I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding some questions pertaining to Relations and Functions. To be specific, I'm stuck on the following practice question:
Let $P_1 = (3~4~1~2~5)$ and $P_2 = (3~5~1~2~4)$ be two permutations.
Find the following permutation
$$P_1^{-1} \circ P_2$$
So $P_1^{-1}$ could be $(4~2~3~1~5)$ and we leave $P_2$ at $(3~5~1~2~4)$.
Would it work to do it in this manner:
$$P_1^{-1}(P_2(1)) = P_1^{-1}(3) = 3$$
I don't really understand how to do composition, how would I find the resulting group? Any help is appreciated! Thanks.
Best Answer
I strongly suspect you're learning permutations using the so-called "list notation". If this is the case, then writing $P_1 = (3~4~1~2~5)$ is simply specifying the outputs, like you'd initially thought;
$$(3~4~1~2~5) = \big(P_1(1)~P_1(2)~P_1(3)~P_1(4)~P_1(5)\big).$$
For inverses, I think a picture helps.
So, if we have $P_1(1) = 3$, then we must have $1 = P_1^{-1}(3)$ (if you like, by applying $P_1^{-1}$ to both sides).
In this way, we must have
$$P_1^{-1} = (3~4~1~2~5)$$
by essentially writing down the position each number occurred in $P_1$ (i.e., in $P_1$, we wrote $1$ in the third spot. So in $P_1^{-1}$, we'll write $3$ in the first spot.
Now that that's out of the way, I like your plan for computing the composition. But with the correct $P_1^{-1}$, let's see what happens:
\begin{align*} P_1^{-1}(P_2(1)) &= P_1^{-1}(3) = 1 \\ \\ P_1^{-1}(P_2(2)) &= P_1^{-1}(5) = 5, \text{ etc.} \end{align*}
Alternatively, we could draw!