Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?
In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups?
Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?
In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups?
In any group, a subgroup is normal if and only if it is a union of conjugacy classes.
In $S_n$, the conjugacy classes are very easy: a conjugacy class consists exactly of all permutations of a given cycle structure. These corresponds to all possible partitions of $n$.
So, consider $S_4$. The conjugacy classes in $S_4$ are:
Now, any subgroup that contains all transpositions is the whole group.
So we can consider only subgroups that don't contain the transpositions. Their order must be a divisor of $24$, and since it does not have the transpositions, it is at most $12$. So the order must be $1$, $2$, $3$, $4$, $6$, or $12$. Moreover, the order must the be sum of the sizes of some conjugacy classes, so it must be a sum of some of the numbers $1$, $3$, $8$, and $6$, and must include $1$.
One possibility is the trivial group, order $1$. We cannot get a normal subgroup of orders $2$ or $3$ (in particular, you $H$ cannot possibly be normal). The only way to get a subgroup of order $4$ is to take the class of the identity and the class of the product of two transpositions. This is your $K$; if it is a subgroup, then being a union of conjugacy classes shows that it is normal. So just check if it is a subgroup.
We cannot get a normal subgroup of order $6$, because we can't just take the conjugacy class of $4$-cycles (we need the identity). As for a subgroup of order $12$, we would need to take the identity ($1$ element), the class of products of two transpositions ($3$ elements), and the class of $3$-cycles ($8$ elements). This collection has a very familiar name...
And that's it! You cannot have any other normal subgroups. So, in summary: the trivial group, the whole group, and possibly $K$ (if it is a subgroup), and possibly this last collection (if it happens to be a subgroup). At most $4$, at least $2$.
As suggested by Babak Sorouh, the answer can be found easily using GAP using the SONATA library. Here's the code:
G:=SymmetricGroup(4); S:=Filtered(Subgroups(G),H->IsNormal(G,H)); for H in S do Print(StructureDescription(H),"\n"); od;
So as to not spoil Arturo Magidin's answer, here's the output if I replace G:=SymmetricGroup(4);
with G:=DihedralGroup(32);
(the dihedral group of order $32$)
1 C2 C4 C8 D16 D16 C16 D32