Finite Number of Groups of Given Finite Order

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then there exists a finite number of types of group of order $n$.

Proof
For any group $\left({G, \circ}\right)$ of order $n$ and any set of $n$ elements, $X$ can be the underlying set of a group which is isomorphic to $\left({G, \circ}\right)$, as follows:

Choose a bijection $\phi: G \to X$.

Define the group product $*$ on $X$ by the rule:
 * $\phi \left({g_1}\right) * \phi \left({g_2}\right) = \phi \left({g_1 \circ g_2}\right)$

for all $g_1, g_2 \in G$.

By definition, $\phi$ is an isomorphism.

From Isomorphism Preserves Groups it follows that $\left({X, *}\right)$ is a group.

Thus all groups of order $n$ of all possible types appear among all possible arrangements of binary operation.

But from Count of Binary Operations on Set, this count is $n^{\left({n^2}\right)}$.

This is the upper bound to the number of types of group of order $n$.