Symmetric group $S_3=\{(),(1,2),(2,3),(1,3),(1,2,3),(1,3,2)\}$, I understand that $H=\{e,(1,2,3),(1,3,2)\}$ is the normal subgroup of S3 ($H\lhd S_3$) because:
$$ gH=Hg, \forall g\in S_3$$
e.g. let $g=(1,2)$ we have
$$gH=\{(1,2),(1,2)(1,2,3),(1,2)(1,3,2) \}=\{(1,2),(2,3),(1,3)\}$$
$$Hg=\{(1,2),(1,2,3)(1,2),(1,3,2)1,2) \}=\{(1,2),(1,3),(2,3)\}$$
or let $g=(1,2,3)$ we have
$$gH=\{(1,2,3),(1,3,2),()\}$$
$$Hg=\{(1,2,3),(1,3,2),()\}$$
Q1: how one can find such $H\lhd S_3$ (proper, non-trivial normal subgroup of maximum order) by a step-by-tep procedure? Finding the normal subgroup of S4 or large symmetric groups seems tedious.
Q2: Is there a way to find the order of the group without constructing it?
Best Answer
The order of $A_n=\frac{n!}{2}$ where $A_n $ is the alternating group on $n$ symbols To construct an alternating group on $n$ symbols just collect all the even permutation of $S_n$ For your case $S_3$ the even permutations are $(1),(123), (132)$ and hence the set set consisting of these three permutations only forms your alternating group of$S_3$