Prove that $(1, 2, \ldots ,n)^{-1}=(n, \ldots, 2, 1)$ within the symmetric group.

Question

Show that the cycles $(1, 2, \ldots, n)$, $(n, \ldots, 2, 1)$ are inverse permutations.

Answer 1

Let $\sigma = (1, \ldots, n)$ and $\tau= (n , \ldots , 1)$. We need to show for every $k= 1, \ldots , n$ we have that $$ \sigma(\tau(k)) = \tau(\sigma(k)) = k. $$ If $k < n$ then $\tau(\sigma(k)) = \tau(k+1) = k$. If $k = n$ then $\tau(\sigma(n)) = \tau(1) = n$. An identical argument shows that $\sigma(\tau(k)) = k$, and we are done.

Answer 2

Suppose that the inverse of your permutation $\begin{pmatrix}1 & 2 & 3 &\cdots& n \end{pmatrix}$ be $\begin{pmatrix}1 & 2 & 3 &\cdots& n \\ a_1 & a_2 & a_3 &\cdots& a_n\end{pmatrix},$ where $a_1,\, a_2,\, a_3,\ldots, a_n$ are to be determined. Since the composition is an identity permutation, so $\begin{pmatrix}1 & 2 & 3 &\cdots& n \end{pmatrix} \circ \begin{pmatrix}1 & 2 & 3 &\cdots& n \\ a_1 & a_2 & a_3 &\cdots& a_n\end{pmatrix}=\begin{pmatrix}1 & 2 & 3 &\cdots& n \\ a_1 & a_2 & a_3 &\cdots& a_n\end{pmatrix} \circ \begin{pmatrix}1 & 2 & 3 &\cdots& n \end{pmatrix}=(1) \circ (2) \circ (3)\circ \cdots \circ (n).$

Now the only choice for the above equation to be true is $a_1=n,\, a_2=1,\,a_3=2,\,a_4=3,\ldots,a_n=n-1.$

Therefore the required inverse is $\begin{pmatrix}n & n-1 & n-2 & \cdots &1 \end{pmatrix}.$