If a group mod its center is cyclic, then the group is abelian

Let G be a group. Prove that if G/Z(G) is cyclic then G is abelian.

Let G be a group and suppose G/Z(G) = \langle x Z(G) \rangle is cyclic.

Note that for all g \in G, we have g \in g Z(G) = x^k Z(G) for some integer k. In particular, g = x^k z for some integer k and some z \in Z(G).

Now let g,h \in G, where g = x^a z and h = x^b w and z,w \in Z(G). We have gh = x^azx^bw = x^{a+b}zw = x^{b+a}wz = x^bwx^az = hg. Thus G is abelian.

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  • no  On February 23, 2011 at 12:29 pm

    If G is abelian then the center Z of G is all of G so G/Z is the trivial group which is indeed cyclic.

    • nbloomf  On February 23, 2011 at 3:31 pm

      True, but here we are looking at the converse question. Given that G/Z(G) is cyclic, must G be abelian?

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