1. ## isomorphic groups

If isomorphic groups are regarded as being the same, prove, for each positive integer $n$, that there are only finitely many distinct groups with exactly $n$ elements.

2. Originally Posted by Biscaim
If isomorphic groups are regarded as being the same, prove, for each positive integer $n$, that there are only finitely many distinct groups with exactly $n$ elements.

every group of order $n$ is isomorphic to a subgroup of $S_n.$ (Cayley's theorem)

3. Originally Posted by NonCommAlg
every group of order $n$ is isomorphic to a subgroup of $S_n.$ (Cayley's theorem)
OK. This is the exercise 1.41 (page 18) of An Introduction to the Theory of Groups by Rotman. The Cayley's Theorem is in the page 52. I think that must be another way to prove this exercise.

Thanks again.

4. Originally Posted by Biscaim

OK. This is the exercise 1.41 (page 18) of An Introduction to the Theory of Groups by Rotman. The Cayley's Theorem is in the page 52. I think that must be another way to prove this exercise.

Thanks again.
another way: let $A=\{1,2, \cdots , n \}$ and let $G=\{g_1, g_2, \cdots , g_n\}$ be a group of order $n.$ define multiplication $*$ in $A$ by $i*j=k,$ where $g_ig_j=g_k.$ then $(A,*)$ is a group and $G \cong (A, *).$

so every group of order $n$ defines a map $*: A \times A \longrightarrow A.$ note that if two groups $G_1,G_2$ define the same $*$ on $A,$ then $G_1 \cong (A,*) \cong G_2.$ thus the number of groups of order $n$ is at

most the number of maps $*$ which can be defined from $A \times A$ to $A,$ which is $n^{n^2}.$

an easier way to get this upper bound (this just came to my mind!) is to look at the $n \times n$ multiplication table of a group of order $n.$ each of $n^2$ entries have at most $n$ pssibilities!!

5. Another (not so nice) approach.
Given a finite set $G = \{g_1,...,g_n\}$ we need to define $*:G\times G\to G$.
Therefore, $*\subseteq \mathcal{P}((G\times G)\times G)$.
Thus, I get $\left( n^3\right)!$ as an upper bound.
Which is a horrible upper bound .