Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 2

Proof
Proof by Counterexample:

Consider the symmetric group $S_4$.

Then the order of the alternating group $A_4$ is $12$.

We list the subgroups of $A_4$:

Now $6$ divides $12$.

But there is no subgroup of $A_4$ of order $6$.