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

C is defined as { | ax = xa for all }

- If I assume G/C is cyclic, then G/C is generated by for any .

- 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 .

- Also, since C is a normal subgroup of G, gC = Cg for all .

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