A very useful group to understand is $$S_4$$, the group of all permutations of $$\{1,2,3,4\}$$ under function composition. The group $$S_3$$ is not very interesting because it is too easy, and higher size symmetric groups are too complicated, but for n=4 the symmetric group is not terribly complicated. We will classifly all the subgroups of $$S_4$$.

By Lagrange's theorem the subgroups of $$S_4$$ must have order dividing $$4!=24$$. Therefore, a subgroup must have one of the following orders: 1,2,3,4,6,8,12,24. The order 1 case is simply the trivial subgroup and the order 24 case is simply the improper subgroup. We can easily disregard these basic cases and consider all the other divisors.

Let us derive the class equation for $$S_4$$. Two permutations are conjugate if and only if they have the same cycle structure, that is, iff when expressed as a product of disjoint permutations the number of disjoint permutations is equal and the cycles can be paired with equal lengths. The possible cycle structures for permutations are: 1-cycle, 2-cycle, 3-cycle, 4-cycle, product of a 2-cycle with a 2-cycle. The 1-cycle is simply the identity and therefore it just conjugate to itself. The 2-cycles are conjugate with themselves and there are $${4\choose 2}=6$$ of them. The 3-cycles are conjugate with themselves and there are $$2!{4\choose 3}=8$$ of them. The 4-cycles are conjugate with themselves and there are $$3!{4\choose 4}=6$$ of them. The 2,2-cycles are conjugate with themselves and there are $$\tfrac{1}{2}{4\choose 2}{2\choose 2}=3$$ of them. Therefore, the class equation for $$S_4$$ is given by: $$1+3+6+6+8=24$$.

A normal subgroup must be a union of conjugate class including the class $$\{\text{id}\}$$. Note among the summands, $$1,3,6,6,8$$ the only ones when paired with $$1$$ give a divisor of $$24$$ are $$1+3$$ and $$1+3+8$$. Therefore, the only possible normal subgroups must consist of a union of the 1-cycle and 2,2-cycles; or a union of the 1-cycle, 2,2-cycles, and 3-cycles. It is indeed the case that {id,(12)(34),(13)(24),(14)(23)} is a subgroup and so it must be normal, we call this subgroup V, it is the Klein four group. The other union is easily seen to contain all the even permutations, so it must be the alternating subgroup $$A_4$$.

Thus, we have shown that the only proper non-trivial normal subgroups of $$S_4$$ are $$V$$ and $$A_4$$. This information immediately classifies all the subgroups of order 12. Because any subgroup of order 12 must immediately be normal for it has index 2 in the full group. However, we determined that $$A_4$$ is the only normal subgroup of order 12, thus, there are no other subgroups of order 12.

Now let us move to the next possibly highest order, subgroups of order 8. Notice that $$2^3$$ is the highest power of two dividing 24, therefore by Sylow's first theorem will there be subgroups of order 8, namely the Sylow 2-subgroups. By Sylow's second theorem these subgroups all conjugate to one another, so there would be a unique Sylow 2-subgroup of order 8 if and only if it was a normal subgroup, however by the class equation we determined there are no normal subgroups of order 8, thus, the number of Sylow 2-subgroups must exceed 1. By Sylow's third theorem the number of Sylow 2-subgroups must be a divisor of 24 and congrunent to 1 modulo 2. The only such numbers are 1 and 3, but it must exceede 1, so the number of subgroups of order 8 must be precisely 3.

All we have to do now is determine the subgroups of order 8. This is easier than it seems because we already know one subgroup of order 8, namely the dihedral group $$D_4$$. The other subgroups are just the conjugates of $$D_4$$. Conjugate subgroups are clearly isomorphic and so we see that the three subgroups of order 8 are just copies of $$D_8$$.

Now we need to figure out the subgroups of order 6. An abelian group of order 6 must be isomorphic to $$\mathbb{Z}_6$$. Since there are no elements of order 6 in $$S_4$$ it follows that if $$S_4$$ has a subgroup of order 6 then it must be isomorphic to the non-abelian group of order 6, that is, isomorphic to $$S_3$$. A group is isomorphic to $$S_3$$ if and only if it has two generators, x and y where x has order 3 and y has order 2 which satisfy $$yx=x^2y$$. There are 4 elements of $$S_4$$ which have order 3. It is not hard to see that for each element we can find a corresponding element of order 2 which satisfies the relation $$yx=x^2y$$. Furthermore, each of these generated subgroups are different. And so there are four subgroups of order 6 each isomorphic to $$S_6$$. In fact, it is easy to describe these subgroups. Let $$S(t)$$ denote the permutations that fix t in the set 1,2,3,4. Then clearly $$S(t)\simeq S_3$$ for each t. And these four values of t determine these four subgroups.

This time we work with subgroups of order 4. These are either cyclic subgroups, generated by a 4-cycle or the Klein four groups. There are six 4-cycles but one can easily check that half of the generate the same subgroup, i.e. if x is a generator then so is $$x^3$$. Thus, there are 3 copies of $$\mathbb{Z}_4$$. The Klein four subgroups are generated by two elements of order 2. These include either two 2-2 cycles, or two 2-cycles which commute. The case with two 2-2 cycles was already considered, it was the subgroup we named as V. The other ones are generated by two 2-cycles. There are three other such subgroups. So we see there are 4 copies of $$\mathbb{Z}_2\times \mathbb{Z}_2$$.

The subgroups of order three are generated by a 3-cycle. There are eight 3-cycles but exactly half of them generate the same subgroup i.e. if x is a generator then so it $$x^2$$. So there will be precisely four copies of $$\mathbb{Z}_3$$.

Finally the subgroups of order two are generated by a 2-cycle or a 2-2 cycle. There are six 2-cycles and there are three 2-2 cycles. Thus, in total there are 9 copies of $$\mathbb{Z}_2$$.

To summarize our results: 9 copies of cyclic two groups, 4 copies of cyclic three groups, 3 copies of cyclic four groups, 4 copies of Klein four groups, 4 copies of symmetric three groups, 3 copies of dihedral four groups, 1 copies of alternating four group. And there are two proper non-trivial normal subgroups which are the alternating subgroup and V.

**Exercise:**Use the structure of $$S_4$$ to come up with a counter-example to the following statement: if H and N are subgroups of a group G and N is normal then HN is a normal subgroup also.

## No comments:

## Post a Comment