$A_{4}$ unique subgroup of $S_4$ of order $12$

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.

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.