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Thread: Prove G is abelian

  1. #1
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    Prove G is abelian

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

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

    - If I assume G/C is cyclic, then G/C is generated by $\displaystyle <gC>$ for any $\displaystyle g \in G$.
    - 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 $\displaystyle a \in C $.
    - Also, since C is a normal subgroup of G, gC = Cg for all $\displaystyle 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 $\displaystyle Z(G) = C$. Prove that if G/C is cyclic, then G is abelian.

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

    - If I assume G/C is cyclic, then G/C is generated by $\displaystyle <gC>$ for any $\displaystyle g \in G$.
    - 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 $\displaystyle a \in C $.
    - Also, since C is a normal subgroup of G, gC = Cg for all $\displaystyle g \in G $.

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

    Spoiler:


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

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