TITLE: Order of the smallest group containing all groups of order $n$ as subgroups.
QUESTION [17 upvotes]: Let $n\in \Bbb N$ be fixed and $m\in \Bbb N$ be the least number such that there exists a group of order $m$ in which all groups of order $n$ can be (isomorphically) embedded.
Can we deduce $n!=m$?

REPLY [7 votes]: For a counterexample outside $n=p$ prime (where obviously $m=p$, since every order $p$ group is isomorphic to $\mathbb{Z}/p\mathbb{Z}$), consider $n=15$.
Then every order $15$ group is Abelian and is actually isomorphic to $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}=\mathbb{Z}/15\mathbb{Z}$.
So $m=15$ is the answer in this case.
Edit: of course, each time there is only one group of order $n$, the answer is $m=n$. See here for the beginning of a list of these cases.