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Math Help - Prove G is abelian

  1. #1
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    Prove G is abelian

    Let G be a group with center Z(G) = C. Prove that if G/C is cyclic, then G is abelian.

    C is defined as { a \in G | ax = xa for all x\in G}

    - If I assume G/C is cyclic, then G/C is generated by <gC> for any g \in G.
    - I also know that C is a normal subgroup of G from an earlier proof.
    - G/C is the set of all distinct cosets of C in G.
    - Since C is a subgroup of G, aC = C = Ca for any  a \in C .
    - Also, since C is a normal subgroup of G, gC = Cg for all  g \in G .

    Mmmmm...I think I'm working with too much...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MissMousey View Post
    Let G be a group with center Z(G) = C. Prove that if G/C is cyclic, then G is abelian.

    C is defined as { a \in G | ax = xa for all x\in G}

    - If I assume G/C is cyclic, then G/C is generated by <gC> for any g \in G.
    - I also know that C is a normal subgroup of G from an earlier proof.
    - G/C is the set of all distinct cosets of C in G.
    - Since C is a subgroup of G, aC = C = Ca for any  a \in C .
    - Also, since C is a normal subgroup of G, gC = Cg for all  g \in G .

    Mmmmm...I think I'm working with too much...
    Hint:

    Spoiler:


    Let x,y\in G then x\mathcal{Z}(G),y\mathcal{Z}(G)\in G/\mathcal{Z}(G) but by assumption G/\mathcal{Z}(G)=\left\langle g\mathcal{Z}(G)\right\rangle and so x\mathcal{Z}(G)=g^n\mathcal{Z}(G),y\mathcal{Z}(G)=  g^m\mathcal{Z}(G) and so x=g^nz,y=g^mz' for some z,z'\in\mathcal{Z}(G) and so xy=\cdots

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