (\alpha 1, \alpha 2, \alpha 3)(\alpha 4, \alpha 5)(\alpha 6)$$
Lastly, let's look at something like

*

*$\beta = (1,2,3)(4,5)(6)$

*$\alpha = (2,4,6)(1,3)$
Again, you should try this yourself before reading the answer!

 $\alpha \beta \alpha^{-1} = (3,4,1)(6,5)(2)$

And as a little edge case, what about

*

*$\beta = (1,2,3)(4,5)(6)$

*$\alpha = (4,5)$

 $\alpha \beta \alpha^{-1} = (1,2,3)(5,4)(6)$. Notice this actually equals $\beta$ since $(4,5) = (5,4)$! That doesn't matter, though -- the process still works.

If you understand all of this, then it seems to me like you understand the proposition! That's all there is to it.

I hope this helps ^_^
