The G/Z Theorem

Theorem: Let G be a group and let Z(G) be the center of G. If G/Z(G) is cyclic, then G is Abelian.

Proof: Suppose that G/Z(G) is cyclic. Then there is some aZ(G)\in G/Z(G) such that \left<aZ(G)\right>=G/Z(G).

Let x,y\in G. We wish to show that xy=yx. It follows that there exist integers j,k such that xZ(G)=(aZ(G))^j and yZ(G)=(aZ(G))^k. So there exist z_1,z_2\in Z(G) such that x=a^jz_1 and y=a^kz_2. Now consider their product:

xy=(a^jz_1)(a^kz_2)

=a^j(z_1a^k)z_2

=(a^ja^k)(z_1z_2)

=(a^ka^j)(z_2z_1)

=(a^kz_2)(a^jz_1)=yx

Note that in the third line above we are able to commute a^k and z_1 since z_1\in Z(G). Similarly we are able to commute from line four to five.

Thus, we have shown that G is Abelian.

\Box

Reflection: It is necessary that your subgroup is the center of G, otherwise we wouldn’t be able to commute and conclude that G is Abelian. The proof itself followed mainly from definitions and helpful manipulation.

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This entry was posted in Abelian, Algebra, Center, Cyclic, Group. Bookmark the permalink.

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