Let $G =A_{4}$. Then 6 divides the order of G but G does not have a subgroup of order 6.

Proof: Note that a 2-cycle is always an odd permutation and any m-cycle can be written as a product of (m-1) 2-cycles and so an m-cycle is odd if and only if m is even, this tells us the shapes of cycles contained in $A_{4}: id, (ab)(cd), (abc)$. The number of p-cycles in $S_{n}$ is given by $(p-1)!\binom{n}{p}$ and in particular there are 8 cycles of length 3 in $A_{4}$.

Now suppose G has a subgroup H of order 6. Then since $[G: H] = \frac{12}{6} = 2$, H is normal in G. Since G has 8 elements of order 3 not all of them are contained in H so we pick $\sigma$ which is not in H. $K = \langle \sigma \rangle$ is a subgroup of G.

Since H is normal, HK is a subgroup of G and we have the important formula: $|HK| = \frac{|H||K|}{|H \cap K|}$ giving us that HK is a subgroup of G of order 18 but this contradicts the LaGrange's Theorem and so such an H cannot exist.