The set-up is this: Let $G$ be a finite group, and $H$ a subgroup. We are given an irreducible representation of $H, \rho: H\rightarrow GL_n(K)$ (I will notationally identify $\rho$ with its character). I want to decompose $Ind^G_H(\rho)$ into irreducibles. I am given character tables of both $G$ and $H$.

If $K$ were $\mathbb{C}$, Frobenius reciprocity (https://planetmath.org/FrobeniusReciprocity) will do the trick. However, I am in the modular case; meaning: $char(K)||H|$. I still have all the character tables, except now they are Brauer character tables for the correct characteristic.

Is there a method for decomposing $Ind^G_H(\rho)$ into irreducible (Brauer) characters?

Edit: I wanted to make clear that since in the modular case we don't have Maschke's theorem, the ``decomposition'' into irreducibles would be in the Grothendieck group of Brauer characters of $G$. (representations of $G$ wouldn't nec. be direct sums of irreducible representations)

  • $\begingroup$ I don't quite see what do you mean when you say that if K were C Frobenius reciprocity does the trick. Easy example: if you take G abelian, and H the trivial subgroup, $\rho$ the trivial representation and induce it to the whole group. What you get is a representation which is not irreducible, but how do you compute the irreducible factors? In this case you know them a priori (since G is abelian), and Frobenius reciprocity will only tell you that each of them appears once in the induce representation, but says nothing about who they are $\endgroup$
    – A. Pacetti
    Jun 18, 2011 at 2:58
  • $\begingroup$ I'm not sure what you mean... I am given all the character tables. Here's what I mean: $Ind^G_H(\rho)=\sum a_i\chi_i$. $(Ind^G_H(\rho),\chi_i)=a_i$, and so if we can compute all the $(Ind^G_H(\rho),\chi_i)'s$ then we're done. Frobenius reciprocity does indeed allow to compute those. I know what all the $\chi_i$'s are because I'm given the character tables of $G$ and $H$. Am I missing something? $\endgroup$ Jun 18, 2011 at 16:04

2 Answers 2


Frobenius reciprocity for Brauer characters is a little more complicated (a lot more complicated if you don't have complete tables). You need the projective characters to compute the multiplicities. I don't use anything special about the character being induced, though sometimes you can leverage that information (especially if you don't have complete tables).

In GAP, this is easily done:

# Let g be the O'Nan simple group
chi:=Sum([1..10],i->Random(Irr(g mod 2)));;
# chi is a random reducible Brauer character
ipr:=Irr(g)*DecompositionMatrix(g mod 2);;
# ipr is the list of projective Brauer characters
# vec is the multiplicity of each Brauer character in chi
chi = vec * Irr( g mod 2 );
# should be true: we have decomposed it correctly.

To get the decomposition matrix, you must not only have the Brauer table but also the ordinary table. You don't need much from the subgroup H, just a character to induce and the element fusion from H into G.

This is theorem 2.13 on page 25 of Navarro's textbook on Characters and Blocks of Finite Groups.

By projective, I mean in the module theory sense, an indecomposable direct summand of the regular representation, usually these are denoted Φi corresponding to a Brauer character φi. I don't mean representation into projective groups, like PGL.


You don't need Frobenius reciprocity: there's a formula for the character of an induced representation which is valid for both ordinary and Brauer characters (called "the Frobenius formula" on Wikipedia); thus, writing the induced character as a sum of irreducibles is just solving a system of linear equations. It's arguably easier for Brauer characters, since there are fewer equations to solve.


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