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    Prove that a group of order 108 must have a normal subgroup of order 9 or 27.

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    Prove that a group of order 108 must have a normal subgroup of order 9 or 27.

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    Modern Algebra
    Group Theory (CXIII)
    Groups of Order having Power of a Prime
    Another Counting Principle

    By:- Thokchom Sarojkumar Sinha

    Prove that a group of order 108 must have a normal subgroup of order 9 or 27.

    Solution:- Let be a group of order 108 . Then

    The number of 3-Sylow subgroups of is of the form and

    For
    which imply that

    If then there exists only one 3-sylow subgroup of , which is of order 27 and hence it is normal in .

    If , then then there exists ...

    Solution Summary

    It is proven that a group of order 108 must have a normal subgroup of order 9 or 27. The solution is detailed and well presented.

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