Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help. 
I want to write the following permutations like : 
$p=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&?&?&?&2\end{pmatrix}$
$q=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&3&1&5&6&8&7&4\end{pmatrix}$
$r=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&8&3&6&5&4&7\end{pmatrix}$
$s=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 3&?&4&1&?&?&?&?\end{pmatrix}$
can you help me plese to fill the $?$ mark. Is there another method to compute $qpq^{-1}$? 
thanks:) 
 A: There is a quick way to compute both of your permutations, and it is by using this fact:
Suppose $\tau, \sigma \in S_n$ and in cycle notation, $\sigma = (a_1, a_2, \cdots a_p) (b_1, b_2, \cdots, b_q) \cdots (z_1, z_2, \cdots, z_l)$ say. Then $$ \tau \sigma \tau^{-1} = (\tau(a_1), \tau(a_2), \cdots \tau(a_p)) (\tau(b_1), \tau(b_2), \cdots, \tau(b_q)) \cdots (\tau(z_1), \tau(z_2), \cdots, \tau(z_l))$$
That is, to conjugate by $\tau$, replace the elements in the cycles of $\sigma$ by their images under $\tau.$ Exercise: Prove this.
A: I assume right to left associativity, as is usually the case. Note that 
$$(\alpha \beta)^{-1} = {\beta}^{-1} {\alpha}^{-1}, \text{ and}$$
$$ (a_1 a_2 \dots a_n)^{-1} = (a_1 a_n \dots a_2).$$
These are standard facts, and easily shown if you like to. Hence,
$$qpq^{-1} = (1 2 3)(4 5 6 8)(1 4 3 8 2)(1 2)(1 5)(4 8 6 5)(1 3 2).$$
I know no smart way to do this, and write the changes down one cycle at a time, right to left. For instance, after the first step, $(1 3 2)$, you get:
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&1&4&5&6&7&8\end{pmatrix}.$$
After the next step, $(4 8 6 5)$, you get:
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&1&5&6&8&7&4\end{pmatrix}.$$
If you keep going with the remaining cycles, you get
$$qpq^{-1} = \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&6&4&3&2&8&7&1\end{pmatrix},$$
if I didn't mess a step up. You read off the resulting permutation, written as cycles:
$$qpq^{-1} = (15268)(34).$$
You calculate $r^{-2}sr^2$ similarly. It's grindwork mostly.  
Edit: took it temporarily down to correct some typos. 
Edit 2 - write $p$ (per comment below): 
You apply the $3$ cycles of $p = (14382)(12)(15)$ one step at a time. Applying $(15)$ first, you get
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&2&3&4&1&6&7&8\end{pmatrix}.$$
Applying $(12)$ next, we get
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&5&3&4&1&6&7&8\end{pmatrix}.$$
Finally applying the last cycle $(14382)$, we get
$$p = \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&8&4&2&1&6&7&3\end{pmatrix}.$$
A: There's actually a quicker way to do this.  As an exercise, prove the following lemma:

Lemma. Suppose that $\alpha$ is an arbitrary permutation and $\beta$ is an $n$-cycle, written $\beta=(\beta_1\hspace{5pt} \beta_2 \hspace{5pt}\ldots \hspace{5pt}\beta_n)$.  Then $$\alpha^{-1}\beta \alpha=(\alpha(\beta_1)\hspace{5pt}\alpha(\beta_2)\hspace{5pt}\ldots \hspace{5pt}\alpha(\beta_n)).$$  In other words, $\alpha^{-1}\beta\alpha$ is the $n$-cycle with $\alpha$ applied to its letters.

Since any permutation can be decomposed into the product of disjoint cycles, say $\gamma=a_1a_2\ldots a_k$ where $a_i$ are disjoint cycles, we can write $$\alpha^{-1}\gamma\alpha=\alpha^{-1}a_1a_2\ldots a_k\alpha=(\alpha^{-1}a_1\alpha)(\alpha^{-1}a_2\alpha)(\alpha^{-1}\ldots \alpha )(\alpha^{-1}a_k\alpha)$$
so by induction the lemma shows that $\alpha^{-1}\gamma\alpha$ is the permutation with the same cycle structure as $\gamma$ with $\alpha$ applied to the letters in each of its disjoint cycles.
Now, $p$ in disjoint cycles is just $p=(1,4,3,8,5)$ and $s$ in disjoint cycles is just $s=(1,5,7,2,3,6)(4,8)$.
So, we apply $q$ to the letters in $p$ and obtain $q^{-1}pq=(1,4,6,2,5)$, then $r^2$ to the letters in each disjoint cycle of $s$ and obtain $r^{-2}sr^2=(1,8,7,6,3,5)(2,4)$.
I'll leave it to you to put these back into permutation notation (which is easy, since the cycles are disjoint!)
A: If you write all right then
$$
p= \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&5&6&7&2
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&1&3&4&5&6&7&8
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\5&2&3&4&1&6&7&8
\end{pmatrix}
$$
and so on (sorry, I write badly the formula). So you have first to calculate $p,q,r,s$.
