I have to prove that
$$\pi(x_1\cdots x_n)\pi^{-1}=(\pi(x_1)\cdots\pi(x_n))$$
but I can't understand what this means. My book doesn't defines what a permutation and cycle product would be. So, for example, if I have a permutation that sends $1$ to $3$ and I multiply it with the cycle $(4,5,3)$, what happens?
How to prove such thing? I'd be very glad <3
My book doesn't define what a product of a permutation and a cycle would be.
Cycles are permutations! Surely your book told you what it means to compose one permutation with another permutation? Surely your book indicated that cycles are in fact permutations?
Please take some time to review the basic concepts before moving forward!
Take the cycle $\alpha=(123)$ in $S_4$. What does this mean? This refers to the permutation that sends $1\mapsto 2$, which sends $2\mapsto 3$, which sends $3\mapsto 1$, and which leaves $4$ unaffected. That's a permutation! In one-line notation, this is $(\color{Purple}{2~3~1~4})$. In two line notation, it is
$$\begin{pmatrix} \color{Blue}{1} & \color{Blue}{2} & \color{Blue}{3} & \color{Blue}{4} \\ \color{Purple}{2} & \color{Purple}{3} & \color{Purple}{1} & \color{Purple}{4} \end{pmatrix}.$$
Explicitly, $\alpha$ is defined by the following:
Now, given two permutations, how do we figure out what their product is?
Say we have cycles $(124)$ and $(352)$ in $S_5$. Then
How do we figure out where $(124)(352)$ sends every element (i.e. what permutation it is)? Well, we just compute where it sends every element! We know how to do these computations: to plug $x$ into $f\circ g$, first you plug $x$ into $g$ to get $g(x)$ and then you plug $g(x)$ into $f$ to get $f(g(x))$. So for example above we plugged $\color{Blue}{1}$ into $(352)$, which yields $\color{Green}{1}$ since $(352)$ doesn't do anything to $\color{Blue}{1}$, and then we plug $\color{Green}{1}$ into $(124)$ to get $\color{Purple}{2}$. Therefore, the two-line notation is
$$(124)(352)=\begin{pmatrix}\color{Blue}{1} & \color{Blue}{2} & \color{Blue}{3} & \color{Blue}{4} & \color{Blue}{5} \\ \color{Purple}{2} & \color{Purple}{3} & \color{Purple}{5} & \color{Purple}{1} & \color{Purple}{4}\end{pmatrix}.$$
One thing you should be learning about soon is disjoint cycle notation. Let's define a term first: the support of a permutation is the set of elements that it actually moves. The complement of the support is the set of fixed points ($x$ is a fixed point of $\pi$ if $\pi(x)=x$).
$\sf \color{Blue}{Exercise~A}$. Prove that if two permutations have disjoint (i.e. trivially intersecting, or nonoverlapping) support, then they commute (i.e. $\alpha\beta=\beta\alpha$). Hint: show that the two permutations do the same thing to every element, by splitting into three cases:
It is then a theorem that for every permutation, there exists a representation of it as a product of disjoint cycles, and moreover that this representation is unique up to order (the disjoint cycles commute anyway, so it doesn't matter what order you compose them in).
Here's the idea: say we have a permutation and a starting element, then the permutation sends us to a new element, and then a newer one, and so on until we cycle back to the original. This determines one of the cycles. We can do this with every element, and the resulting collection of cycles will be disjoint - the permutation will be the product of these cycles. To see why this is true intuitively, think about using this process to construct the cycle graph of the permutation: the vertices of this graph are the objects that are being permuted, and we draw a directed edge from one object to another if the permutation sends that first object to the next.
Let's consider our previous example, which is $(2~3~5~1~4)$ in one-line notation. Starting at $1$, we are sent off to $2$, and then from $2$ we are sent to $3$, then from $3$ we are sent to $5$, then from $5$ we are sent to $4$, and finally from $4$ we are sent back to $1$. This touched all of the elements, so our permutation is the cycle $(12354)$ (and this is its cycle notation, as it is itself a cycle). We could have just as easily started at $2$ and ended up writing down $(23541)$, or started at $3$ and written $(35412)$, or started at $4$ and written $(41235)$, or started at $5$ and written $(54123)$. These all represent the same cycle, the same permutation. They all have the same numbers in the same order, but cyclically shifted!
$\sf \color{DarkCyan}{Exercise~B}$. Convert the disjoint cycle notation to the two-line notation below.
$$\begin{array}{ccc} \textrm{cycle notation} & & \textrm{two-line notation} \\[5pt] (135)(24) & \longrightarrow & \begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 2 & 1\end{pmatrix} \end{array} $$
$\sf \color{DarkOrange}{Exercise~C}$. Convert the one-line notation to disjoint cycle notation below.
$$\begin{array}{ccc} \textrm{one-line notation} & & \textrm{cycle notation} \\ (4~5~2~1~3) & \longrightarrow & (14)(253) \end{array} $$
$\sf \color{Red}{Exercise~D}$. Draw the cycle graphs of the two permutations in exercises A and B. (Solutions.) Make up your own permutation using one-line notation and then draw its cycle graph. What do you draw if it has a fixed point? Can you write down the cycle notation using the cycle graph?
Now let's finally consider $\pi(x_1x_2\cdots x_n)\pi^{-1}=(\pi(x_1)\pi(x_2)\cdots\pi(x_n))$. Do you understand that the support of the permutation $(x_1x_2\cdots x_n)$ is $\{x_1,x_2,\cdots,x_n\}$? Good. For convenience, consider the indices to be integers mod $n$ (so in particular $n+1$ is to be interpreted as $1$).
$\sf \color{Magenta}{Exercise~E}$. Now prove:
Do you understand why this means $\pi(x_1x_2\cdots x_n)\pi^{-1}$ must be the cycle $(\pi(x_1)\pi(x_2)\cdots\pi(x_n))$?
$\sf \color{Purple}{Exercise~G}$. How does conjugating by $\pi$ affect cycle notation in general? In other words, what is
$$\pi (x_1x_2\cdots x_n)(y_1y_2\cdots y_m)\cdots(z_1z_2\cdots z_l)\pi^{-1}~? $$
Hint: conjugation-by-$\pi$ is homomorphic, so $\pi\,\alpha\beta\,\pi^{-1}=\pi\alpha\pi^{-1}\,\pi\beta\pi^{-1}$ for any two permutations $\alpha,\beta$; generalize this to arbitrary products: $\pi\,\alpha\beta\cdots\gamma\,\pi^{-1}=?$
$\sf \color{DarkBlue}{Exercise~F}$. Given that $\pi(12)(34)\pi^{-1}=(\color{Blue}{\pi(1)}\color{Purple}{\pi(2)})(\color{Red}{\pi(3)}\color{Green}{\pi(4)})$, can you find a $\pi$ for which this equals $(\color{Blue}{5}\color{Purple}{6})(\color{Red}{1}\color{Green}{3})$? Hint: look at how I colored everything. What must $\pi(1),\pi(2),\pi(3),\pi(4)$ be? Can you find any other solutions $\pi$? Hint: color the things cycle entries differently. How many can you find? There should be $8$ solutions in total.
From these facts, we can conclude that a conjugacy class in $S_m$ is precisely a set of all permutations with a given cycle type! One can even go further and compute centralizers of elements!
Let's look look what happens to each element of $\{1,2,3,\dots,k\}$ under the permutation:
$\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}$
(here, of course, we must have $n \leq k$).
Let's denote $\pi(a_i) = b_i, i = 1,2,\dots,n$.
Suppose $j \not\in \{b_1,b_2,\dots,b_n\}$.
Then, since $\pi$ is bijective, $\pi^{-1}(j) \not\in \{a_1,a_2,\dots,a_n\}$.
Thus $(a_1\ a_2\ \dots\ a_n)\pi^{-1}(j) = \pi^{-1}(j)$, and so:
$\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}(j) = \pi\pi^{-1}(j) = j$.
Now suppose $j = b_n$.
Then $\pi^{-1}(j) = \pi^{-1}(b_n) = a_n$, and:
$(a_1\ a_2\ \dots\ a_n)(\pi^{-1}(j)) = (a_1\ a_2\ \dots\ a_n)(a_n) = a_1$
and so $\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}(b_n) = \pi(a_1) = b_1$.
Finally, suppose $j = b_i$ for $i < n$.
Then $\pi^{-1}(b_i) = a_i$, and:
$(a_1\ a_2\ \dots\ a_n)(\pi^{-1}(j)) = (a_1\ a_2\ \dots\ a_n)(a_i) = a_{i+1}$
and thus:
$\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}(j) = \pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}(b_i)$
$= \pi(a_1\ a_2\ \dots\ a_n)(\pi^{-1}(b_i)) = \pi(a_1\ a_2\ \dots\ a_n)(a_i)$
$= \pi(a_{i+1}) = b_{i+1}$.
In short, $\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}$ is the $n$-cycle $(b_1\ b_2\ \dots\ b_n) = (\pi(a_1)\ \pi(a_2)\ \dots\ \pi(a_n))$.
Let's work this out element-by-element for $k = 5$, and $n = 4$, with:
$\pi = (1\ 3)(2\ 4\ 5)$, and $a_1 = 2, a_2 = 4, a_3 = 3, a_4 = 5$, so our $4$-cycle is $(2\ 4\ 3\ 5)$.
The short way (using our theorem):
$\pi(2) = 4\\ \pi(4) = 5\\ \pi(3) = 1\\ \pi(5) = 2.$
Thus $b_1 = 4, b_2 = 5, b_3 = 1, b_4 = 2$.
According to what we did above, we will expect $(1\ 3)(2\ 4\ 5)(2\ 4\ 3\ 5)[(1\ 3)(2\ 4\ 5)]^{-1}$ to fix $3$, since it is not in $\{b_1,b_2,b_3,b_4\}$, and to wind up with the $4$-cycle:
$(4\ 5\ 1\ 2) = (1\ 2\ 4\ 5)$.
The long way, composing functions and tracking each element:
First we apply $[(1\ 3)(2\ 4\ 5)]^{-1} = (1\ 3)(2\ 5\ 4)$, then $(2\ 4\ 3\ 5)$ and then $(1\ 3)(2\ 4\ 5)$:
$1 \to 3 \to 5 \to 2\\ 2 \to 5 \to 2 \to 4\\ 3 \to 1 \to 1 \to 3\\ 4 \to 2 \to 4 \to 5\\ 5 \to 4 \to 3 \to 1$
the net result is that $3 \to 3$, so $3$ is a fixed element, and on the other $4$, we have:
$1 \to 2 \to 4 \to 5 \to 1$, which is indeed the $4$-cycle $(1\ 2\ 4\ 5)$.
In the first case, they mean a composition of permutations. In the second case, they are thinking of $\pi$ as a function as well. Consider $\pi \in S_5$ where $\pi = (1~3)$. Then $\pi(1) = 3, \pi(2) = 2, \pi(3) = 1, \pi(4) = 4$ and $\pi(5) = 5$. Now, lets see if the problem works in your example: on the one hand, $$\pi(4~5~3)\pi^{-1} = (1~3)(4~5~3)(1~3) = (1~4~5).$$
On the other hand $$\pi(4~5~3)\pi^{-1} = (\pi(4)~\pi(5)~\pi(3)) = (4~5~1) = (1~4~5).$$
