Permutation inverse form

Question

Given:

$A=\{1,2,3,4,5,6\}$,

$P_1=\begin{pmatrix} 1 &2& 3& 4& 5& 6\\ 2& 3& 4& 1& 5& 6\end{pmatrix}$,

$P_2=\begin{pmatrix}1 &2 &3 &4& 5 &6\\ 1 &3 &5& 4& 6& 2\end{pmatrix}$.

(i) Find $P_2^2$.

My answer: $$ P^2_2 = \begin{pmatrix}1& 2& 3& 4 &5 &6\\ 1& 5& 6& 4& 2& 3\end{pmatrix}. $$ Is this correct?

(ii) $(P_1∘P_2)^{-1}=\begin{pmatrix} 1 &2 &3 &4 &5 &6\\ 2 &4 &5 &1 &6 &3\end{pmatrix}^{-1}$

How to complete it? That is, how do I find the inverse?

Answer 1

(i) is correct.

And for the second one you just see from the co-domain to domain..

that is,

\begin{align} (P_1\circ P_2)^{-1}&=\begin{pmatrix}1&2&3&4&5&6\\ 2&4&5&1&6&3\end{pmatrix}^{-1}\\ &=\begin{pmatrix} 2&4&5&1&6&3\\ 1&2&3&4&5&6\end{pmatrix}\\ &=\begin{pmatrix} 1&2&3&4&5&6\\ 4&1&6&2&3&5\end{pmatrix}\end{align}