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?

  • $\begingroup$ My naive first thought would be that if $\chi:G\to\mathbb{C}$ is the character for the representation $\rho:G\to\mathrm{GL}(V)$, then the function $\chi^m:G\to \mathbb{C}$ is the character for the "representation" $\rho\circ \varphi$, where $\varphi:G\to G$ is the $m$th power map. Except, of course, $\varphi$ is not going to be a homomorphism in general. Perhaps something could still be made of this? $\endgroup$ Feb 20 '14 at 6:35
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    $\begingroup$ $\chi(g^m)=\sigma\chi(g)$ where $\sigma\in{\rm Gal}(\Bbb Q(\zeta_n)/\Bbb Q)$ extends $\zeta\mapsto\zeta^m$. One wants to show one can write $\rho$ in coordinates on which $\sigma$ can act to get a representation with character $\sigma\chi$. $\endgroup$
    – anon
    Feb 20 '14 at 7:34
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    $\begingroup$ Once you have shown it is a character, checking whether it is irreducible is easier by just looking at $\langle\chi^m,\chi^m\rangle$ and comparing this to the same for $\chi$ (since it will be $1$ iff the character is irreducible). $\endgroup$ Feb 20 '14 at 8:14
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    $\begingroup$ $\langle \chi^m,\chi^m \rangle = 1$ follows from the fact that $g \mapsto g^m$ is a bijection of $G$. $\endgroup$
    – Derek Holt
    Feb 20 '14 at 8:43

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.

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    $\begingroup$ @LeonLampret: I think this follows with some Field/Galois theory, unless I've made some mistake.. Any character is afforded by a rep over the algebraic integers, so assume $S$ is a finite set of algebraic integers. So there is a Galois extension $K / \mathbb{Q}(\zeta)$ such that $K$ contains $S$. Now there is the following standard lemma (Thm 4.4 in Jacobson, Basic Algebra I): if $E$ is a splitting field over $F$ and $\sigma: F \rightarrow F$ is a field isomorphism, then $\sigma$ can be extended to a field isomorphism $E \rightarrow E$. $\endgroup$
    – spin
    Feb 20 '14 at 17:53
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    $\begingroup$ Okay, I looked it up and extension to $\mathbb{C}$ is possible. See Theorem 111 (Automorphism Extension Theorem) in these notes by Pete Clark: math.uga.edu/~pete/FieldTheory.pdf $\endgroup$
    – spin
    Feb 20 '14 at 18:14
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    $\begingroup$ @LeonLampret: There is an explanation in Isaacs' character theory book, pg. 22. The irreducible characters of $G$ over $\mathbb{C}$ are exactly the same as irreducible characters of $G$ over the algebraic integers $E$. The idea is that you could construct character theory over $E$ just like you do over $\mathbb{C}$, and you get the same results. An character $\chi$ of an representation of $G$ over $E$ is irreducible if and only if $\langle \chi, \chi \rangle = 1$. So $\operatorname{Irr}_E(G) \subseteq \operatorname{Irr}_\mathbb{C}(G)$. $\endgroup$
    – spin
    Feb 20 '14 at 19:40
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    $\begingroup$ By Isaacs p.22 our $\chi$ is the character of some $\rho\!: G\to GL_d(\overline{\mathbb{Q}})$ and $\mathbb{Q}\leq\mathbb{Q}(\epsilon)\leq\overline{\mathbb{Q}}$. By Grillet Abstract Algebra 4.4.7 $\sigma$ extends to an automorphism $\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$ and also $GL_d(\overline{\mathbb{Q}})\to GL_d(\overline{\mathbb{Q}})$. Then $\rho^\sigma\!: g\!\mapsto\!\sigma(\rho(g))$ has character $\chi^\sigma$. Yes? $\endgroup$
    – Leo
    Feb 21 '14 at 20:15
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    $\begingroup$ By the way, this is related. If $G$ is a finite group of exponent $n$, then it is a 1945 result of Brauer (JSTOR) that any complex irreducible character of $G$ can be realized over $\mathbb{Q}(\zeta)$, where $\zeta$ is an $n$th root of unity. This gives you the result without any need for this extension stuff, but then again the proof of Brauer's result is not so easy (I think you need some modular character theory). $\endgroup$
    – spin
    Feb 21 '14 at 21:44

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