Problem 4.6, I. Martin Isaacs' Character Theory

Question

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?

Answer 1

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 $C = H$.)

Answer 2

Here is a more direct solution, which is likely the intended one. For the sake of completeness, I also include solutions to (a) and (b).

(a) Just replace $n$ by $n^k$ for sufficiently large $k$.

(b) It's clear that $A=\{g\in G\mid g^n\in\bigcap_{\chi\in\operatorname{Irr}(G)}\chi=1\}$. In particular, by Cauchy's theorem, $|A|$ has the same prime factors as $n$. Moreover, $|A|$ is coprime with $|G|/|A|$ since, if $p\mid|G/A|,n$ and $gA\in G/A$ is an element of order $p$, $g$ is of order $k$ dividing $pn$ but not $n$. This means that $v_p(k)=v_p(n)+1$, which is impossible since $k\mid|G|$. It is easy to see that this implies $|A|=d$.

(c) So far we have only used that the $\chi^{(n)}$ were characters and not that they were irreducible ones. Clearly, if $\chi^{(n)}=1$, $\chi$ is an irreducible character of $G/H$. Conversely, since $H$ contains all $n$th powers, if $\chi$ is an irreducible character of $G/H$, i.e. $H\subseteq\ker\chi$, we must have $\chi^{(n)}=\chi(1)$. Since $\chi^{(n)}$ is irreducible, this means $\chi^{(n)}=1$.

Thus, we conclude that $\operatorname{Irr}(G/H)=\{\chi\in\operatorname{Irr}(g)\mid\chi^{(n)}=1\}$. In particular, $G/H$ is abelian since all its characters are linear. Finally, we prove that $\{\chi\in\operatorname{Irr}(g)\mid\chi^{(n)}=1\}$ has cardinality $d=|A|$. Recall that the class function $\theta_n(g)=|\{h\in G\mid h^n=g\}|$ can be written as $\sum_{\chi\in\operatorname{Irr}(G)}\nu_n(\chi)\chi$ where $\nu_n(\chi)=[\chi^{(n)},1]$. Since the $\chi^{(n)}$ are all irreducible (this is where we really use the hypothesis), this is $1$ when $\chi^{(n)}=1$ and $0$ otherwise. We conclude that $$|A|=\theta_n(1)=\sum_{\chi^{(n)}=1}\chi(1)=|\{\chi\in\operatorname{Irr}(G)\mid\chi^{(n)}=1\}=|\widehat{G/H}|=|G/H|$$ as wanted.

To finish, $A$ and $H$ have coprime cardinality so $A\cap H=1$, and $|A|\cdot|H|=|G|$ so $AH=G$. Since they are both normal in $G$, it follows that $G\simeq H\times A$. Moreover, $A\simeq G/H$ is indeed abelian.