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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. –  user641 Jan 16 '13 at 12:15

2 Answers 2

up vote 5 down vote accepted

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! –  Geoff Robinson Dec 27 '12 at 7:51
What's $\Phi(G)?$ –  Ivan Dec 27 '12 at 7:56
@Ivan That's the Frattini subgroup of $G$. –  Alexander Gruber Dec 27 '12 at 8:02

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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