What is the inverse cycle of permutation? Given cyclic permutations, for example,
$σ = (123)$, $σ_{2} = (45)$,
what are the inverse cycles $σ^{-1}$, $σ_2^{-1}$?
Regards.
 A: So, you have that $\sigma_1$ is the cycle such that,
$$\begin{align} 1 \mapsto 2 \\ 2 \mapsto 3 \\ 3 \mapsto 1\end{align}$$
It's inverse, $\sigma_1^{-1}$ is a cycle such that composition, $ \sigma_1 \circ \sigma_1^{-1}=\sigma_1^{-1} \circ \sigma_1$ is identity. So, the inverse cycle should look like :
$$\begin{align} 2 \mapsto 1 \\ 3 \mapsto 2 \\ 1 \mapsto 3\end{align}$$
What is this in cycle notation? 

$\sigma_1^{-1} \equiv(213)$

I'll let you try the other one. 

A particularly easy way of doing this, once you understand what the inverses do is: just to write the cycle backwards! 

Note that for $(123)$, this is just $(321)$. Now, recall, that set of all permutations form a group. In a group, inverses are unique. So, can you tell me why $(321)$ and $(213)$ are the same?

A: To elaborate on my comment:
To find the inverse of a permutation just write it backwards. If $\tau = (1243)(67)$ then $\tau^{-1}=(76)(3421)$ which can then be rewritten as $\tau^{-1}=(1342)(67)$.
For for the above question: $(123)^{-1}=(321)=(132)$.
How does one prove this?
First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the
inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$.
Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_k \mapsto a_1$,
 then $\sigma^{-1}$: $a_k \mapsto a_{k-1} \mapsto a_{k-2} \mapsto \cdots \mapsto a_1 \mapsto a_k$.
This is precisely the cycle  $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards. 
Now what about a list of cycles? Say $\sigma=\sigma_1\cdots \sigma_\ell$. Recall that $\sigma^{-1}=(\sigma_1\cdots \sigma_\ell)^{-1}=\sigma_\ell^{-1}\cdots \sigma_1^{-1}$. So we reverse the list of cycles and then write each one backwards -- thus the inverse is just the whole thing written backwards.
One thing to note: This still works even if $\sigma$ is not written in terms of disjoint cycles.
A: Another way to look at Bill Cook's fifth paragraph in his answer:
First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the
inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$. Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_{k - 1} \mapsto a_k \mapsto a_1$, then  $\sigma^{-1}$:
$a_1 \color{aqua}{\leftarrow} a_2 \color{aqua}{\leftarrow} a_3 \color{aqua}{\leftarrow}\cdots   \color{aqua}{\leftarrow} a_{k - 1} \color{aqua}{\leftarrow} a_k \color{aqua}{\leftarrow} a_1 \iff$
$a_k \color{aqua}{\mapsto} a_{k-1} \color{aqua}{\mapsto}  a_{k-2} \color{aqua}{\mapsto}  \cdots \color{aqua}{\leftarrow} a_2 \color{aqua}{\mapsto}  a_1 \color{aqua}{\mapsto}  a_k$.
This is precisely the cycle $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards. 
A: Every permutation n>1 can be expressed as a product of  2-cycles.
And every 2-cycle (transposition) is inverse of itself.
Therefore the inverse of a permutations is 
Just reverse products of its 2-cycles
(ab)^-1 = b^-1 a^-1
