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Let $G$ be a finite group and $G'$ the commutator group of $G$.

  • What can I say about $G' \cap Z(G)$?

Could you be as specific as possible about p-Groups?

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    $\begingroup$ This question lets me quote a result I really like, but I don't think I have ever seen used seriously. Let $n=[G:G'\cap Z(G)]$. Then $g^n=1$ for all $g\in G$. This is the last result in Isaacs's Finite Group Theory. $\endgroup$
    – user641
    Jan 16, 2013 at 12:15

2 Answers 2

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Well, this is sort of a broad question, but here a couple random facts.

  • $G'\cap Z(G)\leqslant \Phi(G)$ for all groups $G$.

  • Special $p$-groups are an interesting case of $p$-groups in which $G'\cap Z(G)= Z(G)=G'$.

There is also some kind of theorem about $[G:G'\cap Z(G)]^2$ dividing the degrees of a group's irreducible characters, which I believe I saw in Huppert, but I cannot remember the exact statement at the moment.

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    $\begingroup$ What's $\Phi(G)?$ $\endgroup$
    – Ivan Di Liberti
    Dec 27, 2012 at 7:56
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    $\begingroup$ @Ivan That's the Frattini subgroup of $G$. $\endgroup$
    – Alexander Gruber ♦
    Dec 27, 2012 at 8:02
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    $\begingroup$ @user Sure. If not, let $H$ be a maximal subgroup not containing $G^\prime \cap Z(G)=K$. Then every element of $G$ has the form $g=hk$, so $H^g=H^{hk}=H^k=H\unlhd G$. But then $G/H$ is cyclic of prime order, so $G^\prime\unlhd H$. $\endgroup$
    – Alexander Gruber ♦
    Sep 10, 2015 at 2:03
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    $\begingroup$ Since $K$ is not contained in $H$, $HK$ must be bigger than just $H$. $H$ is a maximal subgroup, though, so the only bigger subgroup of $G$ is $G$ itself. Therefore, $HK$ must be all of $G$. So, any element of $G$ can be written as an element of $HK$, and thus in the form $hk$ for some $h\in H$, $k\in K$. $\endgroup$
    – Alexander Gruber ♦
    Sep 11, 2015 at 6:56
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    $\begingroup$ $H^g$ is group theorist shorthand for $g^{-1}Hg$. $H\unlhd G$ ("$H$ is normal in $G$") because we showed that $H^g=H$. $\endgroup$
    – Alexander Gruber ♦
    Sep 11, 2015 at 6:58
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There is a well-known theorem (I think due to Grün) which asserts that if $P$ is a Sylow $p$-subgroup of a finite group $G,$ then $P \cap G^{\prime} \cap Z(G) \leq P^{\prime}.$ If $G$ itself is a $p$-group, then I am not sure how much you can expect to say. In that case, if $G$ is non-Abelian, then $G^{\prime} \cap Z(G)$ is always non-trivial, but it may well have order $p.$ On the other hand, when $p$ is odd, there is always an $n$-generator finite $p$-group $G$ of nilpotent class $2$ and exponent $p$ such that $G^{\prime} = Z(G)$ is elementary Abelian of order $p^{\frac{n(n-1)}{2}}$ and $[G:G^{\prime}] = p^{n}.$

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