# find normal subgroup of symmetric group

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?

• How is the alternating group defined for you? Perhaps the standard definition is of "odd" and "even" permutations, but then this gives you the elements! Is this how it is defined for you? – user1729 Mar 18 '16 at 9:53
• I think I made a mistake. In fact, I mean finding the normal subgroup of a symmetric group,i.e., $H\lhd S_3$ the definition is gH = Hg for all g in G. – whitegreen Mar 18 '16 at 10:05
• Why do you refer to 'the' normal subgroup? For $n \geq 3$, $S_n$ always has at least $3$ normal subgroups, namely $S_n$, $\{id\}$ and $A_n$. If you require proper, then there are at least still two. – sqtrat Mar 18 '16 at 10:42
• @whitegreen I am not understanding your question. Is it "Find the unique proper, non-trivial subgroup of $S_n$?" – user1729 Mar 18 '16 at 10:47
• @user1729, yes, I want to find the proper normal subgroup of $S_n$ of maximum order. – whitegreen Mar 18 '16 at 10:54

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$S_3$