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 $S_4$?
[Math] Prove $S_4$ has only 1 subgroup of order 12
group-theorypermutationssymmetric-groups
Related Solutions
By $D_n$ I assume you mean the dihedral group of order $2n$.
Note that a Sylow-2 of $S_4$ has order $8$, so we know $D_4$ is a Sylow-2 of $S_4$. Since any two Sylow-$p$ are conjugate, they are isomorphic. Thus all subgroups of order $8$ in $S_4$ have to be isomorphic to $D_4$.
Note that $n_2\equiv 1\pmod{2}$, $n_2\mid \#S_4$, so either $n_2=1$ or $n_2=3$. But the number of 4-cycles in $S_4$ is 6, and an $D_4$ only has two element of order 4. Since any group of order 4 appears as a subgroup of a Sylow-2, we cannot have $n_2=1$ and hence $n_2=3$.
For the remark that $S_6$ contains a subgroup ismorphic to $S_4\times S_2$, I believe the neatest way of saying it is that the standard embedding of $S_4$ in $S_6$ (i.e. permutations fixing 5 and 6) commutes elementwise with $H=\langle (56)\rangle\simeq S_2$ and their intersection is trivial, so that their product $S_4\cdot H=\{\sigma\tau\ |\ \sigma\in S_4, \tau\in H\}$ is indeed isomorphic to their direct product.
The rest of the argument is fine in itself: moreover, you will have to do some handiwork when dealing with concrete groups. For example, knowing that $D_8$ has the right order is not enough. Even if you exclude abelian groups of order 8 (which seem unlikely to be 2-Sylows of $S_4$), you are left with two options, namely $D_8$ or $Q_8$ (quaternions), and it is concrete exploration of $S_4$ to give you the answer on which of the two is the correct Sylow group.
Surely, though, there are ways to pick the right permutations that are more clever than others: in this case, moreover, there is a way you can get (somewhat) naturally to the answer, i.e. costruct a $2$-Sylow of $S_n$ inductively on $n$. For a positive integer $n$, call $\mu_2(n)$ the exponent of 2 in the prime factorization of $n!$, $P_n$ the (isomorphism class) of a 2-Sylow of $S_n$.
Clearly, $P_2\simeq C_2$ (where $C_n$ is the cyclic group of order $n$), and this holds for $P_3$ as well since $S_2$ embeds in $S_3$ and $\mu_2(3)=\mu_2(2)$.
As $\mu_2(4)=\mu_2(2)+2$, you must enlarge $\langle(12)\rangle<S_4$ by a factor 4. One factor 2 comes naturally by embedding $S_2$ in $S_4$ as $\langle(34)\rangle$: for the same argument used in the first paragraph, you may clearly see that $$H=\langle(12)\rangle\cdot\langle(34)\rangle\simeq C_2\times C_2.$$ Now, you have no more disjoint $C_2$ to multiply to your $H$, but you can still exchange $(12)$ and $(34)$, namely by conjugating by the double transposition $(13)(24)$. If $K=\langle(13)(24)\rangle$, this gives the product $HK$ (which is a subgroup, since $H$ and $K$ commute) a natural structure of semidirect product $H\rtimes K$ or, if you prefer, the isomorphism structure of $D_8$, so that $P_4\simeq (C_2\times C_2)\rtimes C_2$.
At this point, since $\mu_2(6)=\mu_2(4)+1$, just multiplying $P_4<S_6$ by the disjoint $C_2$ given by $\langle (56)\rangle$ does the trick, and leaves you with $P_6\simeq P_4\times P_2$.
Also, this construction has the advantage of showing a pattern. In fact, if you compute $\mu_2(n)$, you will easily see that:
- $P_{2^{n+1}}\simeq (P_{2^{n}}\times P_{2^{n}})\rtimes C_2$ as in the case of $P_4$, which is normally written $P_{2^{n}}\wr C_2$ and called a wreath product;
- if $n=a_0+a_1\cdot 2+\cdots+a_k\cdot 2^k$ in base 2, $P_n\simeq \prod_{j=1}^k (P_{2^i})^{a_i}$, as with $P_6$;
- this generalizes to other primes in the obvious way.
Best Answer
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.