1. ## Permutation Group

Let $A=\{1.2.3\}$ and $(S_A,\circ)$ be the symmetric group. Then $(S_A,\circ)$ has 3!=6 elements.

Let $S_3=\{\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5 ,\alpha_6\}$ and

$
\alpha_1=\left(\begin{array}{ccc}1&2&3\\1&2&3\end{ array}\right)\:\\\alpha_2=\left(\begin{array}{ccc} 1&2&3\\1&3&2\end{array}\right)\;\alpha_3=\left(\be gin{array}{ccc}1&2&3\\3&2&1\end{array}\right)
$

$
\alpha_4=\left(\begin{array}{ccc}1&2&3\\2&1&3\end{ array}\right)\:\alpha_5=\left(\begin{array}{ccc}1& 2&3\\2&3&1\end{array}\right)\;\alpha_6=\left(\begi n{array}{ccc}1&2&3\\3&1&2\end{array}\right)
$

The following are the inverses:
$\alpha_1^{-1}=\alpha_1,$
$\alpha_2^{-1}=\alpha_2,$
$\alpha_3^{-1}=\alpha_3,$
$\alpha_4^{-1}=\alpha_4,$
$\alpha_5^{-1}=\alpha_6,$ and $\alpha_6^{-1}=\alpha_5$

I have noticed that $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\al pha_6\}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,\alpha_3^{-1},\alpha_4^{-1},\alpha_5^{-1},\alpha_6^{-1}\}$, although not in the same order.

Question 1: Is it always true that $\{\alpha_1,\alpha_2,...,\alpha_n\}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$?

Question 2: While $\{\alpha_1,\alpha_2,...,\alpha_n\}$ is labeled as S_n, how do we label $\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$?

Can we write them like this?

$
S_n=\{\alpha_1,\alpha_2,...,\alpha_n\}
$
and

$S_{n^{-1}}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$

or can they be written like this?

$S_{\alpha}=\{\alpha_1,\alpha_2,...,\alpha_n\}
$
and

$S_{\alpha^{-1}}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$

2. Answer 1. For every finite group $G$, every element has its own inverse, so yes, that equality is true. I think this should also hold for infinite groups.

Answer 2. The $n$ in $S_n$ signifies the number of elements that we are permuting, so $S_n$ has $n!$ elements, so all these notations are incorrect. In fact, $S_n = \{a_1,a_2,\ldots,a_{n!}\} = \{a_1^{-1},a_2^{-1},\ldots,a_{n!}^{-1}\}$.