non-isomorphic groups of given orderNonIsomorphicGroupsOfGivenOrder2009-05-25 14:40:272009-05-25 14:45:43Theoremorder (of a group)Landau's theorem% this is the default PlanetMath preamble. as your knowledge
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\textbf{Theorem.}\, For every positive integer $n$, there exists only a finite amount of non-isomorphic groups of order $n$.\\
This assertion follows from Cayley's theorem, according to which any group of order $n$ is isomorphic with a subgroup of the symmetric group $\mathfrak{S}_n$.\, The number of non-isomorphic subgroups of $\mathfrak{S}_n$ cannot be greater than
$${n!\!-\!1 \choose n\!-\!1}.$$\\
The above theorem may be used in proving the following Landau's theorem:
\textbf{Theorem (Landau).}\, For every positive integer $n$, there exists only a finite amount of finite non-isomorphic groups which contain exactly $n$ conjugacy classes of elements.\\
One needs also the
\textbf{Lemma.}\, If\, $n \in \mathbb{Z}_+$\, and\, $0 < r \in \mathbb{R}$,\, then there is at most a finite amount of the vectors \,$(m_1,\,m_2,\,\ldots,\,m_n)$\, consisting of positive integers such that
$$\sum_{j=1}^n\frac{1}{m_j} \;=\; r.$$
The lemma is easily proved by induction on $n$.