Explore BrainMass

Explore BrainMass

    abelian subsets

    Not what you're looking for? Search our solutions OR ask your own Custom question.

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

    Let G be a group and S any subset of G. Prove that C_G (S) = {g in G such that gs = sg for all s in S} is a subgroup of G. Prove that Z (G) (center of G) = C_G (S) is abelian and is a normal subgroup of G.

    © BrainMass Inc. brainmass.com November 30, 2021, 5:04 am ad1c9bdddf
    https://brainmass.com/math/group-theory/abelian-subsets-422475

    Solution Preview

    Proof:
    First, I show that C_G(S) is a subgroup of G.
    We consider any g, h in C_G(S), then for any s in S, we have
    gs = sg and hs = sh
    Then hsh^(-1) = s and thus sh^(-1) = ...

    Solution Summary

    This solution proves that Z (G) (center of G) = C_G (S) is abelian and is a normal subgroup of G.

    $2.49

    ADVERTISEMENT