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

Thanks in advance.

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- Apr 14th 2009, 03:23 PMBiscaimisomorphic groups
If isomorphic groups are regarded as being the same, prove, for each positive integer $\displaystyle n$, that there are only finitely many distinct groups with exactly $\displaystyle n$ elements.

Thanks in advance. - Apr 14th 2009, 03:41 PMNonCommAlg
- Apr 14th 2009, 04:09 PMBiscaim
- Apr 14th 2009, 06:59 PMNonCommAlg
another way: let $\displaystyle A=\{1,2, \cdots , n \}$ and let $\displaystyle G=\{g_1, g_2, \cdots , g_n\}$ be a group of order $\displaystyle n.$ define multiplication $\displaystyle *$ in $\displaystyle A$ by $\displaystyle i*j=k,$ where $\displaystyle g_ig_j=g_k.$ then $\displaystyle (A,*)$ is a group and $\displaystyle G \cong (A, *).$

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

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

an easier way to get this upper bound (this just came to my mind!) is to look at the $\displaystyle n \times n$ multiplication table of a group of order $\displaystyle n.$ each of $\displaystyle n^2$ entries have at most $\displaystyle n$ pssibilities!! (Nod) - Apr 14th 2009, 09:26 PMThePerfectHacker
Another (not so nice) approach.

Given a finite set $\displaystyle G = \{g_1,...,g_n\}$ we need to define $\displaystyle *:G\times G\to G$.

Therefore, $\displaystyle *\subseteq \mathcal{P}((G\times G)\times G)$.

Thus, I get $\displaystyle \left( n^3\right)! $ as an upper bound.

Which is a horrible upper bound (Puke).