What does it means to multiply a permutation by a cycle? $\pi(x_1\cdots x_n)\pi^{-1}=(\pi(x_1)\cdots\pi(x_n))$ 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
 A: 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 $(1\ 2\ 4\ 5)$.
A: 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 $$\pi(4~5~3)\pi^{-1} = (\pi(4)~\pi(5)~\pi(3)) = (4~5~1) = (1~4~5).$$
