# Finding the inverse of an element of $S_n$ and it's order [duplicate]

I have two questions,

1) What are the ways to find the inverse of an element of $S_n$?

2) What are the ways to find the order of an element of $S_n$?

• Regarding 2) see the discussion here, here, or here. Regarding 1), try first the tranpositions $(i,j)$. – Dietrich Burde Feb 11 '15 at 15:51
• On this site, if you have two questions, ask two questions. Please do not bundle them. Link them when they're related. – Lord_Farin Feb 11 '15 at 21:14

Let $\sigma \in S_{n}$, then $\sigma = \tau_{1} \tau_{2} \ldots \tau_{k}$ where $\tau_{i}$ is a transposition for each $i= 1,2 \ldots k$. Then $\sigma^{-1} = \tau_{k} \tau_{k-1} \ldots \tau_{2} \tau_{1}$. There are many ways to find the order of an arbitrary element of $S_{n}$, it's sometimes useful to write $\sigma$ as a product of disjoint cycles. Say $\sigma = \mu_{1} \mu_{2} \ldots \mu_{j}$. As disjoint cycles commute, if $\mu_{i}$ has length $l_{i}$, then the order of $\sigma$ is the least common multiple of the $l_{i}$'s.
If you write the elements of $S_n$ as matrices as follows: $$\sigma = \begin{pmatrix} 1 &2 &3& 4 \\ 3& 4& 1 &2 \end{pmatrix},$$
(wher of course $\sigma(1)=3, \sigma(2)=4$, and so on), then you can find the inverse just by turning the matrix upside down: $$\sigma^{-1} = \begin{pmatrix} 3 &4 &1& 2 \\ 1& 2& 3&4 \end{pmatrix}.$$
Of course the ordering is different, but you can still read off that ,for example $\sigma^{-1}(3)=1, \sigma^{-1}(4)=2$, and so on.