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I am a bit confused on where to start with this one. The question given was

Let $G$ be a non abelian finite group. Prove $|Z(G)| \leq \frac {1}{4} |G|$.

Am I supposed to use some property of centers or am I just blanking on something really basic

lulu
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Brandon Klein
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    Think about the quotient $G / Z(G)$. – pjs36 Oct 06 '15 at 01:11
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    This provides another good answer for my question http://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point. – lhf Oct 06 '15 at 01:14
  • Duplicate of http://math.stackexchange.com/questions/367184/g-is-a-nonabelian-finite-group-then-zg-leq-frac14g. – lhf Oct 06 '15 at 01:22

1 Answers1

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Hint: If $G/Z(G)$ is cyclic, then $G$ is abelian.

lhf
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  • Using this approach, $1/4$ is the best bound you can get of the form $1/n$. See http://math.stackexchange.com/questions/227962/a-group-with-order-exactly-4-times-the-order-of-its-center – lhf Oct 06 '15 at 01:18