# isomorphic groups

• Apr 14th 2009, 03:23 PM
Biscaim
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.

• Apr 14th 2009, 03:41 PM
NonCommAlg
Quote:

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)
• Apr 14th 2009, 04:09 PM
Biscaim
Quote:

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.
• Apr 14th 2009, 06:59 PM
NonCommAlg
Quote:

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!! (Nod)
• Apr 14th 2009, 09:26 PM
ThePerfectHacker
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 (Puke).