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**MissMousey** 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...