<s>[INST] 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:) 
 [/INST] 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!)
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