Explore BrainMass

Explore BrainMass

    Subgroup proofsno

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    I would love to have a usual word format answer with detailed proofs.
    The problems are 6 in total, the numbers are: from attached page #1: #2.82; 2.86; 2.87, from attached page #2: # 2.97; 2.98; 2.99. Thanks in advance.
    The student.

    © BrainMass Inc. brainmass.com March 4, 2021, 8:32 pm ad1c9bdddf


    Solution Preview

    Please see the attachment.

    Page #1
    Problem 2.82
    First, I show that is a subgroup. For any , we have and for any , then we have

    Thus . So is a subgroup.
    (a) I show that .
    For any , we have , then for any , we can find some , such that . Now for any , we have

    Therefore, . Thus .
    (b) We consider the map defined as , where for any . Then we have

    Thus . So is a group homomorphism.
    If , then .
    If , then .
    Therefore, .
    By the group homomorphism theorem, we have . Since is a subgroup of and , then is isomorphic to a subgroup of .

    Problem 2.86
    (a) " ": If is a maximal normal subgroup of , I claim that is simple. If not, we can find some proper normal subgroup in , then we have . This means that is not maximal. We get a contradiction. Therefore, is simple.
    " ": If is simple, I claim that is a maximal normal subgroup of . If not, we can find some proper normal subgroup of , such that , then is a proper subgroup of . ...

    Solution Summary

    This is a series of proofs regarding groups and subgroups.