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    Proof that There Are No Simple Groups of Order 56 or 148

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    A simple group is a group with nontrivial normal subgroups. Show that there are no simple groups of order 148 or of order 56.

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    Solution Preview

    First suppose |G| = 56 = 2^3 * 7. We know from Sylow's third theorem that the number of Sylow-2 subgroups of G is congruent to 1 modulo 2 and divides 7, whence there is are either one or seven Sylow-2 subgroups of G, We also know that the number of Sylow-7 subgroups of G is congruent to 1 modulo 7 and ...

    Solution Summary

    In this solution, we use Sylow's theorems to prove that no simple groups exist of order 56 or 148.

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