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Is there such a group? I just learned that for a non-abelian group $G$, the order of its center $Z$ is at most $1/4$ of the order of $G$, but I can't think of any group for which the equality hold. Could it be that the inequality is strict?

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The reason this is true is that if $G/Z$ were cyclic, it would be abelian, and we would have $G=Z$ (consider a generator of the cyclic group). So we can't have $G/Z=p$ for a prime $p$ because all groups of prime order are cyclic. The first non-cyclic candidate for $G/Z$ has order 4 - and with centre of order 2, the lowest possible order is 8. –  Mark Bennet Nov 3 '12 at 5:38

2 Answers 2

The non-abelian groups of order 8 give you two examples.

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The center of the dihedral group of order $8$ is cyclic of order $2$.

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And the same is true of the quaternion group of order $8$, by the way. –  Mariano Suárez-Alvarez Nov 3 '12 at 4:41

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