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This is an exercise from Artin's Algebra 2nd Ed. The exercise refers to the following example from the text:

The three partitions of the set of four indices $\{\mathbf{1},\mathbf{2},\mathbf{3},\mathbf{4}\}$ into pairs of subsets of order two are :$$\Pi_{1}\colon\{\mathbf{1},\mathbf{2}\}\cup\{\mathbf{3},\mathbf{4}\},\quad\Pi_{2}\colon\{\mathbf{1},\mathbf{3}\}\cup\{\mathbf{2},\mathbf{4}\},\quad\Pi_{3}\colon\{\mathbf{1},\mathbf{4}\}\cup\{\mathbf{2},\mathbf{3}\}.$$ An element of the symmetric group $S_{4}$ permutes the four indices, and by doing so it also permutes these three partions. This defines a map $\varphi$ from $S_{4}$ to the group of permutations of the set $\{\Pi_{1},\Pi_{2},\Pi_{3}\}$, which is just the symmetric group $S_{3}$.

The example goes on to show that $\varphi$ is a homomorphism and that

Its kernel can be computed. It is the subgroup of $S_{4}$ consisting of the identity and the three products of disjoint transpositions: $$K=\left\{1,~(\mathbf{1}~\mathbf{2})(\mathbf{3}~\mathbf{4}),~(\mathbf{1}~\mathbf{3})(\mathbf{2}~\mathbf{4}),~(\mathbf{1}~\mathbf{4})(\mathbf{2}~\mathbf{3})\right\}.$$

The exercise then asks:

Determine the six subgroups of $S_{4}$ that contain $K$.

I have two difficulties with this exercise. How was $K$ computed? And how can I determine the subgroups of $S_{4}$ that contain $K$ without having to just use brute force to compute all the possible cycle permutations?

So far, my naive approach has been to compose elements from $S_{4}$ in an effort to find a composition that gives an element of $K$. While I think this would work eventually, I know there must be an easier way.

I should mention that I have always had a difficult time working in $S_{n}$. I'm not familiar with the subgroups of $S_{n}$ and anytime I've had to work with a symmetric group I inevitably have to sit down and write out a table for each permutation and/or the composition of permutations.

I haven't tagged this as homework because I'm self studying. (Trying to improve my grasp of permutations and permutation groups.)

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up vote 3 down vote accepted

A permutation of $S_4$ will preserve a partition $\{a,b\}\cup \{c,d\}$ under the following conditions:

  • It preserves the two sets $\{a,b\}$ and $\{c,d\}$ individually, or
  • It shuffles the two sets as $\{a,b\}\leftrightarrow\{c,d\}$.

We then look at the stabilizer of $\Pi_1=\{1,2\}\cup\{3,4\}$. Based on the above facts, it is comprised of

  • $(),(12),(34),(12)(34);$
  • $(13)(24),(14)(23),(1423),(1324)$.

This has $8$ elements, but $K=[S_4:S_3]=4$, so we have four extra ones. Which permutations above do not fix $\Pi_2=\{1,3\}\cup\{2,4\}$ and $\Pi_3=\{1,4\}\cup\{2,3\}$? We compute the other stabilizers:

  • $\mathrm{Stab}(\Pi_1)=(),(12),(34),(12)(34),(13)(24),(14)(23),(1423),(1324)$;
  • $\mathrm{Stab}(\Pi_2)=(),(13),(24),(13)(24),(12)(34),(14)(32),(1432),(1234)$;
  • $\mathrm{Stab}(\Pi_3)=(),(14),(23),(14)(23),(12)(43),(13)(42),(1342),(1243)$.

These may each be computed by first checking where $1$ may go, then where $1$'s partner may go etc.

The intersection of the stabilizers is by inspection $K=\{(),(12)(34),(13)(24),(14)(23)\}$.

Each of these contains $K$. Two other groups are $K\le K$ and $K\le S_4$, so we need one more; note that every permutation in $K$ is a product of two transpositions, an even number, so $K\le A_4$.

In conclusion, the six groups are $\{K,\,\mathrm{Stab}(\Pi_1),\,\mathrm{Stab}(\Pi_2),\,\mathrm{Stab}(\Pi_3),\,A_4,\,S_4\}$

Alternatively, recall that if $K\le G\le S_4$ and $K\triangleleft S_4$ (kernels are normal), then by the lattice aka correspondence theorem $K\triangleleft G$ and $G/K\le S_4/K\cong S_3$, and the six subgroups of $S_3$ are...

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Checking that the stabilizers and alternating subgroup are overgroups of $K$ is quite easy, but this tactic involves more cleverness and serendipity than churning out preimages. – anon Sep 21 '12 at 16:12
Without the correspondence theorem, did you prove that there aren't more than 6 subgroups containing $K$? Apart from that I prefer your solution to the one I suggested. – Michalis Sep 21 '12 at 16:20
@Michalis No, I took it as a given in the problem. [There is an obvious hybrid of the two methods: use correspondence to know a priori there are six intermediate groups but make the above observations to skirt preimage computations.] – anon Sep 21 '12 at 18:22
If $K$ has 4 elements and $K \subseteq \mathop{\text{Stab}}(\Pi_1)$, the latter can't have 6 elements. Also, $(12) \cdot (13)(24) = (1423)$. – Hurkyl Sep 22 '12 at 7:47
@Hurkyl Actually it's $(1324)$, but I still need to amend my answer, yikes! – anon Sep 22 '12 at 7:55

Let $1\neq x\in K$ written in cycle form, such that $x(1)=2$. Then $x(2)$ must be 1, otherwise $x$ wouldn't fix $\Pi_1$. $x(3)=4$, otherwise $\Pi_2$ wouldn't get fixed, so $x=(12)(34)$, which is in $K$. Since the definition of $K$ is symmetric in the 4 elements you are permuting you can now conclude that $K$ consists of the identity element and all the $2,2$-cycles. Otherwise you can continue with "take an element with $x(1)=3$, $x(1)=4)$, $x(2)=3, \ldots$ and draw the same conclusions.

To determine the 6 subgroups containing $K$ you can start by noticing that those subgroups are in 1 to 1 correspondence with subgroups of $S_4/K$. Since $\phi$ is surjective (for this you only have to show that the image contains the elements $(12)$ and $(23)$ in $S_3$) we have $$ S_4/K\cong S_3 $$ Now you probably know the $6$ subgroups of $S_3$, which are the identity, the 3 subgroups generated by transpositions, the 3-cycles and $S_3$. To get the corresponding subgroups in $S_4$ you have to translate the generators in $S_3$ to elements of $S_4/K$ and then pull them back to $S_4$. If you need help with this last part write me a comment and I'll fill it in tonight.

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Here is one way you can do it directly by hand: Write down explicitly (I know this may be a painful process but I've done it before) where each element of $S_4$ goes into $S_3$ under the homomorphism. Your six subgroups of $S_4$ that contain the kernel $K$ by the correspondence theorem are precisely the preimages of subgroups of $S_3$. To compute the preimages of subgroups of $S_3$, take for example the subgroup


If you already computed where each element in $S_4$ went in $S_3$, then the preimage of this subgroup is precisely the union of the preimages of each cycle in the list above.

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