How to find the inverse of a permutation function from its 2-cycle decomposition? Could I quickly spot the inverse of a permutation from its 2-cycle composition?
For example, given that $\rho=(1 \ 9)(1 \ 4)(1 \ 5)(1 \ 8)(2 \ 10)(2 \ 3)(2 \ 6)(2 \ 7)$, how to find its inverse from this 2-cycle decomposition?
 A: Yes, it is easy to write down the inverse of a permutation in this form.
It is also not much harder to do this for any permutation given in cycle form. 
There are two main points to note: 


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*If you have $c_1 \dots c_n $ then the inverse is given by $c_n^{-1} \dots c_1^{-1}$. Where $c_i^{-1}$ is the inverse of $c_i$. This is true in any group, not only for permutations. 

*A $2$-cycle is its own inverse. So if each $c_i$ above is a two cycle then   $c_n^{-1} \dots c_1^{-1}= c_n \dots c_1$. 
If you want something similar without the condition that all cycles are $2$-cycles, it suffices to determine the inverse of a cycle can be obtained by reversing the order of the elements in the cycle, so if you have a cycle  $c = (x_1 \ x_2  \dots x_k)$ then its inverse is $(x_k   \dots x_2 \ x_1)$.
Note this is not a contradiction to the assertion that a $2$-cycle is its own inverse as $(x \ y)$  and $(y \ x)$ are the same permutation. 
A: Hint: Use $(ab)^{-1}=b^{-1}a^{-1}$ and $(i\, j)=(i\, j)^{-1}$.
Where ab is the composite permutation $(a\circ b)(k)$, $k$ is an element of the set being permuted, and $(i\, j)$ represents the $2$-cycle of the elements $i$ and $j$.
