Prove $S_4$ has only 1 subgroup of order 12

Question

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 $S_4$?

Answer 1

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 definition is welldefined.)

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

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

$5)$ To show that $A_n$ is the uniqe group with index $2$. Let us assume that $H$ is 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 $n\ge 5$ we must assume that $n\le4$.

Since $A_3$ has order $3$ the only possible value is $n=4$. We need to show that $A_4$ has no subgroup of order $6$. It is a classic problem and you can see a solution here:

http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/A4noindex2.pdf

And you can find the first four parts in almost any basic abstract algebra book.

Answer 2

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.

Answer 3

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 $2$. .

Answer 4

A subgroup of index two is normal, so it is the union of some conjugacy classes. The conjugacy classes of $S_4$ are of sizes $1,6,8,3,6$, and $1+3+8$ is the only way to sum $12$ with these numbers.

Answer 5

This argument doesn't even mention $A_4$ (as such).

How do I go about showing that a subgroup of order $12$ does indeed exist in $S_4$?

Note that: $$H:=\{(),(12)(34),(13)(24),(14)(23)\}$$ is a closed subset of $S_4$, hence a subgroup of $S_4$. Moreover, its nontrivial elements exhaust all the double transpositions of $S_4$, hence conjugating $H$ by every element of $S_4$ still returns $H$, namely $H\unlhd S_4$. Now, take e.g. $K:=\langle (123)\rangle$; then $HK$ is a subgroup of $S_4$ of order $|HK|=\frac{|H||K|}{|H\cap K|}=$ $\frac{4\cdot 3}{1}=12$.

How do I show that this group of order $12$ is the only group of order $12$ in $S_4$?

Suppose $A$, $B$ are two subgroups of order $12$. Since $[S_4:A]=$ $[S_4:B]=$ $2$, both $A,B\unlhd S_4$. By a counting argument, $|A\cap B|=6$; moreover, $A\cap B\unlhd S_4$. But $S_4$ hasn't got any normal subgroup of order $6$: contradiction. So, $A=B$.