$\chi(g)$ group character $\Rightarrow$ $\chi(g^m)$ group character

Question

Let $G$ be a group of order $n$ and and $\gcd(m,n)=1$. Let $\chi:G\rightarrow\mathbb{C}$ be a class function and define $\chi^m\!: g\mapsto\chi(g^m)$. How can one show that $\chi^m$ is a character iff $\chi$ is a character and that $\chi^m$ is irreducible iff $\chi$ is irreducible?

It is already a class function, so it is a $\mathbb{C}$-linear combination of irreducible characters $\chi_1,\ldots,\chi_r$. Thus it suffices to show that for $\langle\chi,\chi_i\rangle=\frac{1}{n}\sum_{g\in G}\chi(g)\chi_i(g^{-1})$ we have:

1. $\forall i\!: \langle\chi,\chi_i\rangle\in\mathbb{N}$ iff $\forall i\!: \langle\chi^m,\chi_i\rangle\in\mathbb{N}$;
2. $\forall i\!: \langle\chi,\chi_i\rangle\in\{0,1\}$ iff $\forall i\!: \langle\chi^m,\chi_i\rangle\in\{0,1\}$.

Since $\gcd(m,n)=1$, there exist $k,l$ with $km+ln=1$, so the map $G\to G,g\mapsto g^m$ is surjective ($g=g^{km+ln}=g^{km}=(g^{k})^{m}$) hence bijective (since $G$ is finite) and permutes conjugacy classes.

Any suggestions?

Answer 1

Let $\rho: G \rightarrow \operatorname{GL}_r(\mathbb{C})$ be a representation affording the character $\chi$.

As anon notes in the comments, since $(m,n) = 1$, there exists $\sigma\in{\rm Gal}(\Bbb Q(\zeta)/\Bbb Q)$ such that $\sigma(\zeta) = \zeta^m$, where $\zeta$ is a primitive $n$th root of unity. We can further extend $\sigma$ into a field automorphism of a field that contains each entry of $\rho(g)$ for all $g \in G$. From this, we can define a homomorphism $\sigma: \operatorname{Im}(\rho) \rightarrow \operatorname{GL}_r(\mathbb{C})$ by $\sigma(A)_{ij} = \sigma(A_{ij})$ for all $A \in \operatorname{GL}_r(\mathbb{C})$.

Then $\sigma \circ \rho$ is a representation that affords the character $\chi^m$.

This proves that $\chi^m$ is a character when $\chi$ is a character, and the converse follows from the same result.

Since $g \mapsto g^m$ is a bijection $G \rightarrow G$, we get that $\langle \chi, \chi \rangle = 1$ if and only if $\langle \chi^m, \chi^m \rangle = 1$. In other words, $\chi$ is an irreducible character if and only if $\chi^m$ is an irreducible character.