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Let $n>0$ and assume that $\chi^{(n)} \in \mathrm{Irr}(G)$ for every $\chi \in \mathrm{Irr}(G)$. Show that $G = H \times A$, where $A$ is abelian and $(|H|, n) = 1$.

$\chi^{(n)}$ is defined by $\chi^{(n)}(g) = \chi(g^n)$. $\mathrm{Irr}(G)$ is the set of all irreducible characters of $G$.

Here are the hints in the book:

  1. Let $d = (|G|,n)$. Show that it is no loss to assume that $(|G|/d,n) = 1$.

  2. Let $A = \bigcap_{\chi \in \mathrm{Irr}(G)} \mathrm{ker} \chi^{(n)}$. Show that $A = \{g \in G | g^n=1\}$ and $|A| = d$.

  3. Let $H = \bigcap \{\mathrm{ker} \chi | \chi \in \mathrm{Irr}(G), \chi^{(n)} = 1_G\}$. Show $|G : H| = d$.

Hint 1 and 2 are easy, and it is easy to deduce the result from the hints. As for hint 3, I have proven that $\{ g \in G | (o(g), n) = 1 \} \subseteq H$, and that $|G : H| \mid d$. And I am stuck here.

Can anyone help me?

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This seems difficult! I think an argument along the following lines might work. Unfortunately it involves induced characters, which are covered only in Chapter 5 of Isaacs' book, so it cannot be the intended solution.

It is enough to prove that $|H \cap A|=1$.

Since $(|A|,|G|/|A|)=1$, by the Schur-Zassnhaus Theorem $A$ has a complement $C$ in $G$, The irreducible characters $\chi^{(n)}$ with $\chi \in {\rm Irr}(G)$ have $A$ in their kernels, so they correspond to irreducible characters of $G/A$. Since $C \cong G/A$, the character $\chi^{(n)}_C$ corresponds to $\chi^{(n)}$ on $G/A$ and hence is an irreducible character of $C$.

Consider the induced character $1_C^G$ and let $\chi$ be an irreducible constituent of it. Then by Frobenius reciprocity, $1_C$ is a constituent of $\chi_C$. Now, since $(|C|,n)=1$, for $\psi \in {\rm Irr}(C)$, we have $\psi^{(n)} \in {\rm Irr}(C)$, so $1_C$ is also a constituent of $\chi^{(n)}_C$. But $\chi^{(n)}_C$ is irreducible, so $\chi^{(n)}_C = 1_C$ and hence $\chi^{(n)} = 1_G$.

Since no nontrivial element of $A$ is in the kernel of $1_C^G$, for each $1 \ne g \in A$, there is a constituent $\chi$ of $1_C^G$ with $g \not\in \ker \chi$, and hence $g \not\in H$, so $H \cap A = 1$ as claimed. (So in fact $C = H$.)

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