Prove $S_4$ has only 1 subgroup of order 12

The subgroup in $S_4$ that I know has order 12 is the subgroup of all even permutations, otherwise known as the alternating group $A_4$. However, I know this from a fact and not because I am able to show a subgroup of order 12 exists in $S_4$ in the first place. If I had not been told there existed a subgroup of order 12 in $S_4$, I would not have known there was one. So I guess this is actually a two-parter for me: 1) How do I go about showing that a subgroup of order 12 does indeed exist in $S_4$? I know by Lagrange that if there exists a subgroup, then the order of that subgroup must divide the order of the group. However, this doesn't say anything about the existence of the subgroup. And Sylow only verifies subgroups of a prime to some power order, which 12 is not. I also know that there might be a cyclic subgroup of order 12, but without manually multiplying every possible permutation in $S_4$, I don't know any other way to check the existence of this or if this subgroup is isomorphic to $A_4$ because I don't know how to write these permutations as functions in order to check if there is a homomorphism (I know that a permutation is bijective by definition). And 2) How do I show that this group of order 12 is the only group of order 12 in $S_4$?

• The group $S_4$ has $24$ elements. It is a good exercise to write out all subgroups of $S_4$, and you will immediately have your answer. You can simply check all combinations of generators, and you will soon see that there are not that many subgroups :) – Servaes Mar 22 '14 at 11:50
• I don't know how you go about finding something that isn't there. – Derek Holt Mar 22 '14 at 11:57
• A subgroup $G \subset S_4$ of order $12$ will have index $2$, hence $G$ is normal with quotient isomorphic to the unique group of order $2$, which is, in particular, abelian. If you are familiar with the fact that a quotient group $A/B$ is abelian if and only if $B$ contains the commutator subgroup $[A,A]$, then you need only figure out what the commutator subgroup is, then you can utilize that to figure out what the index $2$ subgroups are. – Dustan Levenstein Mar 22 '14 at 12:52
• Cauchy's theorem says that if the order of $G$ is divisible by a prime $p$, then $G$ contains an element of order $p$. – MJD Mar 22 '14 at 13:05
• Well if you note that in general, in the symmetric group, that permutations are either odd or even, and an even perm. composed with an even perm. is again even, then they form a group. Also the identity is even. And you can show that there is an equal number of odd and even permutations. If you know that $|S_n|=n!$ then $|A_n|=n!/2$. That's the existence bit at least. – snulty Mar 22 '14 at 13:06

Let $H\leq S_n$ be a subgroup of order 12. Then $S_4/H$ is a group of order 2, hence $S_4/H \cong C_2$, which is abelian. Therefore, $\left[S_4,S_4\right] \leq H$, where $\left[S_4,S_4\right]$ denotes the commutator subgroup (smallest normal subgroup of $S_n$ with abelian quotient).

Now if we can show that $[S_4,S_4] = A_4$, we have $A_4 \leq H$, $\left|A_4\right| = \left|H\right|$ and therefore $A_4 = H$. Firstly, $S_4/A_4 \cong C_2$ as above, so $[S_4,S_4]\leq A_4$. For any 3-cycle $(i,j,k) \in S_4$:

\begin{align*}(i,j,k) = (i,k,j)^2 = ((i,k)(i,j))^2 &= (i,k)(i,j)(i,k)(i,j)\\ &= (i,k)(i,j)(i,k)^{-1}(i,j)^{-1} = \left[(i,j),(i,k)\right] \in \left[S_4,S_4\right].\end{align*}

The set of all 3-cycles in $S_4$ generate $A_4$, so $A_4 \leq \left[S_4,S_4\right]$ and the result follows.

If you want to show that $A_4$ really is a subgroup of index 2, define $\delta : S_n \rightarrow C_2$ by $$\delta(\sigma) = \begin{cases} 1, &\sigma\; \text{even}\\ -1, &\sigma\; \text{odd}\end{cases}$$ and show it's an epimorphism.

Here is a link which will help you with the question and the comment of Derek: $A_n$ is the only subgroup of $S_n$ of index $2$. .

You need following steps to see the answer.

$1.)$ Define even and odd permutation.

$2)$ Show that for any $\sigma\in S_n$ if it can be written as odd number of cycles then it can't be written as even number of cycles.(this shows that above defination is welldefined)

$3)$ define $\phi: S_n\to Z_2$ by $\phi(\sigma)=0$ if $\sigma$ is even and $1$ otherwise.

$4)$Show that above function is epimorphism and conclude that $Ker(\phi)$ is a subgroup of $S_n$ with index $2$ and is set of all even permutation of $S_n$.

$5)$To show that $A_n$ is the uniqe group with index $2$.Let Assume that $H$ be another subgroup of $S_n$ with index $2$.

Then we must have $HA_n=S_n\implies \frac{|H||A_n|}{|A_n\cap H|}=|S_n|\implies R=A_n\cap H$ is subgroup of $A_n$ with index $2$.So,$R$ is normal in $A_n$.

Since $A_n$ is simple for $5\leq n$ we must assume that $n<5$.

Since $A_3$ has order $3$ only possible value is $4$. We need show that $A_4$ has no subgroup of order $6$.I is very clasic problem you can see the solution

And you can find first $4$ part in almost any basic abstract algebra book.