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I am trying to understand Theorem 6.9 in Isaac's Character Theory of finite groups:

Suppose $\chi(1)$ is a power of the prime $p$ for all irreducible characters $\chi$ of $G$. Then $G$ has a normal abelian $p$-complement. According to the proof, proceeding by induction on $|G| = p^km$, it suffices to show that $G$ has a normal subgroup $N$ of index $p$. The stated reasoning is that $N$ will have a normal abelian $p$-complement by the inductive hypothesis, and this same subgroup will be a normal abelian $p$-complement for $G$. However, I can't seem to prove that it is indeed normal in $G$.

Any ideas?

Fibonacci Cube K
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1 Answers1

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If $A$ is a normal abelian $p$-complement of $N$, then gcd$(|A|,|N:A|)=1$, implying that $A$ is characteristic in $N$. So $A$ char $N \lhd G$, hence $A \lhd G$. And $|G:A|$ is a $p$-power, $p \nmid |A|$, so at this point you could apply the theorem of Schur-Zassenhaus, providing a complement in $G$ to $A$. But you do not have to invoke that heavy theorem: $G/A$ is a $p$-group, so if $S \in Syl_p(G)$, then $SA/A=G/A$. Hence $G=SA$ and of course $S \cap A=1$.

Nicky Hekster
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