# Prove G is abelian

• May 4th 2010, 04:15 PM
MissMousey
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 $$ 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...
• May 4th 2010, 04:23 PM
Drexel28
Quote:

Originally Posted by MissMousey
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 $$ 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 $xy=\cdots$