Let $G$ be finite abelian group and $K$ a field such that $char(K)$ does not divide the order $r$ of $G$. For each divisor $d$ of $r$ let $\omega_d$ be a primitive $d$-root of unity and $a_d:=\frac{\mid \{ a\in G \mid o(a)=d \} \mid}{dim_K(K(\omega_d))}$. By a theorem of Perlis and Walker (see e.g. Perlis/Walker) the group algebra $KG$ is isomorphic to $\bigoplus\limits_{d\mid r} K(\omega_d)^{a_d}$. Hence for each $d\mid r$ there are $a_d$ irreducible characters of dimension $dim_K(K(\omega_d))$.

My question is how these irreducible characters can be constructed? The background of my question is to compute the central idempotents $e_i:=\frac{\chi_i(1)}{r} \sum\limits_{g\in G} \chi_i(g^{-1})g$ by using the constructed irreducible characters $\chi_1,\cdots ,\chi_h$.


1 Answer 1


This is all fairly standard, but here goes. If $M$ is an irreducible $KG$-module, then (since $G$ is Abelian) we obtain a homomorphism $\theta: G \to {\rm End}_{KG}(M)^{\times}$, so $\theta(G)$ is a finite Abelian subgroup of the group of units of a division algebra (using Schur's Lemma). Hence ${\rm Im} \theta$ is cyclic. In other words, the problem is now reduced to proving the result in the case $G$ cyclic. Let us recalibrate notation, and assume that $|G| =r$ and $G = \langle g \rangle $ is cyclic. Let us consider out irreducible $KG$ module $M$ and $\theta$ as before. Let $\theta(g)$ have order $d$ ( we could actually assume that $d = r$ at this point, given what has gone before, but let us work in greater generality). Let $f(x) \in K[x]$ be the minimum polynomial of $\theta(g)$. Then $f(x)$ must be irreducible, for if $p(x) \in K[x]$ is an irreducible factor of $f(x)$, then $p(\theta(g))$ is not invertible, so must be the zero matrix, as it commutes with all of $\theta(G)$. Also, $f(x)$ divides $x^{d}-1$, since $\theta(g)$ has order $d$, and $f(x)$ does not divide $x^{h}-1$ ( hence is coprime to it) for $0 < h < d.$ Each eigenvalue of $\theta(g)$ ( in a suitable extension of $K$) is therefore a primitive $d$-th root of unity. Let $\omega_{d}$ be one of these. By the theory of the rational canonical form, we may choose a $K$-basis for $M$ such that the matrix $X$ representing $\theta(g)$ with respect to that basis is the companion matrix for $f(x)$. The size of the matrix for $\theta(g)$ is thus $[K(\omega_{d}):K] \times [K(\omega_{d}):K]$. For each positive integer $a$ coprime to $d$, and less than $d$, we may define a new matrix representation $\psi$ of $\langle g \rangle$ by setting $\psi(g) = X^{a}$ (and following the previous constructions carefully, (up to similarity) there are no other choices for which $g$ can act as a matrix of order exactly $d$ on an irreducible $KG$-module). This explains how to construct all $\phi(d)$ inequivalent irreducible representations of $\langle g \rangle$ of degree $[K(\omega_{d}):K]$ over $K$ ( where $\phi$ is Euler's function).

(actually, if all you are interested in is primitive idempotents of $KG$, you don't need to explicitly construct the representations. Just use a little Galois theory, starting from what the idempotents look like over $\overline{K}$, the algebraic closure of $K$).

Later edit: Let me clarify this last comment a little. Fix $G$ (now finite Abelian of order $r$, but not necessarily cyclic). Let $L = K[\omega_{r}]$. The primitive idempotents of $LG$ all have the form $e_{\lambda} = \frac{1}{|G|} \sum_{g \in G} \lambda(g^{-1})g$ where $\lambda$ ranges over the homomorphisms from $G$ to $L^{\times}$ (there are exactly $r$ of these). These $e_{\lambda}$ are permuted by ${\rm Gal}(L/K)$, just as the linear characters $\lambda$ are permuted by that Galois group. The primitive idempotents of $KG$ are all of the form $\sum_{\lambda \in \Omega} e_{\lambda}$, where $\Omega$ ranges through the orbits of ${\rm Gal}(L/K)$ on these irreducible ($L$)-characters.

  • $\begingroup$ Thank you Geoff, indeed, I am mainly interested in computing the primitive idempotents. How is the constructing starting from the ones for the algebraic closure? $\endgroup$ Commented Jun 8, 2015 at 10:49
  • $\begingroup$ I have expanded the end of my answer to address this. $\endgroup$ Commented Jun 8, 2015 at 11:42
  • $\begingroup$ Thank you very much, during my time at the university I had only a lesson about the case of the algebraic closure. $\endgroup$ Commented Jun 8, 2015 at 11:54

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