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I am trying to solve the exercise 2.13 in Isaacs' Character Theory Book. However I met some difficulties, let me sketch out what I am thinking so that you may tell me a hint.

The problem 2.13 is stated as follows : Let $|G'|=p$, a prime. Assume that $G'\subseteq Z(G)$. Show that $\chi(1)^2=|G:Z(G)|$ for every nonlinear $\chi\in Irr(G)$.

I proceed as follows : Let $\chi$ be a nonlinear character of $G$. Since $G'\subseteq Z(G)=\cap Z(\chi)$ then $G'\subseteq Z(\chi)$. Therefore $G/Z(\chi)$ is abelian. By THeorem 2.31, we have that $$\chi(1)^2=|G:Z(\chi)| $$ We would like to prove that $Z(G)=Z(\chi)$. We just need to prove the converse. Assume that $g\in Z(\chi)$, we need to prove that $g\in Z(G)$,i.e., commutes with all $h\in G$. Let $h\in G$, then we need to prove that $[g,h]=1$. Since $[g,h]\in G'\subset Z(\chi)$, then $[g,h]^p=1$ since $|G'|=p$. Moreover, $g\in Z(\chi)$ implies that $[g,h]\in Ker\chi$, and so $\chi([g,h])=\chi(1)>1$, since $\chi$ is nonlinear.

But then, at this stage, I dont know how to proceed to get $[g,h]=1$. Could you give me some hints then.

Thanks a lot in advance.

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    $\begingroup$ You assume a lot of notation that isn't necessarily used by everyone who could help you. $\endgroup$
    – Phira
    Commented Oct 11, 2012 at 9:42
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    $\begingroup$ the notation is pretty standard within group/character theory and it is also consistent with the one used by Isaacs in his book. $\endgroup$
    – DonAntonio
    Commented Oct 11, 2012 at 10:32
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    $\begingroup$ @john: $Z(\chi) = \{ g \in G : |\chi(g)|=\chi(1) \}$. A character $\chi$ is nonlinear iff $\chi(1) > 1$ $\endgroup$
    – Jack Schmidt
    Commented Oct 11, 2012 at 17:02
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    $\begingroup$ @9999: Since $\chi$ is nonlinear, $G'$ is not a subset of the kernel, so $\ker(\chi) \cap G' = 1$ since $G'$ is so small. Hence $[g,h]=1$ as in your argument. $\endgroup$
    – Jack Schmidt
    Commented Oct 11, 2012 at 17:05
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    $\begingroup$ @JackSchmidt You might consider writing an answer: your comment solved the problem. $\endgroup$
    – awllower
    Commented Mar 6, 2013 at 14:53

1 Answer 1

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Since $\chi$ is nonlinear, $G'$ is not a subset of the kernel, so $\ker(\chi) \cap G' = 1$ since $|G'|=p$ is so small. Hence $[g,h]=1$ and your argument is complete.

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