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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?

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    $\begingroup$ Hint: look at conjugacy classes of $S_4$. Convince yourself a normal subgroup of $S_4$ must either contain all of a given conjugacy class, or none of it. What possibilities does that give you? $\endgroup$ – David Wheeler Jun 13 '12 at 17:01
  • $\begingroup$ Use the GAP. It help us in these situations. $\endgroup$ – mrs Jun 13 '12 at 17:06
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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:

  1. The class of the $4$-cycles, cycle structure $(abcd)$, corresponding to the partition $4$. There are $6$ elements in this class.
  2. The class of the $3$-cycles, cycle structure $(abc)(d)$, corresponding to the partition $3+1$. There are $8$ elements in this class.
  3. The class of the product of two transpositions, cycle structure $(ab)(cd)$, corresponding to the partition $2+2$. There are $3$ elements in this class.
  4. The class the transpositions, cycle structure $(ab)(c)(d)$, corresponding to the partition $2+1+1$. There are $6$ elements in this class.
  5. The class of the identity, cycle structure $(a)(b)(c)(d)$, corresponding to the partition $1+1+1+1$. There is a single element in this class.

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 $2$.

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    $\begingroup$ Great..Explanation...I was struggling to understand this concept in book...but small doubt, you have mentioned "since it does not have transpositions, it is atmost 12" I cant understand this part..pls explain $\endgroup$ – Sam Christopher May 24 '15 at 6:52
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    $\begingroup$ @SamChristopher If it has transpositions, it is the whole group - a case in which we are not interested. Now, any number $n \in \{13, 14, ..., 23 \}$ is not a divisor of $24$. $\endgroup$ – Danilo Gregorin Oct 24 '17 at 14:57
  • $\begingroup$ You have written "So the order must be $1,2,3,4,6$ or $12$." What about $8$? $8$ is also a divisor of $24$. $\endgroup$ – ZFR Jan 20 '18 at 18:55
  • $\begingroup$ Dear, Arturo! Indeed very nice and simple explanation :) Also I have learnt the fact that "the subgroup is normal iff it is the union of conj. classes" which I have never met in my book. Thanks a lot for such post! +1 :) $\endgroup$ – ZFR Jan 20 '18 at 19:04
  • $\begingroup$ @RFZ: I think I was thinking divisors of 12 at that point. Hard to say, since it was five a half years ago, but you may be right. $\endgroup$ – Arturo Magidin Jan 21 '18 at 1:12
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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
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