Inverse of Permutations I'm having problems understanding inverting permutations, but I believe I understand how to do compositions.  I may be second guessing myself, but some answers do not make sense to me.  I've tried to use two different approaches (graphically and P ∘ P^(-1) = identity)  Below is my question:
P = (5 1 4 2 3),  find the inverse of P
P ∘ P^(-1)=(5 1 4 2 3)∘(a b c d e)=(1 2 3 4 5)
P(P^(-1)(1))=P(a)=1⇔a=2
P(P^(-1)(2))=P(b)=2⇔b=4
P(P^(-1)(3))=P(c)=3⇔c=5
P(P^(-1)(4))=P(d)=4⇔d=3
P(P^(-1)(5))=P(e)=5⇔e=1
∴ P^(-1)=(2 4 5 3 1)  
On a similar question I tried to find the inverse of (4 2 5 1 3), graphically using an arrow diagram and found that it's inverse is (4 2 5 1 3)!  Which doesn't make sense to me... unless that is suppose to be the correct answer...
 A: I guess you are using the cycle notation. Then the identity is not $(1\,2\,3\,4\,5)$, because that also denotes a cycle, rather the permutation with empty cycle decomposition, which might be denoted by $()$.
The cycle notation $(5\,1\,4\,2\,3)$ means the cyclic permutation that acts as following:
$$5\mapsto 1\mapsto 4\mapsto 2\mapsto 3\mapsto 5$$
So, right in your first line, $P(a)=1$ would rather imply $a=5$. 
Anyway, its inverse acts simply backwards, i.e. as:
$$5\mapsto 3\mapsto 2\mapsto 4\mapsto 1\mapsto 5$$
which is the cycle $(5\,3\,2\,4\,1)\ $ [which equals to $(3\,2\,4\,1\,5)$, etc.]
Similarly, the inverse of $(4\,2\,5\,1\,3)$ is not itself, but $(3\,1\,5\,2\,4)$.

Edit: Having reread the exercise, it seems that the notation is not the cycle decomposition, but $[5,1,4,2,3]$ means the permutation $1\mapsto 5,\ 2\mapsto 1,\ 3\mapsto 4,\ 4\mapsto 2,\ 5\mapsto 3$.  
But, you can start out from any element and follow its orbit to find the cycle decomposition: $[5,1,4,2,3]=(1\,5\,3\,4\,2)$. 
In this setting, the identity is indeed $[1,2,3,4,5]$.
The other permutation you ask: $[4,2,5,1,3]$ will decompose as $(1\,4)(3\,5)$ as it does nothing but swaps $1$ with $4$ and $3$ with $5$. And indeed, it is its own inverse.
