Let $G$ be a non-trivial finite group with order $n$ and let $p$ be the smallest prime factor of $n$.

Can the order of the center be $\frac{n}{p}$ ?

I did not find an example with GAP until $n=255$. I thought of using the fact that every subgroup of $G$ with order $\frac{n}{p}$ is normal, but I do not see how I can show this way that a group with at least $\frac{n}{p}$ central elements must be abelian, which would show that there is no group with the desired property.

  • 2
    $\begingroup$ No, if $G/Z(G)$ is cyclic, then $G$ is abelian. So $\lvert G/Z(G)\rvert$ can never be a prime. $\endgroup$ – Daniel Fischer Feb 15 '16 at 21:52
  • $\begingroup$ I just had this idea. Thanks anyway for your comment :) $\endgroup$ – Peter Feb 15 '16 at 21:53

No, the center can never have prime index, because $Z(G)$ is normal. Thus $G/Z(G)$ would be a cyclic group. Suppose $gZ(G)$ is a generator of $G/Z(G)$. Then every element in $G$ can be written as $g^nx=xg^n$ with $x\in Z(G)$, but this implies every element commutes with $g$. So $g\in Z(G)$ and $G=Z(G)$ is abelian.


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