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Math Help - isomorphic groups

  1. #1
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    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.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Biscaim View Post
    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.

    Thanks in advance.
    every group of order n is isomorphic to a subgroup of S_n. (Cayley's theorem)
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    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.
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  4. #4
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    Quote Originally Posted by Biscaim View Post

    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!!
    Last edited by NonCommAlg; April 14th 2009 at 09:49 PM.
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  5. #5
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    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 .
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