Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 1

Proof
Proof by Counterexample:

Consider $S_5$, the symmetric group on $5$ letters.

By Order of Symmetric Group, $\order {S_5} = 5! = 120$.

We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$.

However, Symmetric Group on 5 Letters has no Subgroup of Order 15.