Inverse Permutations from $S_7$

Question

Would someone mind giving an explanation of how to find the inverse permutation of:

$(1 2 3 5 7)^{-1}$ in $S_7$?

I am not quite understanding how to do this.

Answer 1

You can simply re-write the original permutation in reverse to get $$\text{inverse permutation} = \left(75321\right)$$

Why is this true?

As a permutation is a bijection, then it has an inverse which is also a bijection. The inverse merely undoes what the original function does. Thus if $i\sigma = j$ then its inverse $\sigma_{-1}$ maps $j$ to $i$.

To find the inverse of a permutation that is a cycle all we have to do is write the elements of the cycle in reverse order.

You can always check if your inverse is correct by multiplying your permutation with the inverse and verifying that $\sigma \sigma^{-1} = 1$

Cycle v/s Matrix notation

If $$\sigma = (12357) = \begin{pmatrix} 1&2&3&5&7\\ 2&3&5&7&1 \end{pmatrix}$$

Then $$\sigma^{-1} = (75321) = \begin{pmatrix} 1&2&3&5&7\\ 7&1&2&3&5 \end{pmatrix}$$