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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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    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*.2013-01-16

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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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    Perhaps you are thinking of the theorem that if $G$ is a finite nilpotent group, then for each complex irreducible character $\chi$ of $G,$ we have $\chi(1)^{2}$ divides $[G:Z(G)]$-or, then again, perhaps not!2012-12-27
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    What's $\Phi(G)?$2012-12-27
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    @Ivan That's the Frattini subgroup of $G$.2012-12-27
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    @AlexanderGruber. I'm interesting in your statement that $G'\cap Z(G)\leqslant \Phi(G)$. Would you mind to provide the proof for this?2015-09-08
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    @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 $G^\prime\unlhd H$.2015-09-10
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    Thanks @AlexanderGruber. I'm just beginner in Group theory. Can you explain why every element of $G$ has the form $g=hk$? What does your notation $H^g$ mean? Why $H \unlhd G$ then $G/H$ is cyclic of prime order? Why $G' \unlhd H$? Can you suggest me some materials e.g. textbooks, websites,... Thanks in advance.2015-09-10
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    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 $h\in H$, $k\in K$.2015-09-11
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    $H^g$ is group theorist shorthand for $g^{-1}Hg$. $H\unlhd G$ ("$H$ is normal in $G$") because we showed that $H^g=H$.2015-09-11
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    $G/H$ is cyclic of prime order because $G/H$ must be abelian (why?) and because the quotient of a group by a maximal subgroup must be simple. In any case, this is overkill, all we really need is for it to be abelian, because there is a result (prove this part yourself, it's easy) that $G/H$ is abelian if and only if $G^\prime\subset H$.2015-09-11
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    Far as textbooks go, the best progression I can recommend is Dummit and Foote followed by Isaacs Finite Group Theory. (I *really* like Isaacs.) As far as websites go, this one is really the best. Look up some of my old questions and answers, as well as Arturo Magidin, Jack Schmidt, Geoff Robinson. I learned a substantial portion of my group theory from studying their posts.2015-09-11
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    @AlexanderGruber. This is really helpful guidance. I understood all the steps of your proof.2015-09-14
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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 $[G:G^{\prime}] = p^{n}.$