$A_{4}$ unique subgroup of $S_4$ of order $12$ I was reading the proof that $A_{4}$ is the unique subgroup of order $12$.
So the author counts the number of conjugates:
$1$ cycle of type $()$ , $6$ cycles of type $(1 \ 2)$, 8 cycles of type $(1 \ 2 \ 3)$, $6$ cycles of type $(1 \ 2 \ 3 \ 4)$ and $3$ cycles of type $(1 \ 2)(3 \ 4)$.
Now it says, the only possible way to get $12$ elements is $1 + 3 + 8$. Why is this? can't we have $1$ cycle of type $()$, $2$ cycles of type $(1 \ 2)$, $3$ cycles of type  $(1 \ 2 \ 3)$, $2$ cycles of type $(1 \ 2)(3 \ 4)$ and $4$ cycles of type $(1 \ 2 \ 3 \ 4)$. This also gives you $12$. Why is this impossible though? I don't really understand why the only possibility is $1 + 3 + 8$.
 A: Another way to prove this result.
A subgroup of index $2$ is always normal, so $A_4$ is normal in $S_4$.
Assume by absurd that $H\ne A_4$ is another subgroup of $S_4$ of order $12$.
Using the second theorem of isomorphism, we get:
$$\vert H.A_4\vert =\frac{\vert H\vert \cdot\vert A_4\vert }{\vert H\cap A_4\vert }=\frac {144}{\vert H\cap A_4\vert }.$$
Using Lagrange's theorem, we get that:
$$\frac{144}{\vert H\cap A_4\vert}\text{ divides }\vert S_4\vert=24.\qquad (\star)$$
But since $H\ne A_4$, we have $\vert H\cap A_4\vert<12$.
So with $(\star)$ we get $\vert H\cap A_4\vert=6$.
But $A_4$ can not have a subgroup of order $6$.
If it had one $K$, it would be normal (since of index $2$), and by Cauchy's theorem there would exist $s\in K$ of order $3\mid 6$: $s$ would be a $3$-cycle. 
But the $3$-cycles are conjugated so $K=A_4$ of order $12$ which is absurd.
A: A subgroup of index 2 is always normal. Two elements of $S_n$ are conjugate iff they have the same cycle type. Thus, if your subgroup contains one element of a given cycle type, it contains all of them. Since the subgroup must contain the identity, and there's only one way of getting 11 as a sum from ${3,6,6,8}$, $A_4$ is the only possibility.
