Prove that the group $S_4$ has exactly $4$ distinct subgroups of order $6$.

Question

Prove that the group $S_4$ has exactly $4$ distinct subgroups of order $6$.

My attempt: Since any group of order 6 is either a cyclic group or isomorphic to $S_3$, and there's no element of order 6 in $S_4$, the only subgroups of order 6 must be isomorphic to $S_3$. Intuitively, there must be at least 4 subgroups of order 6, since we can choose 3 numbers out of $\{1,2,3,4\}$, which have 4 ways, to be permuted. However,

I couldn't come up with showing that these are the only ones.

This problem is provided as an extra exercise in the class I'm currently enrolled, but somehow the professor just doesn't give any discussion on it after the term exam. The class now only covers the materials in the first three chapters in Dummit, so I hope anyone can provide or give a hint on how to prove it with at least tools as possible. Thanks.

Answer 1

I may have misunderstood the question, but it seems to me that the author actually answered his own question.

Here is the full reasoning.

1. The group $S_4$ has exactly 8 cycles of length $3$, which means exactly $4$ subgroups of order $3$.

2. Each cycle of length $3$ is contained in exactly one subgroup of order $6$.

It is sufficient to prove this for the cycle $(1,2,3)$.

Let $(1,2,3)\in H$, $F=\{e,(1,2,3),(1,3,2)\}$ and $|H|=6$.

Then $F$ is a normal subgroup in $H$ ($|H:F|=2$).

If $\sigma\in H$, then $\sigma(1,2,3)\sigma^{-1}=(\sigma(1),\sigma(2),\sigma(3))\in F$.

Then $\sigma(4)=4$ and therefore $\sigma\in S_3$. That is, $H\leq S_3$, so $H=S_3$.

Answer 2

I guess a very basic proof with no further knowledge in algebra required:

Consider the multiplicative order of a subgroup element. That one has to divide $6$. Let’s factor a permutation $p$ into disjoint cycles.

A cycle of length $2$ has order $2$, a cycle on length $3$ has order $3$ and a cycle of length $4$ has order $4$. The only possible product of disjoint cycles is a product of two transpositions. So these are all possible permutations.

Now as the order must divide the group order no such subgroup may contain a cycle of length 4. But we want more: We want to show that there is one element that is fixed by all permutations.

Consider: If a subgroup contains $(a,b,c)$ and $(c,d)$ it contains the product, which sends $a\to b,b\to c, c\to d, d\to c\to a$, so $(abc)(cd) = (abcd)$. So the subgroup may not allow this.

Then $(abc)(dab) = (ac)(db)$, $(abc)(dba) = (adc)$ and $(adc)(abc) = (ab)(cd)$.

Then $(abc)(ac)(bd) = (bdc)$ and $(abc)(ab)(cd) = (ac)(cd) = (acd)$.

Thus if the subgroup contains a triplet $(abc)$ and any permutation not fixing $d$ it must not contain any transpositions. Also if a subgroup contains $(ab)$ and $(bc)$ then also $(ab)(cb) = (abc)$. So the only possible subgroups that does not contain triplets is created by $(ab),(cd)$ and contain with $(ab),(cd),(ab)(cd),e$ only four elements.

Now suppose the group contains $(abc)$ and $(abd)$. Then it contains $e,(abc),(acb),(abd),(adb),(ac)(db),(bdc)$, so at least $7$ elements. Similar for $(abc)$ and $(bad)$.

So we get: Any such subgroup must leave one element fixed, so it is thus isomorphic to some subgroup of $S_3$. But since $S_3$ has only $6$ elements it has to be in fact isomorphic to $S_3$.