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Can anyone help me with this problem?

if $G$ has abelian Sylow-p-subgroups, prove that $p$ does not divide the order of $G'\cap Z(G)$, where $G'$ and $Z(G)$ are as usual, the subgrup generated by the set of all commutators and the center, respectively.

thanks a lot. :D

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Taunt's paper on A-groups also has a proof. It is presented in textbook form as 10.1.6 of Robinson's book for a Course in the Theory of Groups, page 280 or so. –  Jack Schmidt Apr 20 '10 at 12:01

2 Answers 2

This can be proved using the transfer homomorphism, which is treated in many texts. A more general (and well-known) fact is that if the finite group G has a Sylow p-subgroup P, then $P \cap G' \cap Z(G) \leq P'$, which answers your question when P is Abelian. The focal subgroup theorem of D.G. Higman is also helpful here.

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There is also a character-theoretic argument. Suppose $G' \cap Z(G)$ has a subgroup $U$ of order $p$. We want a contradiction. Let $\lambda$ be a nonprinciipal linear character of $U$. Since $U \subseteq P$ and $P$ is abelian, $\lambda$ has an extension to $\mu$, a linear character of $P$. The induced character $\mu^G$ has degree $|G:U|$, which is prime to $p$, so some irreducible constituent $\chi$ of $\mu^G$ has degree not divisible by $p$. Then $\mu$ is a constituent of the restriction $\chi_P$ by Frobenius reciprocity, and thus $\lambda$ is a constituent of $\chi_U$. But $U$ is central, so $\chi_U = \chi(1)\lambda$. Now let $\sigma$ be the linear character det$(\chi)$. Then $\sigma_U = \lambda^{\chi(1)}$, which is nontrivial since $p$ does not divide $\chi(1)$. This is a contradiction, however, since $U \subseteq G' \subseteq {\rm ker}(\sigma)$. [Note that transfer proofs can often be replaced by arguments using the determinant of a character.]

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