# 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 :) Commented Mar 22, 2014 at 11:50
• I don't know how you go about finding something that isn't there. Commented Mar 22, 2014 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. Commented Mar 22, 2014 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
Commented Mar 22, 2014 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. Commented Mar 22, 2014 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 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:

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

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.

• How does this answer 1) and/or 2) in the OP? Commented Nov 29, 2023 at 19:52
• @citadel, this answers 2). Specifically, I have shown that if a subgroup of $S_4$ with order $12$ exists, then it is uniquely determined by being a union of some conjugacy classes. Commented Nov 29, 2023 at 20:12
• I think that this to work you have to show also that there are not enough conjugacy classes of size $3$ and/or $8$ to build up a second (non conjugated) subgroup of order $12$. Commented Nov 29, 2023 at 21:24
• @citadel. $S_4$ is the union of five conjugacy classes: $S_4=A\cup B\cup C\cup D\cup E$. I have shown that a subgroup $H\leq S_4$ with $|H|=12$ is the union of three of these conjugacy classes, say $H=A\cup B\cup C$. This uniquely determines $H$. Commented Nov 30, 2023 at 8:55

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, $$A=B$$.